{"id":139,"date":"2019-08-20T17:03:09","date_gmt":"2019-08-20T21:03:09","guid":{"rendered":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/chapter\/sum-and-difference-identities\/"},"modified":"2022-06-01T10:39:32","modified_gmt":"2022-06-01T14:39:32","slug":"sum-and-difference-identities","status":"publish","type":"chapter","link":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/chapter\/sum-and-difference-identities\/","title":{"raw":"Sum and Difference Identities","rendered":"Sum and Difference Identities"},"content":{"raw":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\nIn this section, you will:\n<ul>\n \t<li>Use sum and difference formulas for cosine.<\/li>\n \t<li>Use sum and difference formulas for sine.<\/li>\n \t<li>Use sum and difference formulas for tangent.<\/li>\n \t<li>Use sum and difference formulas for cofunctions.<\/li>\n \t<li>Use sum and difference formulas to verify identities.<\/li>\n<\/ul>\n<\/div>\n<div class=\"small\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143905\/CNX_Precalc_Figure_07_02_001.jpg\" alt=\"Photo of Mt. McKinley.\" width=\"488\" height=\"325\"> <strong>Figure 1. <\/strong>Mount McKinley, in Denali National Park, Alaska, rises 20,237 feet (6,168 m) above sea level. It is the highest peak in North America. (credit: Daniel A. Leifheit, Flickr)[\/caption]\n\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<p id=\"fs-id2463462\">How can the height of a mountain be measured? What about the distance from Earth to the sun? Like many seemingly impossible problems, we rely on mathematical formulas to find the answers. The trigonometric identities, commonly used in mathematical proofs, have had real-world applications for centuries, including their use in calculating long distances.<\/p>\n<p id=\"fs-id1237528\">The trigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. These are special equations or postulates, true for all values input to the equations, and with innumerable applications.<\/p>\n<p id=\"fs-id2285295\">In this section, we will learn techniques that will enable us to solve problems such as the ones presented above. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term <em>formula<\/em> is used synonymously with the word <em>identity<\/em>.<\/p>\n\n<div id=\"fs-id1902399\" class=\"bc-section section\">\n<h3>Using the Sum and Difference Formulas for Cosine<\/h3>\n<p id=\"fs-id2264960\">Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the <span class=\"no-emphasis\">special angles<\/span>, which we can review in the unit circle shown in <a class=\"autogenerated-content\" href=\"#Figure_07_02_008\">(Figure)<\/a>.<\/p>\n\n<div class=\"wp-caption aligncenter\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143908\/CNX_Precalc_Figure_07_01_004.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" width=\"975\" height=\"638\"> <strong>Figure 2. <\/strong>The Unit Circle[\/caption]\n\n<\/div>\n<p id=\"fs-id2618307\">We will begin with the <span class=\"no-emphasis\">sum and difference formulas for cosine<\/span>, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles. See <a class=\"autogenerated-content\" href=\"#Table_07_02_01\">(Figure)<\/a>.<\/p>\n\n<table summary=\"Two rows, two columns. The table has ordered pairs of these row values: (Sum formula for cosine, cos(a+B) = cos(a)cos(B) - sin(a)sin(B)) and (Difference formula for cosine, cos(a-B) = cos(a)cos(B) + sin(a)sin(B)).\">\n<tbody>\n<tr>\n<td><strong>Sum formula for cosine<\/strong><\/td>\n<td>[latex]\\mathrm{cos}\\left(\\alpha +\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Difference formula for cosine<\/strong><\/td>\n<td>[latex]\\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id3007968\">First, we will prove the difference formula for cosines. Let\u2019s consider two points on the unit circle. See <a class=\"autogenerated-content\" href=\"#Figure_07_02_002\">(Figure)<\/a>. Point[latex]\\,P\\,[\/latex]is at an angle[latex]\\,\\alpha \\,[\/latex]from the positive <em>x-<\/em>axis with coordinates[latex]\\,\\left(\\mathrm{cos}\\,\\alpha ,\\mathrm{sin}\\,\\alpha \\right)\\,[\/latex]and point[latex]\\,Q\\,[\/latex]is at an angle of[latex]\\,\\beta \\,[\/latex]from the positive <em>x-<\/em>axis with coordinates[latex]\\,\\left(\\mathrm{cos}\\,\\beta ,\\mathrm{sin}\\,\\beta \\right).\\,[\/latex]Note the measure of angle[latex]\\,POQ\\,[\/latex]is[latex]\\,\\alpha -\\beta .\\,[\/latex]<\/p>\n<p id=\"fs-id1565012\">Label two more points:[latex]\\,A\\,[\/latex]at an angle of[latex]\\,\\left(\\alpha -\\beta \\right)\\,[\/latex]from the positive <em>x-<\/em>axis with coordinates[latex]\\,\\left(\\mathrm{cos}\\left(\\alpha -\\beta \\right),\\mathrm{sin}\\left(\\alpha -\\beta \\right)\\right);\\,[\/latex]and point[latex]\\,B\\,[\/latex]with coordinates[latex]\\,\\left(1,0\\right).\\,[\/latex]Triangle[latex]\\,POQ\\,[\/latex]is a rotation of triangle[latex]\\,AOB\\,[\/latex]and thus the distance from[latex]\\,P\\,[\/latex]to[latex]\\,Q\\,[\/latex]is the same as the distance from[latex]\\,A\\,[\/latex]to[latex]\\,B.[\/latex]<\/p>\n\n<div class=\"small\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143917\/CNX_Precalc_Figure_07_02_002.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" width=\"487\" height=\"365\"> <strong>Figure 3.<\/strong>[\/caption]\n\n<\/div>\n<p id=\"fs-id2152546\">We can find the distance from[latex]\\,P\\,[\/latex]to[latex]\\,Q\\,[\/latex]using the <span class=\"no-emphasis\">distance formula<\/span>.<\/p>\n\n<div id=\"fs-id1096956\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill {d}_{PQ}&amp; =&amp; \\sqrt{{\\left(\\mathrm{cos}\\,\\alpha -\\mathrm{cos}\\,\\beta \\right)}^{2}+{\\left(\\mathrm{sin}\\,\\alpha -\\mathrm{sin}\\,\\beta \\right)}^{2}}\\hfill \\\\ &amp; =&amp; \\sqrt{{\\mathrm{cos}}^{2}\\alpha -2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +{\\mathrm{cos}}^{2}\\beta +{\\mathrm{sin}}^{2}\\alpha -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta +{\\mathrm{sin}}^{2}\\beta }\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id2239334\">Then we apply the <span class=\"no-emphasis\">Pythagorean identity<\/span> and simplify.<\/p>\n\n<div id=\"fs-id1343608\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{cc}=&amp; \\sqrt{\\left({\\mathrm{cos}}^{2}\\alpha +{\\mathrm{sin}}^{2}\\alpha \\right)+\\left({\\mathrm{cos}}^{2}\\beta +{\\mathrm{sin}}^{2}\\beta \\right)-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }\\hfill \\\\ =&amp; \\sqrt{1+1-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }\\hfill \\\\ =&amp; \\sqrt{2-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1342538\">Similarly, using the distance formula we can find the distance from[latex]\\,A\\,[\/latex]to[latex]\\,B.[\/latex]<\/p>\n\n<div id=\"fs-id1094579\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill {d}_{AB}&amp; =&amp; \\sqrt{{\\left(\\mathrm{cos}\\left(\\alpha -\\beta \\right)-1\\right)}^{2}+{\\left(\\mathrm{sin}\\left(\\alpha -\\beta \\right)-0\\right)}^{2}}\\hfill \\\\ &amp; =&amp; \\sqrt{{\\mathrm{cos}}^{2}\\left(\\alpha -\\beta \\right)-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)+1+{\\mathrm{sin}}^{2}\\left(\\alpha -\\beta \\right)}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1783115\">Applying the Pythagorean identity and simplifying we get:<\/p>\n\n<div id=\"fs-id1347258\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{cc}=&amp; \\sqrt{\\left({\\mathrm{cos}}^{2}\\left(\\alpha -\\beta \\right)+{\\mathrm{sin}}^{2}\\left(\\alpha -\\beta \\right)\\right)-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)+1}\\hfill \\\\ =&amp; \\sqrt{1-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)+1}\\hfill \\\\ =&amp; \\sqrt{2-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id2196869\">Because the two distances are the same, we set them equal to each other and simplify.<\/p>\n\n<div id=\"fs-id1729317\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\sqrt{2-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }&amp; =&amp; \\sqrt{2-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)}\\hfill \\\\ \\hfill 2-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta &amp; =&amp; 2-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1411215\">Finally we subtract[latex]\\,2\\,[\/latex]from both sides and divide both sides by[latex]\\,-2.[\/latex]<\/p>\n\n<div id=\"fs-id1491684\" class=\"unnumbered aligncenter\">[latex]\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta =\\mathrm{cos}\\left(\\alpha -\\beta \\right)\\text{ }[\/latex]<\/div>\n<p id=\"fs-id711534\">Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.<\/p>\n\n<div id=\"fs-id1334205\" class=\"textbox key-takeaways\">\n<h3>Sum and Difference Formulas for Cosine<\/h3>\n<p id=\"fs-id1261808\">These formulas can be used to calculate the cosine of sums and differences of angles.<\/p>\n\n<div id=\"Eq_07_02_01\">[latex]\\mathrm{cos}\\left(\\alpha +\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/div>\n<div id=\"Eq_07_02_02\">[latex]\\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/div>\n<\/div>\n<div id=\"fs-id861912\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1728576\"><strong>Given two angles, find the cosine of the difference between the angles.\n<\/strong><\/p>\n\n<ol id=\"fs-id1374215\" type=\"1\">\n \t<li>Write the difference formula for cosine.<\/li>\n \t<li>Substitute the values of the given angles into the formula.<\/li>\n \t<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1733838\">\n<div id=\"fs-id2056629\">\n<h3>Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles<\/h3>\n<p id=\"fs-id1953099\">Using the formula for the cosine of the difference of two angles, find the exact value of[latex]\\,\\mathrm{cos}\\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right).[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1128213\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1128213\"]\n<p id=\"fs-id1128213\">Begin by writing the formula for the cosine of the difference of two angles. Then substitute the given values.<\/p>\n\n<div id=\"fs-id1369035\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\alpha -\\beta \\right)&amp; =&amp; \\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{cos}\\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)&amp; =&amp; \\mathrm{cos}\\left(\\frac{5\\pi }{4}\\right)\\mathrm{cos}\\left(\\frac{\\pi }{6}\\right)+\\mathrm{sin}\\left(\\frac{5\\pi }{4}\\right)\\mathrm{sin}\\left(\\frac{\\pi }{6}\\right)\\hfill \\\\ &amp; =&amp; \\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)-\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{1}{2}\\right)\\hfill \\\\ &amp; =&amp; -\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4}\\hfill \\\\ &amp; =&amp; \\frac{-\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"eip-id1256870\">Keep in mind that we can always check the answer using a graphing calculator in radian mode.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1369787\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id2118816\">\n<p id=\"fs-id1557062\">Find the exact value of[latex]\\,\\mathrm{cos}\\left(\\frac{\\pi }{3}-\\frac{\\pi }{4}\\right).[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1721765\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1721765\"]\n<p id=\"fs-id1721765\">[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id2956024\">\n<div id=\"fs-id1367064\">\n<h3>Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine<\/h3>\n<p id=\"fs-id1161788\">Find the exact value of[latex]\\,\\mathrm{cos}\\left(75\u00b0\\right).[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1711386\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1711386\"]\n<p id=\"fs-id1711386\">As[latex]\\,75\u00b0=45\u00b0+30\u00b0,[\/latex]we can evaluate[latex]\\,\\mathrm{cos}\\left(75\u00b0\\right)\\,[\/latex]as[latex]\\,\\mathrm{cos}\\left(45\u00b0+30\u00b0\\right).\\,[\/latex]<\/p>\n\n<div id=\"fs-id1834810\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\alpha +\\beta \\right)&amp; =&amp; \\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{cos}\\left(45\u00b0+30\u00b0\\right)&amp; =&amp; \\mathrm{cos}\\left(45\u00b0\\right)\\mathrm{cos}\\left(30\u00b0\\right)-\\mathrm{sin}\\left(45\u00b0\\right)\\mathrm{sin}\\left(30\u00b0\\right)\\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right)\\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4}\\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"eip-id1731565\">Keep in mind that we can always check the answer using a graphing calculator in degree mode.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<div id=\"fs-id1905704\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id2283950\">Note that we could have also solved this problem using the fact that[latex]\\,75\u00b0=135\u00b0-60\u00b0.[\/latex]<\/p>\n\n<div id=\"fs-id1924708\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\alpha -\\beta \\right)&amp; =&amp; \\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{cos}\\left(135\u00b0-60\u00b0\\right)&amp; =&amp; \\mathrm{cos}\\left(135\u00b0\\right)\\mathrm{cos}\\left(60\u00b0\\right)+\\mathrm{sin}\\left(135\u00b0\\right)\\mathrm{sin}\\left(60\u00b0\\right)\\hfill \\\\ &amp; =&amp; \\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{1}{2}\\right)+\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ &amp; =&amp; \\left(-\\frac{\\sqrt{2}}{4}\\right)+\\left(\\frac{\\sqrt{6}}{4}\\right)\\hfill \\\\ &amp; =&amp; \\left(\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\right)\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1335750\">\n<p id=\"fs-id1231482\">Find the exact value of[latex]\\,\\mathrm{cos}\\left(105\u00b0\\right).[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1094264\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1094264\"]\n<p id=\"fs-id1094264\">[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1388326\" class=\"bc-section section\">\n<h3>Using the Sum and Difference Formulas for Sine<\/h3>\n<p id=\"fs-id1198942\">The <span class=\"no-emphasis\">sum and difference formulas for sine<\/span> can be derived in the same manner as those for cosine, and they resemble the cosine formulas.<\/p>\n\n<div id=\"fs-id1345043\" class=\"textbox key-takeaways\">\n<h3>Sum and Difference Formulas for Sine<\/h3>\n<p id=\"fs-id2335392\">These formulas can be used to calculate the sines of sums and differences of angles.<\/p>\n\n<div id=\"Eq_07_02_03\">[latex]\\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/div>\n<div id=\"Eq_07_02_04\">[latex]\\mathrm{sin}\\left(\\alpha -\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1274546\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1891863\"><strong>Given two angles, find the sine of the difference between the angles.\n<\/strong><\/p>\n\n<ol id=\"fs-id1352419\" type=\"1\">\n \t<li>Write the difference formula for sine.<\/li>\n \t<li>Substitute the given angles into the formula.<\/li>\n \t<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1368235\">\n<div id=\"fs-id1242560\">\n<h3>Using Sum and Difference Identities to Evaluate the Difference of Angles<\/h3>\n<p id=\"fs-id1304945\">Use the sum and difference identities to evaluate the difference of the angles and show that part <em>a<\/em> equals part <em>b.<\/em><\/p>\n\n<ol id=\"fs-id1978491\" type=\"a\">\n \t<li>[latex]\\mathrm{sin}\\left(45\u00b0-30\u00b0\\right)[\/latex]<\/li>\n \t<li>[latex]\\mathrm{sin}\\left(135\u00b0-120\u00b0\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2291665\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2291665\"]\n<ol id=\"fs-id2291665\" type=\"a\">\n \t<li>Let\u2019s begin by writing the formula and substitute the given angles.\n<div class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha -\\beta \\right)&amp; =&amp; \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{sin}\\left(45\u00b0-30\u00b0\\right)&amp; =&amp; \\mathrm{sin}\\left(45\u00b0\\right)\\mathrm{cos}\\left(30\u00b0\\right)-\\mathrm{cos}\\left(45\u00b0\\right)\\mathrm{sin}\\left(30\u00b0\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1565917\">Next, we need to find the values of the trigonometric expressions.<\/p>\n\n<div id=\"fs-id1213031\" class=\"unnumbered aligncenter\">[latex]\\mathrm{sin}\\left(45\u00b0\\right)=\\frac{\\sqrt{2}}{2}, \\mathrm{cos}\\left(30\u00b0\\right)=\\frac{\\sqrt{3}}{2}, \\mathrm{cos}\\left(45\u00b0\\right)=\\frac{\\sqrt{2}}{2}, \\mathrm{sin}\\left(30\u00b0\\right)=\\frac{1}{2}[\/latex]<\/div>\n<p id=\"fs-id1143054\">Now we can substitute these values into the equation and simplify.<\/p>\n\n<div id=\"fs-id2020743\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(45\u00b0-30\u00b0\\right)&amp; =&amp; \\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right)\\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{array}[\/latex]<\/div><\/li>\n \t<li>Again, we write the formula and substitute the given angles.\n<div id=\"fs-id1135014\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha -\\beta \\right)&amp; =&amp; \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{sin}\\left(135\u00b0-120\u00b0\\right)&amp; =&amp; \\mathrm{sin}\\left(135\u00b0\\right)\\mathrm{cos}\\left(120\u00b0\\right)-\\mathrm{cos}\\left(135\u00b0\\right)\\mathrm{sin}\\left(120\u00b0\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1177711\">Next, we find the values of the trigonometric expressions.<\/p>\n\n<div id=\"fs-id2067561\" class=\"unnumbered aligncenter\">[latex]\\mathrm{sin}\\left(135\u00b0\\right)=\\frac{\\sqrt{2}}{2},\\mathrm{cos}\\left(120\u00b0\\right)=-\\frac{1}{2},\\mathrm{cos}\\left(135\u00b0\\right)=\\frac{\\sqrt{2}}{2},\\mathrm{sin}\\left(120\u00b0\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/div>\n<p id=\"fs-id2011119\">Now we can substitute these values into the equation and simplify.<\/p>\n\n<div id=\"fs-id1922441\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(135\u00b0-120\u00b0\\right)&amp; =&amp; \\frac{\\sqrt{2}}{2}\\left(-\\frac{1}{2}\\right)-\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ &amp; =&amp; \\frac{-\\sqrt{2}+\\sqrt{6}}{4}\\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\\\ \\phantom{\\rule{4em}{0ex}}\\mathrm{sin}\\left(135\u00b0-120\u00b0\\right)\\hfill &amp; =&amp; \\frac{\\sqrt{2}}{2}\\left(-\\frac{1}{2}\\right)-\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ &amp; =&amp; \\frac{-\\sqrt{2}+\\sqrt{6}}{4}\\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1494014\">\n<div id=\"fs-id1494016\">\n<h3>Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function<\/h3>\n<p id=\"fs-id1252839\">Find the exact value of[latex]\\,\\mathrm{sin}\\left({\\mathrm{cos}}^{-1}\\frac{1}{2}+{\\mathrm{sin}}^{-1}\\frac{3}{5}\\right).\\,[\/latex]Then check the answer with a graphing calculator.<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1669695\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1669695\"]\n<p id=\"fs-id1669695\">The pattern displayed in this problem is[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right).\\,[\/latex]Let[latex]\\,\\alpha ={\\mathrm{cos}}^{-1}\\frac{1}{2}\\,[\/latex]and[latex]\\,\\beta ={\\mathrm{sin}}^{-1}\\frac{3}{5}.\\,[\/latex]Then we can write<\/p>\n\n<div id=\"fs-id1868250\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\,\\alpha &amp; =&amp; \\frac{1}{2},0\\le \\alpha \\le \\pi \\hfill \\\\ \\hfill \\mathrm{sin}\\,\\beta &amp; =&amp; \\frac{3}{5},-\\frac{\\pi }{2}\\le \\beta \\le \\frac{\\pi }{2}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1562263\">We will use the Pythagorean identities to find[latex]\\,\\mathrm{sin}\\,\\alpha \\,[\/latex]and[latex]\\,\\mathrm{cos}\\,\\beta .[\/latex]<\/p>\n\n<div id=\"fs-id2044989\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\,\\alpha &amp; =&amp; \\sqrt{1-{\\mathrm{cos}}^{2}\\alpha }\\hfill \\\\ &amp; =&amp; \\sqrt{1-\\frac{1}{4}}\\hfill \\\\ &amp; =&amp; \\sqrt{\\frac{3}{4}}\\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{3}}{2}\\hfill \\\\ \\hfill \\mathrm{cos}\\,\\beta &amp; =&amp; \\sqrt{1-{\\mathrm{sin}}^{2}\\beta }\\hfill \\\\ &amp; =&amp; \\sqrt{1-\\frac{9}{25}}\\hfill \\\\ &amp; =&amp; \\sqrt{\\frac{16}{25}}\\hfill \\\\ &amp; =&amp; \\frac{4}{5}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1155089\">Using the sum formula for sine,<\/p>\n\n<div id=\"fs-id1231462\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left({\\mathrm{cos}}^{-1}\\frac{1}{2}+{\\mathrm{sin}}^{-1}\\frac{3}{5}\\right)&amp; =&amp; \\mathrm{sin}\\left(\\alpha +\\beta \\right)\\hfill \\\\ &amp; =&amp; \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{3}}{2}\\cdot \\frac{4}{5}+\\frac{1}{2}\\cdot \\frac{3}{5}\\hfill \\\\ &amp; =&amp; \\frac{4\\sqrt{3}+3}{10}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2072911\" class=\"bc-section section\">\n<h3>Using the Sum and Difference Formulas for Tangent<\/h3>\n<p id=\"fs-id2780782\">Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.<\/p>\n<p id=\"fs-id2780786\">Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall,[latex]\\,\\mathrm{tan}\\,x=\\frac{\\mathrm{sin}\\,x}{\\mathrm{cos}\\,x},\\mathrm{cos}\\,x\\ne 0.[\/latex]<\/p>\n<p id=\"fs-id1712352\">Let\u2019s derive the sum formula for tangent.<\/p>\n\n<div id=\"fs-id1185845\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{cccc}\\hfill \\mathrm{tan}\\left(\\alpha +\\beta \\right)&amp; =&amp; \\frac{\\mathrm{sin}\\left(\\alpha +\\beta \\right)}{\\mathrm{cos}\\left(\\alpha +\\beta \\right)}\\hfill &amp; \\\\ &amp; =&amp; \\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }\\hfill &amp; \\\\ &amp; =&amp; \\frac{\\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }}{\\frac{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }}\\hfill &amp; \\text{Divide the numerator and denominator by cos}\\,\\alpha \\,\\text{cos}\\,\\beta .\\hfill \\\\ &amp; =&amp; \\frac{\\frac{\\mathrm{sin}\\,\\alpha \\overline{)\\,\\mathrm{cos}\\,\\beta }}{\\mathrm{cos}\\,\\alpha \\overline{)\\,\\mathrm{cos}\\,\\beta }}+\\frac{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\mathrm{sin}\\,\\beta }{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\mathrm{cos}\\,\\beta }}{\\frac{\\overline{)\\mathrm{cos}\\,\\alpha }\\overline{)\\,\\mathrm{cos}\\,\\beta }}{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\overline{)\\mathrm{cos}\\,\\beta }}-\\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }}\\hfill &amp; \\\\ &amp; =&amp; \\frac{\\frac{\\mathrm{sin}\\,\\alpha }{\\mathrm{cos}\\,\\alpha }+\\frac{\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\beta }}{1-\\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }}\\hfill &amp; \\\\ &amp; =&amp; \\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill &amp; \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1775901\">We can derive the difference formula for tangent in a similar way.<\/p>\n\n<div id=\"fs-id2916561\" class=\"textbox key-takeaways\">\n<h3>Sum and Difference Formulas for Tangent<\/h3>\n<p id=\"fs-id2320803\">The <span class=\"no-emphasis\">sum and difference formulas for tangent<\/span> are:<\/p>\n\n<div id=\"Eq_07_02_05\">[latex]\\mathrm{tan}\\left(\\alpha +\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }[\/latex]<\/div>\n<div id=\"Eq_07_02_06\">[latex]\\mathrm{tan}\\left(\\alpha -\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta }{1+\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1567999\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1233471\"><strong>Given two angles, find the tangent of the sum of the angles.\n<\/strong><\/p>\n\n<ol id=\"fs-id1407099\" type=\"1\">\n \t<li>Write the sum formula for tangent.<\/li>\n \t<li>Substitute the given angles into the formula.<\/li>\n \t<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1619329\">\n<div id=\"fs-id2289172\">\n<h3>Finding the Exact Value of an Expression Involving Tangent<\/h3>\n<p id=\"fs-id2289178\">Find the exact value of[latex]\\,\\mathrm{tan}\\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right).[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1378918\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1378918\"]\n<p id=\"fs-id1378918\">Let\u2019s first write the sum formula for tangent and then substitute the given angles into the formula.<\/p>\n\n<div id=\"fs-id1514603\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\alpha +\\beta \\right)&amp; =&amp; \\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\\\ \\hfill \\mathrm{tan}\\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)&amp; =&amp; \\frac{\\mathrm{tan}\\left(\\frac{\\pi }{6}\\right)+\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)}{1-\\left(\\mathrm{tan}\\left(\\frac{\\pi }{6}\\right)\\right)\\left(\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)\\right)}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1378455\">Next, we determine the individual function values within the formula:<\/p>\n\n<div id=\"fs-id1378458\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\left(\\frac{\\pi }{6}\\right)=\\frac{1}{\\sqrt{3}},\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)=1[\/latex]<\/div>\n<p id=\"fs-id2016988\">So we have<\/p>\n\n<div id=\"fs-id1576577\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)&amp; =&amp; \\frac{\\frac{1}{\\sqrt{3}}+1}{1-\\left(\\frac{1}{\\sqrt{3}}\\right)\\left(1\\right)}\\hfill \\\\ &amp; =&amp; \\frac{\\frac{1+\\sqrt{3}}{\\sqrt{3}}}{\\frac{\\sqrt{3}-1}{\\sqrt{3}}}\\hfill \\\\ &amp; =&amp; \\frac{1+\\sqrt{3}}{\\sqrt{3}}\\left(\\frac{\\sqrt{3}}{\\sqrt{3}-1}\\right)\\hfill \\\\ &amp; =&amp; \\frac{\\sqrt{3}+1}{\\sqrt{3}-1}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1541743\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1527860\">\n<p id=\"fs-id1892528\">Find the exact value of[latex]\\,\\mathrm{tan}\\left(\\frac{2\\pi }{3}+\\frac{\\pi }{4}\\right).[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2040850\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2040850\"]\n<p id=\"fs-id2040850\">[latex]\\frac{1-\\sqrt{3}}{1+\\sqrt{3}}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id2927791\">\n<div id=\"fs-id2927793\">\n<h3>Finding Multiple Sums and Differences of Angles<\/h3>\n<p id=\"fs-id1333289\">Given[latex]\\text{ }\\mathrm{sin}\\,\\alpha =\\frac{3}{5},0&lt;\\alpha &lt;\\frac{\\pi }{2},\\mathrm{cos}\\,\\beta =-\\frac{5}{13},\\pi &lt;\\beta &lt;\\frac{3\\pi }{2},[\/latex]find<\/p>\n\n<ol id=\"fs-id2201332\" type=\"a\">\n \t<li>[latex]\\mathrm{sin}\\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n \t<li>[latex]\\mathrm{cos}\\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n \t<li>[latex]\\mathrm{tan}\\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n \t<li>[latex]\\mathrm{tan}\\left(\\alpha -\\beta \\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1483249\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1483249\"]\n<p id=\"fs-id1483249\">We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.<\/p>\n\n<ol id=\"fs-id1440867\" type=\"a\">\n \t<li>To find[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right),[\/latex]we begin with[latex]\\,\\mathrm{sin}\\,\\alpha =\\frac{3}{5}\\,[\/latex]and[latex]\\,0&lt;\\alpha &lt;\\frac{\\pi }{2}.\\,[\/latex]The side opposite[latex]\\,\\alpha \\,[\/latex]has length 3, the hypotenuse has length 5, and[latex]\\,\\alpha \\,[\/latex]is in the first quadrant. See <a class=\"autogenerated-content\" href=\"#Figure_07_02_003\">(Figure)<\/a>. Using the Pythagorean Theorem, we can find the length of side[latex]\\,a\\text{:}[\/latex]\n<div id=\"fs-id2191216\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+{3}^{2}&amp; =&amp; {5}^{2}\\hfill \\\\ \\hfill {a}^{2}&amp; =&amp; 16\\hfill \\\\ \\hfill a&amp; =&amp; 4\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"small\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143920\/CNX_Precalc_Figure_07_02_003.jpg\" alt=\"Diagram of a triangle in the x,y plane. The vertices are at the origin, (4,0), and (4,3). The angle at the origin is alpha degrees, The angle formed by the x-axis and the side from (4,3) to (4,0) is a right angle. The side opposite the right angle has length 5.\" width=\"487\" height=\"252\"> <strong>Figure 4.<\/strong>[\/caption]\n\n<\/div>\n<p id=\"fs-id1342733\">Since[latex]\\,\\mathrm{cos}\\,\\beta =-\\frac{5}{13}\\,[\/latex]and[latex]\\,\\pi &lt;\\beta &lt;\\frac{3\\pi }{2},[\/latex]the side adjacent to[latex]\\,\\beta \\,[\/latex]is[latex]\\,-5,[\/latex]the hypotenuse is 13, and[latex]\\,\\beta \\,[\/latex]is in the third quadrant. See <a class=\"autogenerated-content\" href=\"#Figure_07_02_004\">(Figure)<\/a>. Again, using the Pythagorean Theorem, we have<\/p>\n\n<div id=\"fs-id1944725\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill {\\left(-5\\right)}^{2}+{a}^{2}&amp; =&amp; {13}^{2}\\hfill \\\\ \\hfill 25+{a}^{2}&amp; =&amp; 169\\hfill \\\\ \\hfill {a}^{2}&amp; =&amp; 144\\hfill \\\\ \\hfill a&amp; =&amp; \u00b112\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1928233\">Since[latex]\\,\\beta \\,[\/latex]is in the third quadrant,[latex]\\,a=\u201312.[\/latex]<\/p>\n\n<div class=\"small\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143922\/CNX_Precalc_Figure_07_02_004.jpg\" alt=\"Diagram of a triangle in the x,y plane. The vertices are at the origin, (-5,0), and (-5, -12). The angle at the origin is Beta degrees. The angle formed by the x axis and the side from (-5, -12) to (-5,0) is a right angle. The side opposite the right angle has length 13.\" width=\"487\" height=\"568\"> <strong>Figure 5.<\/strong>[\/caption]\n\n<\/div>\n<p id=\"fs-id1155212\">The next step is finding the cosine of[latex]\\,\\alpha \\,[\/latex]and the sine of[latex]\\,\\beta .\\,[\/latex]The cosine of[latex]\\,\\alpha \\,[\/latex]is the adjacent side over the hypotenuse. We can find it from the triangle in <a class=\"autogenerated-content\" href=\"#Figure_07_02_004\">(Figure)<\/a>:[latex]\\,\\mathrm{cos}\\,\\alpha =\\frac{4}{5}.\\,[\/latex]We can also find the sine of[latex]\\,\\beta \\,[\/latex]from the triangle in <a class=\"autogenerated-content\" href=\"#Figure_07_02_004\">(Figure)<\/a>, as opposite side over the hypotenuse:[latex]\\,\\mathrm{sin}\\,\\beta =-\\frac{12}{13}.\\,[\/latex]Now we are ready to evaluate[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right).[\/latex]<\/p>\n\n<div id=\"fs-id2256453\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha +\\beta \\right)&amp; =&amp; \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ &amp; =&amp; \\left(\\frac{3}{5}\\right)\\left(-\\frac{5}{13}\\right)+\\left(\\frac{4}{5}\\right)\\left(-\\frac{12}{13}\\right)\\hfill \\\\ &amp; =&amp; -\\frac{15}{65}-\\frac{48}{65}\\hfill \\\\ &amp; =&amp; -\\frac{63}{65}\\hfill \\end{array}[\/latex]<\/div><\/li>\n \t<li>We can find[latex]\\,\\mathrm{cos}\\left(\\alpha +\\beta \\right)\\,[\/latex]in a similar manner. We substitute the values according to the formula.\n<div id=\"fs-id1487615\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\alpha +\\beta \\right)&amp; =&amp; \\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ &amp; =&amp; \\left(\\frac{4}{5}\\right)\\left(-\\frac{5}{13}\\right)-\\left(\\frac{3}{5}\\right)\\left(-\\frac{12}{13}\\right)\\hfill \\\\ &amp; =&amp; -\\frac{20}{65}+\\frac{36}{65}\\hfill \\\\ &amp; =&amp; \\frac{16}{65}\\hfill \\end{array}[\/latex]<\/div><\/li>\n \t<li>For[latex]\\,\\mathrm{tan}\\left(\\alpha +\\beta \\right),[\/latex]if[latex]\\,\\mathrm{sin}\\,\\alpha =\\frac{3}{5}\\,[\/latex]and[latex]\\,\\mathrm{cos}\\,\\alpha =\\frac{4}{5},[\/latex] then\n<div id=\"fs-id1512831\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\,\\alpha =\\frac{\\frac{3}{5}}{\\frac{4}{5}}=\\frac{3}{4}[\/latex]<\/div>\n<p id=\"fs-id1837927\">If[latex]\\,\\mathrm{sin}\\,\\beta =-\\frac{12}{13}\\,[\/latex]and[latex]\\,\\mathrm{cos}\\,\\beta =-\\frac{5}{13},[\/latex]\nthen<\/p>\n\n<div id=\"fs-id2917183\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\,\\beta =\\frac{\\frac{-12}{13}}{\\frac{-5}{13}}=\\frac{12}{5}[\/latex]<\/div>\n<p id=\"fs-id2900205\">Then,<\/p>\n\n<div id=\"fs-id2900208\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\alpha +\\beta \\right)&amp; =&amp; \\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\\\ &amp; =&amp; \\frac{\\frac{3}{4}+\\frac{12}{5}}{1-\\frac{3}{4}\\left(\\frac{12}{5}\\right)}\\hfill \\\\ &amp; =&amp; \\frac{\\text{ }\\frac{63}{20}}{-\\frac{16}{20}}\\hfill \\\\ &amp; =&amp; -\\frac{63}{16}\\hfill \\end{array}[\/latex]<\/div><\/li>\n \t<li>To find[latex]\\,\\mathrm{tan}\\left(\\alpha -\\beta \\right),[\/latex] we have the values we need. We can substitute them in and evaluate.\n<div id=\"fs-id2575375\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\alpha -\\beta \\right)&amp; =&amp; \\frac{\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta }{1+\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\\\ &amp; =&amp; \\frac{\\frac{3}{4}-\\frac{12}{5}}{1+\\frac{3}{4}\\left(\\frac{12}{5}\\right)}\\hfill \\\\ &amp; =&amp; \\frac{-\\frac{33}{20}}{\\frac{56}{20}}\\hfill \\\\ &amp; =&amp; -\\frac{33}{56}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div><\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id2714729\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id2714734\">A common mistake when addressing problems such as this one is that we may be tempted to think that[latex]\\,\\alpha \\,[\/latex]and[latex]\\,\\beta \\,[\/latex]are angles in the same triangle, which of course, they are not. Also note that<\/p>\n\n<div id=\"fs-id2202966\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\left(\\alpha +\\beta \\right)=\\frac{\\mathrm{sin}\\left(\\alpha +\\beta \\right)}{\\mathrm{cos}\\left(\\alpha +\\beta \\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2050772\" class=\"bc-section section\">\n<h3>Using Sum and Difference Formulas for Cofunctions<\/h3>\n<p id=\"fs-id1527759\">Now that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall from <a class=\"target-chapter\" href=\"\/contents\/7375cfa6-269c-4e0d-a567-bc6a4c66b1f4\">Right Triangle Trigonometry<\/a> that, if the sum of two positive angles is[latex]\\,\\frac{\\pi }{2},[\/latex]those two angles are complements, and the sum of the two acute angles in a right triangle is[latex]\\,\\frac{\\pi }{2},[\/latex]so they are also complements. In <a class=\"autogenerated-content\" href=\"#Figure_07_02_007\">(Figure)<\/a>, notice that if one of the acute angles is labeled as[latex]\\,\\theta ,[\/latex]then the other acute angle must be labeled[latex]\\,\\left(\\frac{\\pi }{2}-\\theta \\right).[\/latex]<\/p>\n<p id=\"fs-id2045066\">Notice also that[latex]\\,\\mathrm{sin}\\,\\theta =\\mathrm{cos}\\left(\\frac{\\pi }{2}-\\theta \\right),[\/latex]which is opposite over hypotenuse. Thus, when two angles are complementary, we can say that the sine of[latex]\\,\\theta \\,[\/latex]equals the <span class=\"no-emphasis\">cofunction<\/span> of the complement of[latex]\\,\\theta .\\,[\/latex]Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions.<\/p>\n\n<div class=\"small\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143927\/CNX_Precalc_Figure_07_02_007.jpg\" alt=\"Image of a right triangle. The remaining angles are labeled theta and pi\/2 - theta.\" width=\"487\" height=\"268\"> <strong>Figure 6.<\/strong>[\/caption]\n\n<\/div>\n<p id=\"fs-id2338660\">From these relationships, the <span class=\"no-emphasis\">cofunction identities<\/span> are formed. Recall that you first encountered these identities in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/chapter\/introduction-to-the-unit-circle-sine-and-cosine-functions\/\">The Unit Circle: Sine and Cosine Functions<\/a>.<\/p>\n\n<div id=\"fs-id2247036\" class=\"textbox key-takeaways\">\n<h3>Cofunction Identities<\/h3>\n<p id=\"fs-id2247045\">The cofunction identities are summarized in <a class=\"autogenerated-content\" href=\"#Table_07_02_02\">(Figure)<\/a>.<\/p>\n\n<table id=\"Table_07_02_02\" summary=\"Three rows, two columns\/ The table has ordered pairs of these row values: (sin(theta) = cos(pi\/2 - theta), cos(theta) = sin(pi\/2 - theta)), (tan(theta) = cot(pi\/2 - theta), cot(theta) = tan(pi\/2 - theta)), and (sec(theta) = csc(pi\/2 - theta), csc(theta) = sec(pi\/2 - theta)).\"><colgroup> <col> <col><\/colgroup>\n<tbody>\n<tr>\n<td>[latex]\\mathrm{sin}\\,\\theta =\\mathrm{cos}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\mathrm{cos}\\,\\theta =\\mathrm{sin}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\mathrm{tan}\\,\\theta =\\mathrm{cot}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\mathrm{cot}\\,\\theta =\\mathrm{tan}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\mathrm{sec}\\,\\theta =\\mathrm{csc}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\mathrm{csc}\\,\\theta =\\mathrm{sec}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p id=\"fs-id2814240\">Notice that the formulas in the table may also justified algebraically using the sum and difference formulas. For example, using<\/p>\n\n<div id=\"fs-id2814245\" class=\"unnumbered aligncenter\">[latex]\\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\mathrm{sin}\\,\\beta ,[\/latex]<\/div>\n<p id=\"fs-id2021193\">we can write<\/p>\n\n<div id=\"fs-id2427498\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\frac{\\pi }{2}-\\theta \\right)&amp; =&amp; \\mathrm{cos}\\,\\frac{\\pi }{2}\\,\\mathrm{cos}\\,\\theta +\\mathrm{sin}\\,\\frac{\\pi }{2}\\,\\mathrm{sin}\\,\\theta \\hfill \\\\ &amp; =&amp; \\left(0\\right)\\mathrm{cos}\\,\\theta +\\left(1\\right)\\mathrm{sin}\\,\\theta \\hfill \\\\ &amp; =&amp; \\mathrm{sin}\\,\\theta \\hfill \\end{array}[\/latex]<\/div>\n<div id=\"Example_07_02_07\" class=\"textbox examples\">\n<div id=\"fs-id2223313\">\n<div id=\"fs-id2223315\">\n<h3>Finding a Cofunction with the Same Value as the Given Expression<\/h3>\n<p id=\"fs-id2231859\">Write[latex]\\,\\mathrm{tan}\\,\\frac{\\pi }{9}\\,[\/latex]in terms of its cofunction.<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1556117\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1556117\"]\n<p id=\"fs-id1556117\">The cofunction of[latex]\\,\\mathrm{tan}\\,\\theta =\\mathrm{cot}\\left(\\frac{\\pi }{2}-\\theta \\right).\\,[\/latex]Thus,<\/p>\n\n<div id=\"fs-id2274481\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\frac{\\pi }{9}\\right)&amp; =&amp; \\mathrm{cot}\\left(\\frac{\\pi }{2}-\\frac{\\pi }{9}\\right)\\hfill \\\\ &amp; =&amp; \\mathrm{cot}\\left(\\frac{9\\pi }{18}-\\frac{2\\pi }{18}\\right)\\hfill \\\\ &amp; =&amp; \\mathrm{cot}\\left(\\frac{7\\pi }{18}\\right)\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2109946\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1923901\">\n<p id=\"fs-id1923902\">Write[latex]\\,\\mathrm{sin}\\,\\frac{\\pi }{7}\\,[\/latex]in terms of its cofunction.<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1386356\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1386356\"]\n<p id=\"fs-id1386356\">[latex]\\mathrm{cos}\\left(\\frac{5\\pi }{14}\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1926444\" class=\"bc-section section\">\n<h3>Using the Sum and Difference Formulas to Verify Identities<\/h3>\n<p id=\"fs-id2637951\">Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems. Reviewing the general rules presented earlier may help simplify the process of verifying an identity.<\/p>\n\n<div id=\"fs-id2637962\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1863876\"><strong>Given an identity, verify using sum and difference formulas.<\/strong><\/p>\n\n<ol id=\"fs-id1863880\" type=\"1\">\n \t<li>Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.<\/li>\n \t<li>Look for opportunities to use the sum and difference formulas.<\/li>\n \t<li>Rewrite sums or differences of quotients as single quotients.<\/li>\n \t<li>If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_07_02_08\" class=\"textbox examples\">\n<div id=\"fs-id1183047\">\n<div id=\"fs-id1183049\">\n<h3>Verifying an Identity Involving Sine<\/h3>\n<p id=\"fs-id1899828\">Verify the identity[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right)+\\mathrm{sin}\\left(\\alpha -\\beta \\right)=2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta .[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1469824\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1469824\"]\n<p id=\"fs-id1469824\">We see that the left side of the equation includes the sines of the sum and the difference of angles.<\/p>\n\n<div id=\"fs-id1469827\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha +\\beta \\right)&amp; =&amp; \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{sin}\\left(\\alpha -\\beta \\right)&amp; =&amp; \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1154806\">We can rewrite each using the sum and difference formulas.<\/p>\n\n<div id=\"fs-id1154809\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha +\\beta \\right)+\\mathrm{sin}\\left(\\alpha -\\beta \\right)&amp; =&amp; \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ &amp; =&amp; 2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1947264\">We see that the identity is verified.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_02_09\" class=\"textbox examples\">\n<div id=\"fs-id1858950\">\n<div id=\"fs-id1858952\">\n<h3>Verifying an Identity Involving Tangent<\/h3>\n<p id=\"fs-id1858957\">Verify the following identity.<\/p>\n\n<div id=\"fs-id1858960\" class=\"unnumbered aligncenter\">[latex]\\frac{\\mathrm{sin}\\left(\\alpha -\\beta \\right)}{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }=\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta [\/latex]<\/div>\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1636592\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1636592\"]\n<p id=\"fs-id1636592\">We can begin by rewriting the numerator on the left side of the equation.<\/p>\n\n<div id=\"fs-id1636595\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{cccc}\\hfill \\frac{\\mathrm{sin}\\left(\\alpha -\\beta \\right)}{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }&amp; =&amp; \\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\mathrm{cos}\\,\\beta }\\hfill &amp; \\\\ &amp; =&amp; \\frac{\\mathrm{sin}\\,\\alpha \\overline{)\\,\\mathrm{cos}\\,\\beta }}{\\mathrm{cos}\\,\\alpha \\overline{)\\,\\mathrm{cos}\\,\\beta }}-\\frac{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\mathrm{sin}\\,\\beta }{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\mathrm{cos}\\,\\beta }\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Rewrite using a common denominator}.\\hfill \\\\ &amp; =&amp; \\frac{\\mathrm{sin}\\,\\alpha }{\\mathrm{cos}\\,\\alpha }-\\frac{\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\beta }\\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Cancel}.\\hfill \\\\ &amp; =&amp; \\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta \\hfill &amp; \\phantom{\\rule{2em}{0ex}}\\text{Rewrite in terms of tangent}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1260145\">We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.[\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2822685\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1248124\">\n<p id=\"fs-id1248125\">Verify the identity:[latex]\\,\\mathrm{tan}\\left(\\pi -\\theta \\right)=-\\mathrm{tan}\\,\\theta .[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">\n\n[reveal-answer q=\"1152803\"]Show Solution[\/reveal-answer][hidden-answer a=\"1152803\"]\n<div id=\"fs-id1152805\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\pi -\\theta \\right)&amp; =&amp; \\frac{\\mathrm{tan}\\left(\\pi \\right)-\\mathrm{tan}\\,\\theta }{1+\\mathrm{tan}\\left(\\pi \\right)\\mathrm{tan}\\theta }\\hfill \\\\ &amp; =&amp; \\frac{0-\\mathrm{tan}\\,\\theta }{1+0\\cdot \\mathrm{tan}\\,\\theta }\\hfill \\\\ &amp; =&amp; -\\mathrm{tan}\\,\\theta \\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_02_10\" class=\"textbox examples\">\n<div id=\"fs-id2092391\">\n<div id=\"fs-id2092393\">\n<h3>Using Sum and Difference Formulas to Solve an Application Problem<\/h3>\n<p id=\"fs-id2092399\">Let[latex]\\,{L}_{1}\\,[\/latex]and[latex]\\,{L}_{2}\\,[\/latex]denote two non-vertical intersecting lines, and let[latex]\\,\\theta \\,[\/latex]denote the acute angle between[latex]\\,{L}_{1}\\,[\/latex]and[latex]\\,{L}_{2}.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Figure_07_02_005\">(Figure)<\/a>. Show that<\/p>\n\n<div id=\"fs-id1728421\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\,\\theta =\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}[\/latex]<\/div>\n<p id=\"fs-id1924955\">where[latex]\\,{m}_{1}\\,[\/latex]and[latex]\\,{m}_{2}\\,[\/latex]are the slopes of[latex]\\,{L}_{1}\\,[\/latex]and[latex]\\,{L}_{2}\\,[\/latex]respectively. (<strong>Hint:<\/strong> Use the fact that[latex]\\,\\mathrm{tan}\\,{\\theta }_{1}={m}_{1}\\,[\/latex]and[latex]\\,\\mathrm{tan}\\,{\\theta }_{2}={m}_{2}.[\/latex])<\/p>\n\n<div class=\"small\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143941\/CNX_Precalc_Figure_07_02_005.jpg\" alt=\"Diagram of two non-vertical intersecting lines L1 and L2 also intersecting the x-axis. The acute angle formed by the intersection of L1 and L2 is theta. The acute angle formed by L2 and the x-axis is theta 1, and the acute angle formed by the x-axis and L1 is theta 2.\" width=\"487\" height=\"289\"> <strong>Figure 7.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2260061\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2260061\"]\n<p id=\"fs-id2260061\">Using the difference formula for tangent, this problem does not seem as daunting as it might.<\/p>\n\n<div id=\"fs-id2260064\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\,\\theta &amp; =&amp; \\mathrm{tan}\\left({\\theta }_{2}-{\\theta }_{1}\\right)\\hfill \\\\ &amp; =&amp; \\frac{\\mathrm{tan}\\,{\\theta }_{2}-\\mathrm{tan}\\,{\\theta }_{1}}{1+\\mathrm{tan}\\,{\\theta }_{1}\\mathrm{tan}\\,{\\theta }_{2}}\\hfill \\\\ &amp; =&amp; \\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_02_11\" class=\"textbox examples\">\n<div id=\"fs-id1367169\">\n<div id=\"fs-id1784865\">\n<h3>Investigating a Guy-wire Problem<\/h3>\n<p id=\"fs-id1784870\">For a climbing wall, a guy-wire[latex]\\,R\\,[\/latex]is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire[latex]\\,S\\,[\/latex]attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle[latex]\\,\\alpha \\,[\/latex]between the wires. See <a class=\"autogenerated-content\" href=\"#Figure_07_02_006\">(Figure)<\/a>.<\/p>\n\n<div class=\"small\">\n\n[caption id=\"\" align=\"aligncenter\" width=\"590\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143949\/CNX_Precalc_Figure_07_02_006.jpg\" alt=\"Two right triangles. Both share the same base, 50 feet. The first has a height of 40 ft and hypotenuse S. The second has height 47 ft and hypotenuse R. The height sides of the triangles are overlapping. There is a B degree angle between R and the base, and an a degree angle between the two hypotenuses within the B degree angle.\" width=\"590\" height=\"322\"> <strong>Figure 8.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2136278\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2136278\"]\n<p id=\"fs-id2136278\">Let\u2019s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that[latex]\\,\\mathrm{tan}\\,\\beta =\\frac{47}{50},[\/latex]and[latex]\\,\\mathrm{tan}\\left(\\beta -\\alpha \\right)=\\frac{40}{50}=\\frac{4}{5}.\\,[\/latex]We can then use difference formula for tangent.<\/p>\n\n<div id=\"fs-id2177692\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\left(\\beta -\\alpha \\right)=\\frac{\\mathrm{tan}\\,\\beta -\\mathrm{tan}\\,\\alpha }{1+\\mathrm{tan}\\,\\beta \\mathrm{tan}\\,\\alpha }[\/latex]<\/div>\n<p id=\"fs-id2007573\">Now, substituting the values we know into the formula, we have<\/p>\n\n<div id=\"fs-id2007576\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\frac{4}{5}&amp; =&amp; \\frac{\\frac{47}{50}-\\mathrm{tan}\\,\\alpha }{1+\\frac{47}{50}\\mathrm{tan}\\,\\alpha }\\hfill \\\\ \\hfill 4\\left(1+\\frac{47}{50}\\mathrm{tan}\\,\\alpha \\right)&amp; =&amp; 5\\left(\\frac{47}{50}-\\mathrm{tan}\\,\\alpha \\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1933373\">Use the distributive property, and then simplify the functions.<\/p>\n\n<div id=\"fs-id1933377\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill 4\\left(1\\right)+4\\left(\\frac{47}{50}\\right)\\mathrm{tan}\\,\\alpha &amp; =&amp; 5\\left(\\frac{47}{50}\\right)-5\\,\\mathrm{tan}\\,\\alpha \\hfill \\\\ \\hfill 4+3.76\\,\\mathrm{tan}\\,\\alpha &amp; =&amp; 4.7-5\\,\\mathrm{tan}\\,\\alpha \\hfill \\\\ \\hfill 5\\,\\mathrm{tan}\\,\\alpha +3.76\\,\\mathrm{tan}\\,\\alpha &amp; =&amp; 0.7\\hfill \\\\ \\hfill 8.76\\,\\mathrm{tan}\\,\\alpha &amp; =&amp; 0.7\\hfill \\\\ \\hfill \\mathrm{tan}\\,\\alpha &amp; \\approx &amp; 0.07991\\hfill \\\\ \\hfill {\\mathrm{tan}}^{-1}\\left(0.07991\\right)&amp; \\approx &amp; .079741\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1691817\">Now we can calculate the angle in degrees.<\/p>\n\n<div id=\"fs-id1691820\" class=\"unnumbered aligncenter\">[latex]\\alpha \\approx 0.079741\\left(\\frac{180}{\\pi }\\right)\\approx 4.57\u00b0[\/latex][\/hidden-answer]<\/div>\n<\/div>\n<div id=\"fs-id2230083\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id2230090\">Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1243690\" class=\"precalculus media\">\n<p id=\"eip-id3766817\">Access these online resources for additional instruction and practice with sum and difference identities.<\/p>\n\n<ul id=\"fs-id1243697\">\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/sumdifcos\">Sum and Difference Identities for Cosine<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/sumdifsin\">Sum and Difference Identities for Sine<\/a><\/li>\n \t<li><a href=\"http:\/\/openstaxcollege.org\/l\/sumdiftan\">Sum and Difference Identities for Tangent<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id2332636\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"fs-id2332643\" summary=\"..\">\n<tbody>\n<tr>\n<td>Sum Formula for Cosine<\/td>\n<td>[latex]\\mathrm{cos}\\left(\\alpha +\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\mathrm{sin}\\,\\beta [\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Difference Formula for Cosine<\/td>\n<td>[latex]\\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Sum Formula for Sine<\/td>\n<td>[latex]\\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Difference Formula for Sine<\/td>\n<td>[latex]\\mathrm{sin}\\left(\\alpha -\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta [\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Sum Formula for Tangent<\/td>\n<td>[latex]\\mathrm{tan}\\left(\\alpha +\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Difference Formula for Tangent<\/td>\n<td>[latex]\\mathrm{tan}\\left(\\alpha -\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta }{1+\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Cofunction identities<\/td>\n<td>[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\,\\theta &amp; =&amp; \\mathrm{cos}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{cos}\\,\\theta &amp; =&amp; \\mathrm{sin}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{tan}\\,\\theta &amp; =&amp; \\mathrm{cot}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{cot}\\,\\theta &amp; =&amp; \\mathrm{tan}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{sec}\\,\\theta &amp; =&amp; \\mathrm{csc}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{csc}\\,\\theta &amp; =&amp; \\mathrm{sec}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id2224028\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id2224035\">\n \t<li>The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles.<\/li>\n \t<li>The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See <a class=\"autogenerated-content\" href=\"#Example_07_02_01\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_07_02_02\">(Figure)<\/a>.<\/li>\n \t<li>The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See <a class=\"autogenerated-content\" href=\"#Example_07_02_03\">(Figure)<\/a>.<\/li>\n \t<li>The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See <a class=\"autogenerated-content\" href=\"#Example_07_02_04\">(Figure)<\/a>.<\/li>\n \t<li>The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles. See <a class=\"autogenerated-content\" href=\"#Example_07_02_05\">(Figure)<\/a>.<\/li>\n \t<li>The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See <a class=\"autogenerated-content\" href=\"#Example_07_02_06\">(Figure)<\/a>.<\/li>\n \t<li>The cofunction identities apply to complementary angles and pairs of reciprocal functions. See <a class=\"autogenerated-content\" href=\"#Example_07_02_07\">(Figure)<\/a>.<\/li>\n \t<li>Sum and difference formulas are useful in verifying identities. See <a class=\"autogenerated-content\" href=\"#Example_07_02_08\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_07_02_09\">(Figure)<\/a>.<\/li>\n \t<li>Application problems are often easier to solve by using sum and difference formulas. See <a class=\"autogenerated-content\" href=\"#Example_07_02_10\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_07_02_10\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id2633717\" class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div id=\"fs-id2633720\" class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id2633726\">\n<div id=\"fs-id2633727\">\n<p id=\"fs-id2633728\">Explain the basis for the cofunction identities and when they apply.<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1879715\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1879715\"]\n<p id=\"fs-id1879715\">The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures[latex]\\,x,[\/latex]the second angle measures[latex]\\,\\frac{\\pi }{2}-x.\\,[\/latex]Then[latex]\\,\\mathrm{sin}x=\\mathrm{cos}\\left(\\frac{\\pi }{2}-x\\right).\\,[\/latex]The same holds for the other cofunction identities. The key is that the angles are complementary.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1784023\">\n<div id=\"fs-id1784024\">\n<p id=\"fs-id1784025\">Is there only one way to evaluate[latex]\\,\\mathrm{cos}\\left(\\frac{5\\pi }{4}\\right)?\\,[\/latex]Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1943345\">\n<div id=\"fs-id1943346\">\n<p id=\"fs-id1943347\">Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for[latex]\\,f\\left(x\\right)=\\mathrm{sin}\\left(x\\right)\\,[\/latex]and[latex]\\,g\\left(x\\right)=\\mathrm{cos}\\left(x\\right).\\,[\/latex](Hint:[latex]\\,0-x=-x[\/latex])<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2169010\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2169010\"]\n<p id=\"fs-id2169010\">[latex]\\mathrm{sin}\\left(-x\\right)=-\\mathrm{sin}x,[\/latex]so[latex]\\,\\mathrm{sin}x\\,[\/latex] is odd.[latex]\\,\\mathrm{cos}\\left(-x\\right)=\\mathrm{cos}\\left(0-x\\right)=\\mathrm{cos}x,[\/latex]so[latex]\\,\\mathrm{cos}x\\,[\/latex]is even.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2030146\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\n<p id=\"fs-id2030152\">For the following exercises, find the exact value.<\/p>\n\n<div id=\"fs-id1535795\">\n<div id=\"fs-id1535796\">\n<p id=\"fs-id1535797\">[latex]\\mathrm{cos}\\left(\\frac{7\\pi }{12}\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2686776\">\n<div id=\"fs-id2686777\">\n<p id=\"fs-id2686778\">[latex]\\mathrm{cos}\\left(\\frac{\\pi }{12}\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1546201\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1546201\"]\n<p id=\"fs-id1546201\">[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1343104\">\n<div id=\"fs-id1343105\">\n<p id=\"fs-id1343106\">[latex]\\mathrm{sin}\\left(\\frac{5\\pi }{12}\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1662806\">\n<div id=\"fs-id1662807\">\n<p id=\"fs-id1662808\">[latex]\\mathrm{sin}\\left(\\frac{11\\pi }{12}\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1617567\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1617567\"]\n<p id=\"fs-id1617567\">[latex]\\frac{\\sqrt{6}-\\sqrt{2}}{4}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2327970\">\n<div id=\"fs-id2327971\">\n<p id=\"fs-id2327972\">[latex]\\mathrm{tan}\\left(-\\frac{\\pi }{12}\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2641720\">\n<div id=\"fs-id2641721\">\n<p id=\"fs-id2641722\">[latex]\\mathrm{tan}\\left(\\frac{19\\pi }{12}\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2340651\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2340651\"]\n<p id=\"fs-id2340651\">[latex]-2-\\sqrt{3}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<p id=\"fs-id1872453\">For the following exercises, rewrite in terms of[latex]\\,\\mathrm{sin}\\,x\\,[\/latex]and[latex]\\,\\mathrm{cos}\\,x.[\/latex]<\/p>\n\n<div id=\"fs-id1944688\">\n<div id=\"fs-id1944689\">\n<p id=\"fs-id1944690\">[latex]\\mathrm{sin}\\left(x+\\frac{11\\pi }{6}\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1924818\">\n<div id=\"fs-id1924819\">\n<p id=\"fs-id1924820\">[latex]\\mathrm{sin}\\left(x-\\frac{3\\pi }{4}\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2907495\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2907495\"]\n<p id=\"fs-id2907495\">[latex]-\\frac{\\sqrt{2}}{2}\\mathrm{sin}x-\\frac{\\sqrt{2}}{2}\\mathrm{cos}x[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2289260\">\n<div id=\"fs-id2289261\">\n<p id=\"fs-id2289262\">[latex]\\mathrm{cos}\\left(x-\\frac{5\\pi }{6}\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1923485\">\n<div id=\"fs-id1923486\">\n<p id=\"fs-id1923487\">[latex]\\mathrm{cos}\\left(x+\\frac{2\\pi }{3}\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1784096\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1784096\"]\n<p id=\"fs-id1784096\">[latex]-\\frac{1}{2}\\mathrm{cos}x-\\frac{\\sqrt{3}}{2}\\mathrm{sin}x[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<p id=\"fs-id1186242\">For the following exercises, simplify the given expression.<\/p>\n\n<div id=\"fs-id1587348\">\n<div id=\"fs-id1587349\">\n<p id=\"fs-id1587350\">[latex]\\mathrm{csc}\\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1931167\">\n<div id=\"fs-id1931168\">\n<p id=\"fs-id1931169\">[latex]\\mathrm{sec}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1973314\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1973314\"]\n<p id=\"fs-id1973314\">[latex]\\mathrm{csc}\\theta [\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1405117\">\n<div id=\"fs-id1405118\">\n<p id=\"fs-id1405119\">[latex]\\mathrm{cot}\\left(\\frac{\\pi }{2}-x\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1223240\">\n<div id=\"fs-id1223241\">\n<p id=\"fs-id2168811\">[latex]\\mathrm{tan}\\left(\\frac{\\pi }{2}-x\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2780587\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2780587\"]\n<p id=\"fs-id2780587\">[latex]\\mathrm{cot}x[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2306521\">\n<div id=\"fs-id2306522\">\n<p id=\"fs-id2306523\">[latex]\\mathrm{sin}\\left(2x\\right)\\,\\mathrm{cos}\\left(5x\\right)-\\mathrm{sin}\\left(5x\\right)\\,\\mathrm{cos}\\left(2x\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1359485\">\n<div id=\"fs-id2164270\">\n<p id=\"fs-id2164271\">[latex]\\frac{\\mathrm{tan}\\left(\\frac{3}{2}x\\right)-\\mathrm{tan}\\left(\\frac{7}{5}x\\right)}{1+\\mathrm{tan}\\left(\\frac{3}{2}x\\right)\\mathrm{tan}\\left(\\frac{7}{5}x\\right)}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2148686\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2148686\"]\n<p id=\"fs-id2148686\">[latex]\\mathrm{tan}\\left(\\frac{x}{10}\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<p id=\"fs-id1912452\">For the following exercises, find the requested information.<\/p>\n\n<div id=\"fs-id1912455\">\n<div id=\"fs-id1912456\">\n<p id=\"fs-id1912457\">Given that[latex]\\,\\mathrm{sin}\\,a=\\frac{2}{3}\\,[\/latex]and[latex]\\,\\mathrm{cos}\\,b=-\\frac{1}{4},[\/latex]with[latex]\\,a\\,[\/latex]and[latex]\\,b\\,[\/latex]both in the interval[latex]\\,\\left[\\frac{\\pi }{2},\\pi \\right),[\/latex]find[latex]\\,\\mathrm{sin}\\left(a+b\\right)\\,[\/latex]and[latex]\\,\\mathrm{cos}\\left(a-b\\right).[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1256165\">\n<div id=\"fs-id1256166\">\n<p id=\"fs-id1256167\">Given that[latex]\\,\\mathrm{sin}\\,a=\\frac{4}{5},[\/latex]and[latex]\\,\\mathrm{cos}\\,b=\\frac{1}{3},[\/latex]with[latex]\\,a\\,[\/latex]and[latex]\\,b\\,[\/latex]both in the interval[latex]\\,\\left[0,\\frac{\\pi }{2}\\right),[\/latex]find[latex]\\,\\mathrm{sin}\\left(a-b\\right)\\,[\/latex]and[latex]\\,\\mathrm{cos}\\left(a+b\\right).[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2011929\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2011929\"]\n<p id=\"fs-id2011929\">[latex]\\begin{array}{ccccc}\\hfill \\mathrm{sin}\\left(a-b\\right)&amp; =&amp; \\left(\\frac{4}{5}\\right)\\left(\\frac{1}{3}\\right)-\\left(\\frac{3}{5}\\right)\\left(\\frac{2\\sqrt{2}}{3}\\right)\\hfill &amp; =&amp; \\frac{4-6\\sqrt{2}}{15}\\hfill \\\\ \\hfill \\mathrm{cos}\\left(a+b\\right)&amp; =&amp; \\left(\\frac{3}{5}\\right)\\left(\\frac{1}{3}\\right)-\\left(\\frac{4}{5}\\right)\\left(\\frac{2\\sqrt{2}}{3}\\right)\\hfill &amp; =&amp; \\frac{3-8\\sqrt{2}}{15}\\hfill \\end{array}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<p id=\"fs-id2478807\">For the following exercises, find the exact value of each expression.<\/p>\n\n<div id=\"fs-id2478810\">\n<div id=\"fs-id2478811\">\n<p id=\"fs-id2478812\">[latex]\\mathrm{sin}\\left({\\mathrm{cos}}^{-1}\\left(0\\right)-{\\mathrm{cos}}^{-1}\\left(\\frac{1}{2}\\right)\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2126676\">\n<div id=\"fs-id2126677\">\n<p id=\"fs-id2126678\">[latex]\\mathrm{cos}\\left({\\mathrm{cos}}^{-1}\\left(\\frac{\\sqrt{2}}{2}\\right)+{\\mathrm{sin}}^{-1}\\left(\\frac{\\sqrt{3}}{2}\\right)\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2229931\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2229931\"]\n<p id=\"fs-id2229931\">[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2337567\">\n<div id=\"fs-id2336035\">\n<p id=\"fs-id2336036\">[latex]\\mathrm{tan}\\left({\\mathrm{sin}}^{-1}\\left(\\frac{1}{2}\\right)-{\\mathrm{cos}}^{-1}\\left(\\frac{1}{2}\\right)\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1408229\" class=\"bc-section section\">\n<h4>Graphical<\/h4>\n<p id=\"fs-id2259918\">For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. Confirm your answer using a graphing calculator.<\/p>\n\n<div id=\"fs-id2259922\">\n<div id=\"fs-id2259924\">\n<p id=\"fs-id2259925\">[latex]\\mathrm{cos}\\left(\\frac{\\pi }{2}-x\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1411055\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1411055\"]\n<p id=\"fs-id1411055\">[latex]\\mathrm{sin}x[\/latex]<\/p>\n<span id=\"fs-id1615462\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143955\/CNX_Precalc_Figure_07_02_201.jpg\" alt=\"Graph of y=sin(x) from -2pi to 2pi.\"><\/span>[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1555893\">\n<div id=\"fs-id1555894\">\n<p id=\"fs-id1555895\">[latex]\\mathrm{sin}\\left(\\pi -x\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2960991\">\n<div id=\"fs-id2960992\">\n<p id=\"fs-id1430117\">[latex]\\mathrm{tan}\\left(\\frac{\\pi }{3}+x\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2113190\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2113190\"]\n<p id=\"fs-id2113190\">[latex]\\mathrm{cot}\\left(\\frac{\\pi }{6}-x\\right)[\/latex]<\/p>\n<span id=\"fs-id2233832\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19144001\/CNX_Precalc_Figure_07_02_203.jpg\" alt=\"Graph of y=cot(pi\/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi\/6.\"><\/span>[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1502696\">\n<div id=\"fs-id1502697\">\n<p id=\"fs-id1502698\">[latex]\\mathrm{sin}\\left(\\frac{\\pi }{3}+x\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1724458\">\n<div id=\"fs-id1724459\">\n<p id=\"fs-id1724460\">[latex]\\mathrm{tan}\\left(\\frac{\\pi }{4}-x\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2307484\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2307484\"]\n<p id=\"fs-id2307484\">[latex]\\mathrm{cot}\\left(\\frac{\\pi }{4}+x\\right)[\/latex]<\/p>\n<span id=\"fs-id1877975\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19144008\/CNX_Precalc_Figure_07_02_205.jpg\" alt=\"Graph of y=cot(pi\/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi\/4.\"><\/span>[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1252765\">\n<div id=\"fs-id1252766\">\n<p id=\"fs-id1252768\">[latex]\\mathrm{cos}\\left(\\frac{7\\pi }{6}+x\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1681120\">\n<div id=\"fs-id2255437\">\n<p id=\"fs-id2255438\">[latex]\\mathrm{sin}\\left(\\frac{\\pi }{4}+x\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1752329\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1752329\"]\n<p id=\"fs-id1752329\">[latex]\\frac{\\mathrm{sin}x}{\\sqrt{2}}+\\frac{\\mathrm{cos}x}{\\sqrt{2}}[\/latex]<\/p>\n<span id=\"fs-id2199923\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19144020\/CNX_Precalc_Figure_07_02_207.jpg\" alt=\"Graph of y = sin(x) \/ rad2 + cos(x) \/ rad2 - it looks like the sin curve shifted by pi\/4.\"><\/span>[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2199934\">\n<div id=\"fs-id2199935\">\n<p id=\"fs-id1672664\">[latex]\\mathrm{cos}\\left(\\frac{5\\pi }{4}+x\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<p id=\"fs-id2710551\">For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think[latex]\\,2x=x+x.[\/latex]\n)<\/p>\n\n<div id=\"fs-id1493462\">\n<div id=\"fs-id2308858\">\n<p id=\"fs-id2308859\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(4x\\right)-\\mathrm{sin}\\left(3x\\right)\\mathrm{cos}\\,x,g\\left(x\\right)=\\mathrm{sin}\\,x\\,\\mathrm{cos}\\left(3x\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2253381\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2253381\"]\n<p id=\"fs-id2253381\">They are the same.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1086612\">\n<div id=\"fs-id1086613\">\n<p id=\"fs-id1086614\">[latex]f\\left(x\\right)=\\mathrm{cos}\\left(4x\\right)+\\mathrm{sin}\\,x\\,\\mathrm{sin}\\left(3x\\right),g\\left(x\\right)=-\\mathrm{cos}\\,x\\,\\mathrm{cos}\\left(3x\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2102136\">\n<div id=\"fs-id2102137\">\n<p id=\"fs-id2102138\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(3x\\right)\\mathrm{cos}\\left(6x\\right),g\\left(x\\right)=-\\mathrm{sin}\\left(3x\\right)\\mathrm{cos}\\left(6x\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2190228\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2190228\"]\n<p id=\"fs-id2190228\">They are the different, try[latex]\\,g\\left(x\\right)=\\mathrm{sin}\\left(9x\\right)-\\mathrm{cos}\\left(3x\\right)\\mathrm{sin}\\left(6x\\right).[\/latex][\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1367081\">\n<div id=\"fs-id1367082\">\n<p id=\"fs-id1367083\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(4x\\right),g\\left(x\\right)=\\mathrm{sin}\\left(5x\\right)\\mathrm{cos}\\,x-\\mathrm{cos}\\left(5x\\right)\\mathrm{sin}\\,x[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1573766\">\n<div id=\"fs-id1573767\">\n<p id=\"fs-id1573768\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(2x\\right),g\\left(x\\right)=2\\,\\mathrm{sin}\\,x\\,\\mathrm{cos}\\,x[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1544808\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1544808\"]\n<p id=\"fs-id1544808\">They are the same.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1544812\">\n<div id=\"fs-id1544813\">\n<p id=\"fs-id1544814\">[latex]f\\left(\\theta \\right)=\\mathrm{cos}\\left(2\\theta \\right),g\\left(\\theta \\right)={\\mathrm{cos}}^{2}\\theta -{\\mathrm{sin}}^{2}\\theta [\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1687448\">\n<div id=\"fs-id1687449\">\n<p id=\"fs-id1687450\">[latex]f\\left(\\theta \\right)=\\mathrm{tan}\\left(2\\theta \\right),g\\left(\\theta \\right)=\\frac{\\mathrm{tan}\\,\\theta }{1+{\\mathrm{tan}}^{2}\\theta }[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1186012\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1186012\"]\n<p id=\"fs-id1186012\">They are the different, try[latex]\\,g\\left(\\theta \\right)=\\frac{2\\,\\mathrm{tan}\\theta }{1-{\\mathrm{tan}}^{2}\\theta }.[\/latex][\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2888541\">\n<div id=\"fs-id2888542\">\n<p id=\"fs-id2888544\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(3x\\right)\\mathrm{sin}\\,x,g\\left(x\\right)={\\mathrm{sin}}^{2}\\left(2x\\right){\\mathrm{cos}}^{2}x-{\\mathrm{cos}}^{2}\\left(2x\\right){\\mathrm{sin}}^{2}x[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1664714\">\n<div id=\"fs-id1664715\">\n<p id=\"fs-id1664716\">[latex]f\\left(x\\right)=\\mathrm{tan}\\left(-x\\right),g\\left(x\\right)=\\frac{\\mathrm{tan}\\,x-\\mathrm{tan}\\left(2x\\right)}{1-\\mathrm{tan}\\,x\\,\\mathrm{tan}\\left(2x\\right)}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1547118\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1547118\"]\n<p id=\"fs-id1547118\">They are different, try[latex]\\,g\\left(x\\right)=\\frac{\\mathrm{tan}x-\\mathrm{tan}\\left(2x\\right)}{1+\\mathrm{tan}x\\mathrm{tan}\\left(2x\\right)}.[\/latex][\/hidden-answer]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h4>Technology<\/h4>\n<p id=\"fs-id1429674\">For the following exercises, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point.<\/p>\n\n<div id=\"fs-id1429678\">\n<div id=\"fs-id1429679\">\n<p id=\"fs-id1429680\">[latex]\\mathrm{sin}\\left(75\u00b0\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1604725\">\n<div id=\"fs-id1604726\">\n<p id=\"fs-id1604727\">[latex]\\mathrm{sin}\\left(195\u00b0\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1874102\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1874102\"]\n<p id=\"fs-id1874102\">[latex]-\\frac{\\sqrt{3}-1}{2\\sqrt{2}},\\text{or }-0.2588[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2160300\">\n<div id=\"fs-id2160301\">\n<p id=\"fs-id2160302\">[latex]\\mathrm{cos}\\left(165\u00b0\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2613308\">\n<div id=\"fs-id2613309\">\n<p id=\"fs-id2613310\">[latex]\\mathrm{cos}\\left(345\u00b0\\right)[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1676814\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1676814\"]\n<p id=\"fs-id1676814\">[latex]\\frac{1+\\sqrt{3}}{2\\sqrt{2}},[\/latex]or 0.9659<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2291639\">\n<div id=\"fs-id2291640\">\n<p id=\"fs-id2291641\">[latex]\\mathrm{tan}\\left(-15\u00b0\\right)[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2244374\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<p id=\"fs-id2244379\">For the following exercises, prove the identities provided.<\/p>\n\n<div id=\"fs-id2244382\">\n<div id=\"fs-id2244383\">\n<p id=\"fs-id2244384\">[latex]\\mathrm{tan}\\left(x+\\frac{\\pi }{4}\\right)=\\frac{\\mathrm{tan}\\,x+1}{1-\\mathrm{tan}\\,x}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2884049\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2884049\"]\n<p id=\"fs-id2884049\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(x+\\frac{\\pi }{4}\\right)&amp; =&amp; \\\\ \\hfill \\frac{\\mathrm{tan}x+\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)}{1-\\mathrm{tan}x\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)}&amp; =&amp; \\\\ \\hfill \\frac{\\mathrm{tan}x+1}{1-\\mathrm{tan}x\\left(1\\right)}&amp; =&amp; \\frac{\\mathrm{tan}x+1}{1-\\mathrm{tan}x}\\hfill \\end{array}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1743113\">\n<div id=\"fs-id1743114\">\n<p id=\"fs-id1743115\">[latex]\\frac{\\mathrm{tan}\\left(a+b\\right)}{\\mathrm{tan}\\left(a-b\\right)}=\\frac{\\mathrm{sin}\\,a\\,\\mathrm{cos}\\,a+\\mathrm{sin}\\,b\\,\\mathrm{cos}\\,b}{\\mathrm{sin}\\,a\\,\\mathrm{cos}\\,a-\\mathrm{sin}\\,b\\,\\mathrm{cos}\\,b}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2024685\">\n<div id=\"fs-id2024686\">\n<p id=\"fs-id2024687\">[latex]\\frac{\\mathrm{cos}\\left(a+b\\right)}{\\mathrm{cos}\\,a\\,\\mathrm{cos}\\,b}=1-\\mathrm{tan}\\,a\\,\\mathrm{tan}\\,b[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2810960\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2810960\"]\n<p id=\"fs-id2810960\">[latex]\\begin{array}{ccc}\\hfill \\frac{\\mathrm{cos}\\left(a+b\\right)}{\\mathrm{cos}a\\mathrm{cos}b}&amp; =&amp; \\\\ \\hfill \\frac{\\mathrm{cos}a\\mathrm{cos}b}{\\mathrm{cos}a\\mathrm{cos}b}-\\frac{\\mathrm{sin}a\\mathrm{sin}b}{\\mathrm{cos}a\\mathrm{cos}b}&amp; =&amp; 1-\\mathrm{tan}a\\mathrm{tan}b\\hfill \\end{array}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2467728\">\n<div id=\"fs-id2467729\">\n<p id=\"fs-id2467730\">[latex]\\mathrm{cos}\\left(x+y\\right)\\mathrm{cos}\\left(x-y\\right)={\\mathrm{cos}}^{2}x-{\\mathrm{sin}}^{2}y[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1831266\">\n<div id=\"fs-id1831267\">\n<p id=\"fs-id1831268\">[latex]\\frac{\\mathrm{cos}\\left(x+h\\right)-\\mathrm{cos}\\,x}{h}=\\mathrm{cos}\\,x\\frac{\\mathrm{cos}\\,h-1}{h}-\\mathrm{sin}\\,x\\frac{\\mathrm{sin}\\,h}{h}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2167170\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2167170\"]\n<p id=\"fs-id2167170\">[latex]\\begin{array}{ccc}\\hfill \\frac{\\mathrm{cos}\\left(x+h\\right)-\\mathrm{cos}x}{h}&amp; =&amp; \\\\ \\hfill \\frac{\\mathrm{cos}x\\mathrm{cosh}-\\mathrm{sin}x\\mathrm{sinh}-\\mathrm{cos}x}{h}&amp; =&amp; \\\\ \\hfill \\frac{\\mathrm{cos}x\\left(\\mathrm{cosh}-1\\right)-\\mathrm{sin}x\\mathrm{sinh}}{h}&amp; =&amp; \\mathrm{cos}x\\frac{\\mathrm{cos}h-1}{h}-\\mathrm{sin}x\\frac{\\mathrm{sin}h}{h}\\hfill \\end{array}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<p id=\"fs-id1647857\">For the following exercises, prove or disprove the statements.<\/p>\n\n<div id=\"fs-id1647860\">\n<div id=\"fs-id1647861\">\n<p id=\"fs-id1647862\">[latex]\\mathrm{tan}\\left(u+v\\right)=\\frac{\\mathrm{tan}\\,u+\\mathrm{tan}\\,v}{1-\\mathrm{tan}\\,u\\,\\mathrm{tan}\\,v}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id2575246\">\n<div id=\"fs-id2575247\">\n<p id=\"fs-id2575248\">[latex]\\mathrm{tan}\\left(u-v\\right)=\\frac{\\mathrm{tan}\\,u-\\mathrm{tan}\\,v}{1+\\mathrm{tan}\\,u\\,\\mathrm{tan}\\,v}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id1369518\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id1369518\"]\n<p id=\"fs-id1369518\">True<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id1369522\">\n<div id=\"fs-id1369523\">\n<p id=\"fs-id1369524\">[latex]\\frac{\\mathrm{tan}\\left(x+y\\right)}{1+\\mathrm{tan}\\,x\\,\\mathrm{tan}\\,x}=\\frac{\\mathrm{tan}\\,x+\\mathrm{tan}\\,y}{1-{\\mathrm{tan}}^{2}x\\,{\\mathrm{tan}}^{2}y}[\/latex]<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-id1859738\">\n<div id=\"fs-id1859739\">\n<p id=\"fs-id1859740\">If[latex]\\,\\alpha ,\\beta ,[\/latex]and[latex]\\,\\gamma \\,[\/latex]are angles in the same triangle, then prove or disprove[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\,\\gamma .[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox shaded\">[reveal-answer q=\"fs-id2290243\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"fs-id2290243\"]\n<p id=\"fs-id2290243\">True. Note that[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\left(\\pi -\\gamma \\right)\\,[\/latex]and expand the right hand side.<\/p>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div id=\"fs-id2640056\">\n<div id=\"fs-id2640057\">\n<p id=\"fs-id2640058\">If[latex]\\,\\alpha ,\\beta ,[\/latex]and[latex]\\,y\\,[\/latex]are angles in the same triangle, then prove or disprove[latex]\\,\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta +\\mathrm{tan}\\,\\gamma =\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta \\,\\mathrm{tan}\\,\\gamma [\/latex]<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Use sum and difference formulas for cosine.<\/li>\n<li>Use sum and difference formulas for sine.<\/li>\n<li>Use sum and difference formulas for tangent.<\/li>\n<li>Use sum and difference formulas for cofunctions.<\/li>\n<li>Use sum and difference formulas to verify identities.<\/li>\n<\/ul>\n<\/div>\n<div class=\"small\">\n<figure style=\"width: 488px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143905\/CNX_Precalc_Figure_07_02_001.jpg\" alt=\"Photo of Mt. McKinley.\" width=\"488\" height=\"325\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 1. <\/strong>Mount McKinley, in Denali National Park, Alaska, rises 20,237 feet (6,168 m) above sea level. It is the highest peak in North America. (credit: Daniel A. Leifheit, Flickr)<\/figcaption><\/figure>\n<div class=\"wp-caption-text\"><\/div>\n<\/div>\n<p id=\"fs-id2463462\">How can the height of a mountain be measured? What about the distance from Earth to the sun? Like many seemingly impossible problems, we rely on mathematical formulas to find the answers. The trigonometric identities, commonly used in mathematical proofs, have had real-world applications for centuries, including their use in calculating long distances.<\/p>\n<p id=\"fs-id1237528\">The trigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. These are special equations or postulates, true for all values input to the equations, and with innumerable applications.<\/p>\n<p id=\"fs-id2285295\">In this section, we will learn techniques that will enable us to solve problems such as the ones presented above. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term <em>formula<\/em> is used synonymously with the word <em>identity<\/em>.<\/p>\n<div id=\"fs-id1902399\" class=\"bc-section section\">\n<h3>Using the Sum and Difference Formulas for Cosine<\/h3>\n<p id=\"fs-id2264960\">Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the <span class=\"no-emphasis\">special angles<\/span>, which we can review in the unit circle shown in <a class=\"autogenerated-content\" href=\"#Figure_07_02_008\">(Figure)<\/a>.<\/p>\n<div class=\"wp-caption aligncenter\">\n<figure style=\"width: 975px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143908\/CNX_Precalc_Figure_07_01_004.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" width=\"975\" height=\"638\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 2. <\/strong>The Unit Circle<\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id2618307\">We will begin with the <span class=\"no-emphasis\">sum and difference formulas for cosine<\/span>, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles. See <a class=\"autogenerated-content\" href=\"#Table_07_02_01\">(Figure)<\/a>.<\/p>\n<table summary=\"Two rows, two columns. The table has ordered pairs of these row values: (Sum formula for cosine, cos(a+B) = cos(a)cos(B) - sin(a)sin(B)) and (Difference formula for cosine, cos(a-B) = cos(a)cos(B) + sin(a)sin(B)).\">\n<tbody>\n<tr>\n<td><strong>Sum formula for cosine<\/strong><\/td>\n<td>[latex]\\mathrm{cos}\\left(\\alpha +\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><strong>Difference formula for cosine<\/strong><\/td>\n<td>[latex]\\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id3007968\">First, we will prove the difference formula for cosines. Let\u2019s consider two points on the unit circle. See <a class=\"autogenerated-content\" href=\"#Figure_07_02_002\">(Figure)<\/a>. Point[latex]\\,P\\,[\/latex]is at an angle[latex]\\,\\alpha \\,[\/latex]from the positive <em>x-<\/em>axis with coordinates[latex]\\,\\left(\\mathrm{cos}\\,\\alpha ,\\mathrm{sin}\\,\\alpha \\right)\\,[\/latex]and point[latex]\\,Q\\,[\/latex]is at an angle of[latex]\\,\\beta \\,[\/latex]from the positive <em>x-<\/em>axis with coordinates[latex]\\,\\left(\\mathrm{cos}\\,\\beta ,\\mathrm{sin}\\,\\beta \\right).\\,[\/latex]Note the measure of angle[latex]\\,POQ\\,[\/latex]is[latex]\\,\\alpha -\\beta .\\,[\/latex]<\/p>\n<p id=\"fs-id1565012\">Label two more points:[latex]\\,A\\,[\/latex]at an angle of[latex]\\,\\left(\\alpha -\\beta \\right)\\,[\/latex]from the positive <em>x-<\/em>axis with coordinates[latex]\\,\\left(\\mathrm{cos}\\left(\\alpha -\\beta \\right),\\mathrm{sin}\\left(\\alpha -\\beta \\right)\\right);\\,[\/latex]and point[latex]\\,B\\,[\/latex]with coordinates[latex]\\,\\left(1,0\\right).\\,[\/latex]Triangle[latex]\\,POQ\\,[\/latex]is a rotation of triangle[latex]\\,AOB\\,[\/latex]and thus the distance from[latex]\\,P\\,[\/latex]to[latex]\\,Q\\,[\/latex]is the same as the distance from[latex]\\,A\\,[\/latex]to[latex]\\,B.[\/latex]<\/p>\n<div class=\"small\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143917\/CNX_Precalc_Figure_07_02_002.jpg\" alt=\"Diagram of the unit circle with points labeled on its edge. P point is at an angle a from the positive x axis with coordinates (cosa, sina). Point Q is at an angle of B from the positive x axis with coordinates (cosb, sinb). Angle POQ is a - B degrees. Point A is at an angle of (a-B) from the x axis with coordinates (cos(a-B), sin(a-B)). Point B is just at point (1,0). Angle AOB is also a - B degrees. Radii PO, AO, QO, and BO are all 1 unit long and are the legs of triangles POQ and AOB. Triangle POQ is a rotation of triangle AOB, so the distance from P to Q is the same as the distance from A to B.\" width=\"487\" height=\"365\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong><\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id2152546\">We can find the distance from[latex]\\,P\\,[\/latex]to[latex]\\,Q\\,[\/latex]using the <span class=\"no-emphasis\">distance formula<\/span>.<\/p>\n<div id=\"fs-id1096956\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill {d}_{PQ}& =& \\sqrt{{\\left(\\mathrm{cos}\\,\\alpha -\\mathrm{cos}\\,\\beta \\right)}^{2}+{\\left(\\mathrm{sin}\\,\\alpha -\\mathrm{sin}\\,\\beta \\right)}^{2}}\\hfill \\\\ & =& \\sqrt{{\\mathrm{cos}}^{2}\\alpha -2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +{\\mathrm{cos}}^{2}\\beta +{\\mathrm{sin}}^{2}\\alpha -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta +{\\mathrm{sin}}^{2}\\beta }\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id2239334\">Then we apply the <span class=\"no-emphasis\">Pythagorean identity<\/span> and simplify.<\/p>\n<div id=\"fs-id1343608\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{cc}=& \\sqrt{\\left({\\mathrm{cos}}^{2}\\alpha +{\\mathrm{sin}}^{2}\\alpha \\right)+\\left({\\mathrm{cos}}^{2}\\beta +{\\mathrm{sin}}^{2}\\beta \\right)-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }\\hfill \\\\ =& \\sqrt{1+1-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }\\hfill \\\\ =& \\sqrt{2-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1342538\">Similarly, using the distance formula we can find the distance from[latex]\\,A\\,[\/latex]to[latex]\\,B.[\/latex]<\/p>\n<div id=\"fs-id1094579\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill {d}_{AB}& =& \\sqrt{{\\left(\\mathrm{cos}\\left(\\alpha -\\beta \\right)-1\\right)}^{2}+{\\left(\\mathrm{sin}\\left(\\alpha -\\beta \\right)-0\\right)}^{2}}\\hfill \\\\ & =& \\sqrt{{\\mathrm{cos}}^{2}\\left(\\alpha -\\beta \\right)-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)+1+{\\mathrm{sin}}^{2}\\left(\\alpha -\\beta \\right)}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1783115\">Applying the Pythagorean identity and simplifying we get:<\/p>\n<div id=\"fs-id1347258\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{cc}=& \\sqrt{\\left({\\mathrm{cos}}^{2}\\left(\\alpha -\\beta \\right)+{\\mathrm{sin}}^{2}\\left(\\alpha -\\beta \\right)\\right)-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)+1}\\hfill \\\\ =& \\sqrt{1-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)+1}\\hfill \\\\ =& \\sqrt{2-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id2196869\">Because the two distances are the same, we set them equal to each other and simplify.<\/p>\n<div id=\"fs-id1729317\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\sqrt{2-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }& =& \\sqrt{2-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)}\\hfill \\\\ \\hfill 2-2\\,\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta & =& 2-2\\,\\mathrm{cos}\\left(\\alpha -\\beta \\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1411215\">Finally we subtract[latex]\\,2\\,[\/latex]from both sides and divide both sides by[latex]\\,-2.[\/latex]<\/p>\n<div id=\"fs-id1491684\" class=\"unnumbered aligncenter\">[latex]\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta =\\mathrm{cos}\\left(\\alpha -\\beta \\right)\\text{ }[\/latex]<\/div>\n<p id=\"fs-id711534\">Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.<\/p>\n<div id=\"fs-id1334205\" class=\"textbox key-takeaways\">\n<h3>Sum and Difference Formulas for Cosine<\/h3>\n<p id=\"fs-id1261808\">These formulas can be used to calculate the cosine of sums and differences of angles.<\/p>\n<div id=\"Eq_07_02_01\">[latex]\\mathrm{cos}\\left(\\alpha +\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/div>\n<div id=\"Eq_07_02_02\">[latex]\\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id861912\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1728576\"><strong>Given two angles, find the cosine of the difference between the angles.<br \/>\n<\/strong><\/p>\n<ol id=\"fs-id1374215\" type=\"1\">\n<li>Write the difference formula for cosine.<\/li>\n<li>Substitute the values of the given angles into the formula.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1733838\">\n<div id=\"fs-id2056629\">\n<h3>Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles<\/h3>\n<p id=\"fs-id1953099\">Using the formula for the cosine of the difference of two angles, find the exact value of[latex]\\,\\mathrm{cos}\\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1128213\">Begin by writing the formula for the cosine of the difference of two angles. Then substitute the given values.<\/p>\n<div id=\"fs-id1369035\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\alpha -\\beta \\right)& =& \\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{cos}\\left(\\frac{5\\pi }{4}-\\frac{\\pi }{6}\\right)& =& \\mathrm{cos}\\left(\\frac{5\\pi }{4}\\right)\\mathrm{cos}\\left(\\frac{\\pi }{6}\\right)+\\mathrm{sin}\\left(\\frac{5\\pi }{4}\\right)\\mathrm{sin}\\left(\\frac{\\pi }{6}\\right)\\hfill \\\\ & =& \\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)-\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{1}{2}\\right)\\hfill \\\\ & =& -\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4}\\hfill \\\\ & =& \\frac{-\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"eip-id1256870\">Keep in mind that we can always check the answer using a graphing calculator in radian mode.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1369787\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id2118816\">\n<p id=\"fs-id1557062\">Find the exact value of[latex]\\,\\mathrm{cos}\\left(\\frac{\\pi }{3}-\\frac{\\pi }{4}\\right).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1721765\">[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id2956024\">\n<div id=\"fs-id1367064\">\n<h3>Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine<\/h3>\n<p id=\"fs-id1161788\">Find the exact value of[latex]\\,\\mathrm{cos}\\left(75\u00b0\\right).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1711386\">As[latex]\\,75\u00b0=45\u00b0+30\u00b0,[\/latex]we can evaluate[latex]\\,\\mathrm{cos}\\left(75\u00b0\\right)\\,[\/latex]as[latex]\\,\\mathrm{cos}\\left(45\u00b0+30\u00b0\\right).\\,[\/latex]<\/p>\n<div id=\"fs-id1834810\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\alpha +\\beta \\right)& =& \\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{cos}\\left(45\u00b0+30\u00b0\\right)& =& \\mathrm{cos}\\left(45\u00b0\\right)\\mathrm{cos}\\left(30\u00b0\\right)-\\mathrm{sin}\\left(45\u00b0\\right)\\mathrm{sin}\\left(30\u00b0\\right)\\hfill \\\\ & =& \\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right)\\hfill \\\\ & =& \\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{2}}{4}\\hfill \\\\ & =& \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"eip-id1731565\">Keep in mind that we can always check the answer using a graphing calculator in degree mode.<\/p>\n<\/details>\n<\/div>\n<div id=\"fs-id1905704\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id2283950\">Note that we could have also solved this problem using the fact that[latex]\\,75\u00b0=135\u00b0-60\u00b0.[\/latex]<\/p>\n<div id=\"fs-id1924708\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\alpha -\\beta \\right)& =& \\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{cos}\\left(135\u00b0-60\u00b0\\right)& =& \\mathrm{cos}\\left(135\u00b0\\right)\\mathrm{cos}\\left(60\u00b0\\right)+\\mathrm{sin}\\left(135\u00b0\\right)\\mathrm{sin}\\left(60\u00b0\\right)\\hfill \\\\ & =& \\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{1}{2}\\right)+\\left(\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ & =& \\left(-\\frac{\\sqrt{2}}{4}\\right)+\\left(\\frac{\\sqrt{6}}{4}\\right)\\hfill \\\\ & =& \\left(\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\right)\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1335750\">\n<p id=\"fs-id1231482\">Find the exact value of[latex]\\,\\mathrm{cos}\\left(105\u00b0\\right).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1094264\">[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1388326\" class=\"bc-section section\">\n<h3>Using the Sum and Difference Formulas for Sine<\/h3>\n<p id=\"fs-id1198942\">The <span class=\"no-emphasis\">sum and difference formulas for sine<\/span> can be derived in the same manner as those for cosine, and they resemble the cosine formulas.<\/p>\n<div id=\"fs-id1345043\" class=\"textbox key-takeaways\">\n<h3>Sum and Difference Formulas for Sine<\/h3>\n<p id=\"fs-id2335392\">These formulas can be used to calculate the sines of sums and differences of angles.<\/p>\n<div id=\"Eq_07_02_03\">[latex]\\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/div>\n<div id=\"Eq_07_02_04\">[latex]\\mathrm{sin}\\left(\\alpha -\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1274546\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1891863\"><strong>Given two angles, find the sine of the difference between the angles.<br \/>\n<\/strong><\/p>\n<ol id=\"fs-id1352419\" type=\"1\">\n<li>Write the difference formula for sine.<\/li>\n<li>Substitute the given angles into the formula.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1368235\">\n<div id=\"fs-id1242560\">\n<h3>Using Sum and Difference Identities to Evaluate the Difference of Angles<\/h3>\n<p id=\"fs-id1304945\">Use the sum and difference identities to evaluate the difference of the angles and show that part <em>a<\/em> equals part <em>b.<\/em><\/p>\n<ol id=\"fs-id1978491\" type=\"a\">\n<li>[latex]\\mathrm{sin}\\left(45\u00b0-30\u00b0\\right)[\/latex]<\/li>\n<li>[latex]\\mathrm{sin}\\left(135\u00b0-120\u00b0\\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<ol id=\"fs-id2291665\" type=\"a\">\n<li>Let\u2019s begin by writing the formula and substitute the given angles.\n<div class=\"unnumbered\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha -\\beta \\right)& =& \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{sin}\\left(45\u00b0-30\u00b0\\right)& =& \\mathrm{sin}\\left(45\u00b0\\right)\\mathrm{cos}\\left(30\u00b0\\right)-\\mathrm{cos}\\left(45\u00b0\\right)\\mathrm{sin}\\left(30\u00b0\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1565917\">Next, we need to find the values of the trigonometric expressions.<\/p>\n<div id=\"fs-id1213031\" class=\"unnumbered aligncenter\">[latex]\\mathrm{sin}\\left(45\u00b0\\right)=\\frac{\\sqrt{2}}{2}, \\mathrm{cos}\\left(30\u00b0\\right)=\\frac{\\sqrt{3}}{2}, \\mathrm{cos}\\left(45\u00b0\\right)=\\frac{\\sqrt{2}}{2}, \\mathrm{sin}\\left(30\u00b0\\right)=\\frac{1}{2}[\/latex]<\/div>\n<p id=\"fs-id1143054\">Now we can substitute these values into the equation and simplify.<\/p>\n<div id=\"fs-id2020743\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(45\u00b0-30\u00b0\\right)& =& \\frac{\\sqrt{2}}{2}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{\\sqrt{2}}{2}\\left(\\frac{1}{2}\\right)\\hfill \\\\ & =& \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Again, we write the formula and substitute the given angles.\n<div id=\"fs-id1135014\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha -\\beta \\right)& =& \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{sin}\\left(135\u00b0-120\u00b0\\right)& =& \\mathrm{sin}\\left(135\u00b0\\right)\\mathrm{cos}\\left(120\u00b0\\right)-\\mathrm{cos}\\left(135\u00b0\\right)\\mathrm{sin}\\left(120\u00b0\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1177711\">Next, we find the values of the trigonometric expressions.<\/p>\n<div id=\"fs-id2067561\" class=\"unnumbered aligncenter\">[latex]\\mathrm{sin}\\left(135\u00b0\\right)=\\frac{\\sqrt{2}}{2},\\mathrm{cos}\\left(120\u00b0\\right)=-\\frac{1}{2},\\mathrm{cos}\\left(135\u00b0\\right)=\\frac{\\sqrt{2}}{2},\\mathrm{sin}\\left(120\u00b0\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/div>\n<p id=\"fs-id2011119\">Now we can substitute these values into the equation and simplify.<\/p>\n<div id=\"fs-id1922441\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(135\u00b0-120\u00b0\\right)& =& \\frac{\\sqrt{2}}{2}\\left(-\\frac{1}{2}\\right)-\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ & =& \\frac{-\\sqrt{2}+\\sqrt{6}}{4}\\hfill \\\\ & =& \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\\\ \\phantom{\\rule{4em}{0ex}}\\mathrm{sin}\\left(135\u00b0-120\u00b0\\right)\\hfill & =& \\frac{\\sqrt{2}}{2}\\left(-\\frac{1}{2}\\right)-\\left(-\\frac{\\sqrt{2}}{2}\\right)\\left(\\frac{\\sqrt{3}}{2}\\right)\\hfill \\\\ & =& \\frac{-\\sqrt{2}+\\sqrt{6}}{4}\\hfill \\\\ & =& \\frac{\\sqrt{6}-\\sqrt{2}}{4}\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1494014\">\n<div id=\"fs-id1494016\">\n<h3>Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function<\/h3>\n<p id=\"fs-id1252839\">Find the exact value of[latex]\\,\\mathrm{sin}\\left({\\mathrm{cos}}^{-1}\\frac{1}{2}+{\\mathrm{sin}}^{-1}\\frac{3}{5}\\right).\\,[\/latex]Then check the answer with a graphing calculator.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1669695\">The pattern displayed in this problem is[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right).\\,[\/latex]Let[latex]\\,\\alpha ={\\mathrm{cos}}^{-1}\\frac{1}{2}\\,[\/latex]and[latex]\\,\\beta ={\\mathrm{sin}}^{-1}\\frac{3}{5}.\\,[\/latex]Then we can write<\/p>\n<div id=\"fs-id1868250\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\,\\alpha & =& \\frac{1}{2},0\\le \\alpha \\le \\pi \\hfill \\\\ \\hfill \\mathrm{sin}\\,\\beta & =& \\frac{3}{5},-\\frac{\\pi }{2}\\le \\beta \\le \\frac{\\pi }{2}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1562263\">We will use the Pythagorean identities to find[latex]\\,\\mathrm{sin}\\,\\alpha \\,[\/latex]and[latex]\\,\\mathrm{cos}\\,\\beta .[\/latex]<\/p>\n<div id=\"fs-id2044989\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\,\\alpha & =& \\sqrt{1-{\\mathrm{cos}}^{2}\\alpha }\\hfill \\\\ & =& \\sqrt{1-\\frac{1}{4}}\\hfill \\\\ & =& \\sqrt{\\frac{3}{4}}\\hfill \\\\ & =& \\frac{\\sqrt{3}}{2}\\hfill \\\\ \\hfill \\mathrm{cos}\\,\\beta & =& \\sqrt{1-{\\mathrm{sin}}^{2}\\beta }\\hfill \\\\ & =& \\sqrt{1-\\frac{9}{25}}\\hfill \\\\ & =& \\sqrt{\\frac{16}{25}}\\hfill \\\\ & =& \\frac{4}{5}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1155089\">Using the sum formula for sine,<\/p>\n<div id=\"fs-id1231462\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left({\\mathrm{cos}}^{-1}\\frac{1}{2}+{\\mathrm{sin}}^{-1}\\frac{3}{5}\\right)& =& \\mathrm{sin}\\left(\\alpha +\\beta \\right)\\hfill \\\\ & =& \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ & =& \\frac{\\sqrt{3}}{2}\\cdot \\frac{4}{5}+\\frac{1}{2}\\cdot \\frac{3}{5}\\hfill \\\\ & =& \\frac{4\\sqrt{3}+3}{10}\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2072911\" class=\"bc-section section\">\n<h3>Using the Sum and Difference Formulas for Tangent<\/h3>\n<p id=\"fs-id2780782\">Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.<\/p>\n<p id=\"fs-id2780786\">Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall,[latex]\\,\\mathrm{tan}\\,x=\\frac{\\mathrm{sin}\\,x}{\\mathrm{cos}\\,x},\\mathrm{cos}\\,x\\ne 0.[\/latex]<\/p>\n<p id=\"fs-id1712352\">Let\u2019s derive the sum formula for tangent.<\/p>\n<div id=\"fs-id1185845\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{cccc}\\hfill \\mathrm{tan}\\left(\\alpha +\\beta \\right)& =& \\frac{\\mathrm{sin}\\left(\\alpha +\\beta \\right)}{\\mathrm{cos}\\left(\\alpha +\\beta \\right)}\\hfill & \\\\ & =& \\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }\\hfill & \\\\ & =& \\frac{\\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }}{\\frac{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }}\\hfill & \\text{Divide the numerator and denominator by cos}\\,\\alpha \\,\\text{cos}\\,\\beta .\\hfill \\\\ & =& \\frac{\\frac{\\mathrm{sin}\\,\\alpha \\overline{)\\,\\mathrm{cos}\\,\\beta }}{\\mathrm{cos}\\,\\alpha \\overline{)\\,\\mathrm{cos}\\,\\beta }}+\\frac{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\mathrm{sin}\\,\\beta }{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\mathrm{cos}\\,\\beta }}{\\frac{\\overline{)\\mathrm{cos}\\,\\alpha }\\overline{)\\,\\mathrm{cos}\\,\\beta }}{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\overline{)\\mathrm{cos}\\,\\beta }}-\\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }}\\hfill & \\\\ & =& \\frac{\\frac{\\mathrm{sin}\\,\\alpha }{\\mathrm{cos}\\,\\alpha }+\\frac{\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\beta }}{1-\\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }}\\hfill & \\\\ & =& \\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill & \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1775901\">We can derive the difference formula for tangent in a similar way.<\/p>\n<div id=\"fs-id2916561\" class=\"textbox key-takeaways\">\n<h3>Sum and Difference Formulas for Tangent<\/h3>\n<p id=\"fs-id2320803\">The <span class=\"no-emphasis\">sum and difference formulas for tangent<\/span> are:<\/p>\n<div id=\"Eq_07_02_05\">[latex]\\mathrm{tan}\\left(\\alpha +\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }[\/latex]<\/div>\n<div id=\"Eq_07_02_06\">[latex]\\mathrm{tan}\\left(\\alpha -\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta }{1+\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }[\/latex]<\/div>\n<\/div>\n<div id=\"fs-id1567999\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1233471\"><strong>Given two angles, find the tangent of the sum of the angles.<br \/>\n<\/strong><\/p>\n<ol id=\"fs-id1407099\" type=\"1\">\n<li>Write the sum formula for tangent.<\/li>\n<li>Substitute the given angles into the formula.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id1619329\">\n<div id=\"fs-id2289172\">\n<h3>Finding the Exact Value of an Expression Involving Tangent<\/h3>\n<p id=\"fs-id2289178\">Find the exact value of[latex]\\,\\mathrm{tan}\\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1378918\">Let\u2019s first write the sum formula for tangent and then substitute the given angles into the formula.<\/p>\n<div id=\"fs-id1514603\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\alpha +\\beta \\right)& =& \\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\\\ \\hfill \\mathrm{tan}\\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)& =& \\frac{\\mathrm{tan}\\left(\\frac{\\pi }{6}\\right)+\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)}{1-\\left(\\mathrm{tan}\\left(\\frac{\\pi }{6}\\right)\\right)\\left(\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)\\right)}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1378455\">Next, we determine the individual function values within the formula:<\/p>\n<div id=\"fs-id1378458\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\left(\\frac{\\pi }{6}\\right)=\\frac{1}{\\sqrt{3}},\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)=1[\/latex]<\/div>\n<p id=\"fs-id2016988\">So we have<\/p>\n<div id=\"fs-id1576577\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\frac{\\pi }{6}+\\frac{\\pi }{4}\\right)& =& \\frac{\\frac{1}{\\sqrt{3}}+1}{1-\\left(\\frac{1}{\\sqrt{3}}\\right)\\left(1\\right)}\\hfill \\\\ & =& \\frac{\\frac{1+\\sqrt{3}}{\\sqrt{3}}}{\\frac{\\sqrt{3}-1}{\\sqrt{3}}}\\hfill \\\\ & =& \\frac{1+\\sqrt{3}}{\\sqrt{3}}\\left(\\frac{\\sqrt{3}}{\\sqrt{3}-1}\\right)\\hfill \\\\ & =& \\frac{\\sqrt{3}+1}{\\sqrt{3}-1}\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1541743\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1527860\">\n<p id=\"fs-id1892528\">Find the exact value of[latex]\\,\\mathrm{tan}\\left(\\frac{2\\pi }{3}+\\frac{\\pi }{4}\\right).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2040850\">[latex]\\frac{1-\\sqrt{3}}{1+\\sqrt{3}}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id2927791\">\n<div id=\"fs-id2927793\">\n<h3>Finding Multiple Sums and Differences of Angles<\/h3>\n<p id=\"fs-id1333289\">Given[latex]\\text{ }\\mathrm{sin}\\,\\alpha =\\frac{3}{5},0<\\alpha <\\frac{\\pi }{2},\\mathrm{cos}\\,\\beta =-\\frac{5}{13},\\pi <\\beta <\\frac{3\\pi }{2},[\/latex]find<\/p>\n<ol id=\"fs-id2201332\" type=\"a\">\n<li>[latex]\\mathrm{sin}\\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n<li>[latex]\\mathrm{cos}\\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n<li>[latex]\\mathrm{tan}\\left(\\alpha +\\beta \\right)[\/latex]<\/li>\n<li>[latex]\\mathrm{tan}\\left(\\alpha -\\beta \\right)[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1483249\">We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.<\/p>\n<ol id=\"fs-id1440867\" type=\"a\">\n<li>To find[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right),[\/latex]we begin with[latex]\\,\\mathrm{sin}\\,\\alpha =\\frac{3}{5}\\,[\/latex]and[latex]\\,0<\\alpha <\\frac{\\pi }{2}.\\,[\/latex]The side opposite[latex]\\,\\alpha \\,[\/latex]has length 3, the hypotenuse has length 5, and[latex]\\,\\alpha \\,[\/latex]is in the first quadrant. See <a class=\"autogenerated-content\" href=\"#Figure_07_02_003\">(Figure)<\/a>. Using the Pythagorean Theorem, we can find the length of side[latex]\\,a\\text{:}[\/latex]\n<div id=\"fs-id2191216\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+{3}^{2}& =& {5}^{2}\\hfill \\\\ \\hfill {a}^{2}& =& 16\\hfill \\\\ \\hfill a& =& 4\\hfill \\end{array}[\/latex]<\/div>\n<div class=\"small\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143920\/CNX_Precalc_Figure_07_02_003.jpg\" alt=\"Diagram of a triangle in the x,y plane. The vertices are at the origin, (4,0), and (4,3). The angle at the origin is alpha degrees, The angle formed by the x-axis and the side from (4,3) to (4,0) is a right angle. The side opposite the right angle has length 5.\" width=\"487\" height=\"252\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 4.<\/strong><\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id1342733\">Since[latex]\\,\\mathrm{cos}\\,\\beta =-\\frac{5}{13}\\,[\/latex]and[latex]\\,\\pi <\\beta <\\frac{3\\pi }{2},[\/latex]the side adjacent to[latex]\\,\\beta \\,[\/latex]is[latex]\\,-5,[\/latex]the hypotenuse is 13, and[latex]\\,\\beta \\,[\/latex]is in the third quadrant. See <a class=\"autogenerated-content\" href=\"#Figure_07_02_004\">(Figure)<\/a>. Again, using the Pythagorean Theorem, we have<\/p>\n<div id=\"fs-id1944725\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill {\\left(-5\\right)}^{2}+{a}^{2}& =& {13}^{2}\\hfill \\\\ \\hfill 25+{a}^{2}& =& 169\\hfill \\\\ \\hfill {a}^{2}& =& 144\\hfill \\\\ \\hfill a& =& \u00b112\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1928233\">Since[latex]\\,\\beta \\,[\/latex]is in the third quadrant,[latex]\\,a=\u201312.[\/latex]<\/p>\n<div class=\"small\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143922\/CNX_Precalc_Figure_07_02_004.jpg\" alt=\"Diagram of a triangle in the x,y plane. The vertices are at the origin, (-5,0), and (-5, -12). The angle at the origin is Beta degrees. The angle formed by the x axis and the side from (-5, -12) to (-5,0) is a right angle. The side opposite the right angle has length 13.\" width=\"487\" height=\"568\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 5.<\/strong><\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id1155212\">The next step is finding the cosine of[latex]\\,\\alpha \\,[\/latex]and the sine of[latex]\\,\\beta .\\,[\/latex]The cosine of[latex]\\,\\alpha \\,[\/latex]is the adjacent side over the hypotenuse. We can find it from the triangle in <a class=\"autogenerated-content\" href=\"#Figure_07_02_004\">(Figure)<\/a>:[latex]\\,\\mathrm{cos}\\,\\alpha =\\frac{4}{5}.\\,[\/latex]We can also find the sine of[latex]\\,\\beta \\,[\/latex]from the triangle in <a class=\"autogenerated-content\" href=\"#Figure_07_02_004\">(Figure)<\/a>, as opposite side over the hypotenuse:[latex]\\,\\mathrm{sin}\\,\\beta =-\\frac{12}{13}.\\,[\/latex]Now we are ready to evaluate[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right).[\/latex]<\/p>\n<div id=\"fs-id2256453\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha +\\beta \\right)& =& \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ & =& \\left(\\frac{3}{5}\\right)\\left(-\\frac{5}{13}\\right)+\\left(\\frac{4}{5}\\right)\\left(-\\frac{12}{13}\\right)\\hfill \\\\ & =& -\\frac{15}{65}-\\frac{48}{65}\\hfill \\\\ & =& -\\frac{63}{65}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>We can find[latex]\\,\\mathrm{cos}\\left(\\alpha +\\beta \\right)\\,[\/latex]in a similar manner. We substitute the values according to the formula.\n<div id=\"fs-id1487615\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\alpha +\\beta \\right)& =& \\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ & =& \\left(\\frac{4}{5}\\right)\\left(-\\frac{5}{13}\\right)-\\left(\\frac{3}{5}\\right)\\left(-\\frac{12}{13}\\right)\\hfill \\\\ & =& -\\frac{20}{65}+\\frac{36}{65}\\hfill \\\\ & =& \\frac{16}{65}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>For[latex]\\,\\mathrm{tan}\\left(\\alpha +\\beta \\right),[\/latex]if[latex]\\,\\mathrm{sin}\\,\\alpha =\\frac{3}{5}\\,[\/latex]and[latex]\\,\\mathrm{cos}\\,\\alpha =\\frac{4}{5},[\/latex] then\n<div id=\"fs-id1512831\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\,\\alpha =\\frac{\\frac{3}{5}}{\\frac{4}{5}}=\\frac{3}{4}[\/latex]<\/div>\n<p id=\"fs-id1837927\">If[latex]\\,\\mathrm{sin}\\,\\beta =-\\frac{12}{13}\\,[\/latex]and[latex]\\,\\mathrm{cos}\\,\\beta =-\\frac{5}{13},[\/latex]<br \/>\nthen<\/p>\n<div id=\"fs-id2917183\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\,\\beta =\\frac{\\frac{-12}{13}}{\\frac{-5}{13}}=\\frac{12}{5}[\/latex]<\/div>\n<p id=\"fs-id2900205\">Then,<\/p>\n<div id=\"fs-id2900208\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\alpha +\\beta \\right)& =& \\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\\\ & =& \\frac{\\frac{3}{4}+\\frac{12}{5}}{1-\\frac{3}{4}\\left(\\frac{12}{5}\\right)}\\hfill \\\\ & =& \\frac{\\text{ }\\frac{63}{20}}{-\\frac{16}{20}}\\hfill \\\\ & =& -\\frac{63}{16}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<li>To find[latex]\\,\\mathrm{tan}\\left(\\alpha -\\beta \\right),[\/latex] we have the values we need. We can substitute them in and evaluate.\n<div id=\"fs-id2575375\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\alpha -\\beta \\right)& =& \\frac{\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta }{1+\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\\\ & =& \\frac{\\frac{3}{4}-\\frac{12}{5}}{1+\\frac{3}{4}\\left(\\frac{12}{5}\\right)}\\hfill \\\\ & =& \\frac{-\\frac{33}{20}}{\\frac{56}{20}}\\hfill \\\\ & =& -\\frac{33}{56}\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id2714729\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id2714734\">A common mistake when addressing problems such as this one is that we may be tempted to think that[latex]\\,\\alpha \\,[\/latex]and[latex]\\,\\beta \\,[\/latex]are angles in the same triangle, which of course, they are not. Also note that<\/p>\n<div id=\"fs-id2202966\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\left(\\alpha +\\beta \\right)=\\frac{\\mathrm{sin}\\left(\\alpha +\\beta \\right)}{\\mathrm{cos}\\left(\\alpha +\\beta \\right)}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2050772\" class=\"bc-section section\">\n<h3>Using Sum and Difference Formulas for Cofunctions<\/h3>\n<p id=\"fs-id1527759\">Now that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall from <a class=\"target-chapter\" href=\"\/contents\/7375cfa6-269c-4e0d-a567-bc6a4c66b1f4\">Right Triangle Trigonometry<\/a> that, if the sum of two positive angles is[latex]\\,\\frac{\\pi }{2},[\/latex]those two angles are complements, and the sum of the two acute angles in a right triangle is[latex]\\,\\frac{\\pi }{2},[\/latex]so they are also complements. In <a class=\"autogenerated-content\" href=\"#Figure_07_02_007\">(Figure)<\/a>, notice that if one of the acute angles is labeled as[latex]\\,\\theta ,[\/latex]then the other acute angle must be labeled[latex]\\,\\left(\\frac{\\pi }{2}-\\theta \\right).[\/latex]<\/p>\n<p id=\"fs-id2045066\">Notice also that[latex]\\,\\mathrm{sin}\\,\\theta =\\mathrm{cos}\\left(\\frac{\\pi }{2}-\\theta \\right),[\/latex]which is opposite over hypotenuse. Thus, when two angles are complementary, we can say that the sine of[latex]\\,\\theta \\,[\/latex]equals the <span class=\"no-emphasis\">cofunction<\/span> of the complement of[latex]\\,\\theta .\\,[\/latex]Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions.<\/p>\n<div class=\"small\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143927\/CNX_Precalc_Figure_07_02_007.jpg\" alt=\"Image of a right triangle. The remaining angles are labeled theta and pi\/2 - theta.\" width=\"487\" height=\"268\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 6.<\/strong><\/figcaption><\/figure>\n<\/div>\n<p id=\"fs-id2338660\">From these relationships, the <span class=\"no-emphasis\">cofunction identities<\/span> are formed. Recall that you first encountered these identities in <a class=\"target-chapter\" href=\"https:\/\/courses.lumenlearning.com\/suny-osalgebratrig\/chapter\/introduction-to-the-unit-circle-sine-and-cosine-functions\/\">The Unit Circle: Sine and Cosine Functions<\/a>.<\/p>\n<div id=\"fs-id2247036\" class=\"textbox key-takeaways\">\n<h3>Cofunction Identities<\/h3>\n<p id=\"fs-id2247045\">The cofunction identities are summarized in <a class=\"autogenerated-content\" href=\"#Table_07_02_02\">(Figure)<\/a>.<\/p>\n<table id=\"Table_07_02_02\" summary=\"Three rows, two columns\/ The table has ordered pairs of these row values: (sin(theta) = cos(pi\/2 - theta), cos(theta) = sin(pi\/2 - theta)), (tan(theta) = cot(pi\/2 - theta), cot(theta) = tan(pi\/2 - theta)), and (sec(theta) = csc(pi\/2 - theta), csc(theta) = sec(pi\/2 - theta)).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td>[latex]\\mathrm{sin}\\,\\theta =\\mathrm{cos}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\mathrm{cos}\\,\\theta =\\mathrm{sin}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\mathrm{tan}\\,\\theta =\\mathrm{cot}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\mathrm{cot}\\,\\theta =\\mathrm{tan}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\mathrm{sec}\\,\\theta =\\mathrm{csc}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<td>[latex]\\mathrm{csc}\\,\\theta =\\mathrm{sec}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p id=\"fs-id2814240\">Notice that the formulas in the table may also justified algebraically using the sum and difference formulas. For example, using<\/p>\n<div id=\"fs-id2814245\" class=\"unnumbered aligncenter\">[latex]\\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\mathrm{sin}\\,\\beta ,[\/latex]<\/div>\n<p id=\"fs-id2021193\">we can write<\/p>\n<div id=\"fs-id2427498\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{cos}\\left(\\frac{\\pi }{2}-\\theta \\right)& =& \\mathrm{cos}\\,\\frac{\\pi }{2}\\,\\mathrm{cos}\\,\\theta +\\mathrm{sin}\\,\\frac{\\pi }{2}\\,\\mathrm{sin}\\,\\theta \\hfill \\\\ & =& \\left(0\\right)\\mathrm{cos}\\,\\theta +\\left(1\\right)\\mathrm{sin}\\,\\theta \\hfill \\\\ & =& \\mathrm{sin}\\,\\theta \\hfill \\end{array}[\/latex]<\/div>\n<div id=\"Example_07_02_07\" class=\"textbox examples\">\n<div id=\"fs-id2223313\">\n<div id=\"fs-id2223315\">\n<h3>Finding a Cofunction with the Same Value as the Given Expression<\/h3>\n<p id=\"fs-id2231859\">Write[latex]\\,\\mathrm{tan}\\,\\frac{\\pi }{9}\\,[\/latex]in terms of its cofunction.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1556117\">The cofunction of[latex]\\,\\mathrm{tan}\\,\\theta =\\mathrm{cot}\\left(\\frac{\\pi }{2}-\\theta \\right).\\,[\/latex]Thus,<\/p>\n<div id=\"fs-id2274481\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\frac{\\pi }{9}\\right)& =& \\mathrm{cot}\\left(\\frac{\\pi }{2}-\\frac{\\pi }{9}\\right)\\hfill \\\\ & =& \\mathrm{cot}\\left(\\frac{9\\pi }{18}-\\frac{2\\pi }{18}\\right)\\hfill \\\\ & =& \\mathrm{cot}\\left(\\frac{7\\pi }{18}\\right)\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2109946\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1923901\">\n<p id=\"fs-id1923902\">Write[latex]\\,\\mathrm{sin}\\,\\frac{\\pi }{7}\\,[\/latex]in terms of its cofunction.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1386356\">[latex]\\mathrm{cos}\\left(\\frac{5\\pi }{14}\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1926444\" class=\"bc-section section\">\n<h3>Using the Sum and Difference Formulas to Verify Identities<\/h3>\n<p id=\"fs-id2637951\">Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems. Reviewing the general rules presented earlier may help simplify the process of verifying an identity.<\/p>\n<div id=\"fs-id2637962\" class=\"precalculus howto textbox tryit\">\n<h3>How To<\/h3>\n<p id=\"fs-id1863876\"><strong>Given an identity, verify using sum and difference formulas.<\/strong><\/p>\n<ol id=\"fs-id1863880\" type=\"1\">\n<li>Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.<\/li>\n<li>Look for opportunities to use the sum and difference formulas.<\/li>\n<li>Rewrite sums or differences of quotients as single quotients.<\/li>\n<li>If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_07_02_08\" class=\"textbox examples\">\n<div id=\"fs-id1183047\">\n<div id=\"fs-id1183049\">\n<h3>Verifying an Identity Involving Sine<\/h3>\n<p id=\"fs-id1899828\">Verify the identity[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right)+\\mathrm{sin}\\left(\\alpha -\\beta \\right)=2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta .[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1469824\">We see that the left side of the equation includes the sines of the sum and the difference of angles.<\/p>\n<div id=\"fs-id1469827\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha +\\beta \\right)& =& \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\hfill \\mathrm{sin}\\left(\\alpha -\\beta \\right)& =& \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1154806\">We can rewrite each using the sum and difference formulas.<\/p>\n<div id=\"fs-id1154809\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\left(\\alpha +\\beta \\right)+\\mathrm{sin}\\left(\\alpha -\\beta \\right)& =& \\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ & =& 2\\,\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1947264\">We see that the identity is verified.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_02_09\" class=\"textbox examples\">\n<div id=\"fs-id1858950\">\n<div id=\"fs-id1858952\">\n<h3>Verifying an Identity Involving Tangent<\/h3>\n<p id=\"fs-id1858957\">Verify the following identity.<\/p>\n<div id=\"fs-id1858960\" class=\"unnumbered aligncenter\">[latex]\\frac{\\mathrm{sin}\\left(\\alpha -\\beta \\right)}{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }=\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta[\/latex]<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1636592\">We can begin by rewriting the numerator on the left side of the equation.<\/p>\n<div id=\"fs-id1636595\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{cccc}\\hfill \\frac{\\mathrm{sin}\\left(\\alpha -\\beta \\right)}{\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta }& =& \\frac{\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\alpha \\mathrm{cos}\\,\\beta }\\hfill & \\\\ & =& \\frac{\\mathrm{sin}\\,\\alpha \\overline{)\\,\\mathrm{cos}\\,\\beta }}{\\mathrm{cos}\\,\\alpha \\overline{)\\,\\mathrm{cos}\\,\\beta }}-\\frac{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\mathrm{sin}\\,\\beta }{\\overline{)\\mathrm{cos}\\,\\alpha }\\,\\mathrm{cos}\\,\\beta }\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Rewrite using a common denominator}.\\hfill \\\\ & =& \\frac{\\mathrm{sin}\\,\\alpha }{\\mathrm{cos}\\,\\alpha }-\\frac{\\mathrm{sin}\\,\\beta }{\\mathrm{cos}\\,\\beta }\\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Cancel}.\\hfill \\\\ & =& \\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta \\hfill & \\phantom{\\rule{2em}{0ex}}\\text{Rewrite in terms of tangent}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1260145\">We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2822685\" class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<div>\n<div id=\"fs-id1248124\">\n<p id=\"fs-id1248125\">Verify the identity:[latex]\\,\\mathrm{tan}\\left(\\pi -\\theta \\right)=-\\mathrm{tan}\\,\\theta .[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<div id=\"fs-id1152805\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(\\pi -\\theta \\right)& =& \\frac{\\mathrm{tan}\\left(\\pi \\right)-\\mathrm{tan}\\,\\theta }{1+\\mathrm{tan}\\left(\\pi \\right)\\mathrm{tan}\\theta }\\hfill \\\\ & =& \\frac{0-\\mathrm{tan}\\,\\theta }{1+0\\cdot \\mathrm{tan}\\,\\theta }\\hfill \\\\ & =& -\\mathrm{tan}\\,\\theta \\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_02_10\" class=\"textbox examples\">\n<div id=\"fs-id2092391\">\n<div id=\"fs-id2092393\">\n<h3>Using Sum and Difference Formulas to Solve an Application Problem<\/h3>\n<p id=\"fs-id2092399\">Let[latex]\\,{L}_{1}\\,[\/latex]and[latex]\\,{L}_{2}\\,[\/latex]denote two non-vertical intersecting lines, and let[latex]\\,\\theta \\,[\/latex]denote the acute angle between[latex]\\,{L}_{1}\\,[\/latex]and[latex]\\,{L}_{2}.\\,[\/latex]See <a class=\"autogenerated-content\" href=\"#Figure_07_02_005\">(Figure)<\/a>. Show that<\/p>\n<div id=\"fs-id1728421\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\,\\theta =\\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}[\/latex]<\/div>\n<p id=\"fs-id1924955\">where[latex]\\,{m}_{1}\\,[\/latex]and[latex]\\,{m}_{2}\\,[\/latex]are the slopes of[latex]\\,{L}_{1}\\,[\/latex]and[latex]\\,{L}_{2}\\,[\/latex]respectively. (<strong>Hint:<\/strong> Use the fact that[latex]\\,\\mathrm{tan}\\,{\\theta }_{1}={m}_{1}\\,[\/latex]and[latex]\\,\\mathrm{tan}\\,{\\theta }_{2}={m}_{2}.[\/latex])<\/p>\n<div class=\"small\">\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143941\/CNX_Precalc_Figure_07_02_005.jpg\" alt=\"Diagram of two non-vertical intersecting lines L1 and L2 also intersecting the x-axis. The acute angle formed by the intersection of L1 and L2 is theta. The acute angle formed by L2 and the x-axis is theta 1, and the acute angle formed by the x-axis and L1 is theta 2.\" width=\"487\" height=\"289\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 7.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2260061\">Using the difference formula for tangent, this problem does not seem as daunting as it might.<\/p>\n<div id=\"fs-id2260064\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\,\\theta & =& \\mathrm{tan}\\left({\\theta }_{2}-{\\theta }_{1}\\right)\\hfill \\\\ & =& \\frac{\\mathrm{tan}\\,{\\theta }_{2}-\\mathrm{tan}\\,{\\theta }_{1}}{1+\\mathrm{tan}\\,{\\theta }_{1}\\mathrm{tan}\\,{\\theta }_{2}}\\hfill \\\\ & =& \\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}\\hfill \\end{array}[\/latex]<\/details>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_02_11\" class=\"textbox examples\">\n<div id=\"fs-id1367169\">\n<div id=\"fs-id1784865\">\n<h3>Investigating a Guy-wire Problem<\/h3>\n<p id=\"fs-id1784870\">For a climbing wall, a guy-wire[latex]\\,R\\,[\/latex]is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire[latex]\\,S\\,[\/latex]attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle[latex]\\,\\alpha \\,[\/latex]between the wires. See <a class=\"autogenerated-content\" href=\"#Figure_07_02_006\">(Figure)<\/a>.<\/p>\n<div class=\"small\">\n<figure style=\"width: 590px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143949\/CNX_Precalc_Figure_07_02_006.jpg\" alt=\"Two right triangles. Both share the same base, 50 feet. The first has a height of 40 ft and hypotenuse S. The second has height 47 ft and hypotenuse R. The height sides of the triangles are overlapping. There is a B degree angle between R and the base, and an a degree angle between the two hypotenuses within the B degree angle.\" width=\"590\" height=\"322\" \/><figcaption class=\"wp-caption-text\"><strong>Figure 8.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2136278\">Let\u2019s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that[latex]\\,\\mathrm{tan}\\,\\beta =\\frac{47}{50},[\/latex]and[latex]\\,\\mathrm{tan}\\left(\\beta -\\alpha \\right)=\\frac{40}{50}=\\frac{4}{5}.\\,[\/latex]We can then use difference formula for tangent.<\/p>\n<div id=\"fs-id2177692\" class=\"unnumbered aligncenter\">[latex]\\mathrm{tan}\\left(\\beta -\\alpha \\right)=\\frac{\\mathrm{tan}\\,\\beta -\\mathrm{tan}\\,\\alpha }{1+\\mathrm{tan}\\,\\beta \\mathrm{tan}\\,\\alpha }[\/latex]<\/div>\n<p id=\"fs-id2007573\">Now, substituting the values we know into the formula, we have<\/p>\n<div id=\"fs-id2007576\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill \\frac{4}{5}& =& \\frac{\\frac{47}{50}-\\mathrm{tan}\\,\\alpha }{1+\\frac{47}{50}\\mathrm{tan}\\,\\alpha }\\hfill \\\\ \\hfill 4\\left(1+\\frac{47}{50}\\mathrm{tan}\\,\\alpha \\right)& =& 5\\left(\\frac{47}{50}-\\mathrm{tan}\\,\\alpha \\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1933373\">Use the distributive property, and then simplify the functions.<\/p>\n<div id=\"fs-id1933377\" class=\"unnumbered aligncenter\">[latex]\\begin{array}{ccc}\\hfill 4\\left(1\\right)+4\\left(\\frac{47}{50}\\right)\\mathrm{tan}\\,\\alpha & =& 5\\left(\\frac{47}{50}\\right)-5\\,\\mathrm{tan}\\,\\alpha \\hfill \\\\ \\hfill 4+3.76\\,\\mathrm{tan}\\,\\alpha & =& 4.7-5\\,\\mathrm{tan}\\,\\alpha \\hfill \\\\ \\hfill 5\\,\\mathrm{tan}\\,\\alpha +3.76\\,\\mathrm{tan}\\,\\alpha & =& 0.7\\hfill \\\\ \\hfill 8.76\\,\\mathrm{tan}\\,\\alpha & =& 0.7\\hfill \\\\ \\hfill \\mathrm{tan}\\,\\alpha & \\approx & 0.07991\\hfill \\\\ \\hfill {\\mathrm{tan}}^{-1}\\left(0.07991\\right)& \\approx & .079741\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1691817\">Now we can calculate the angle in degrees.<\/p>\n<div id=\"fs-id1691820\" class=\"unnumbered aligncenter\">[latex]\\alpha \\approx 0.079741\\left(\\frac{180}{\\pi }\\right)\\approx 4.57\u00b0[\/latex]<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2230083\">\n<h4>Analysis<\/h4>\n<p id=\"fs-id2230090\">Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1243690\" class=\"precalculus media\">\n<p id=\"eip-id3766817\">Access these online resources for additional instruction and practice with sum and difference identities.<\/p>\n<ul id=\"fs-id1243697\">\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/sumdifcos\">Sum and Difference Identities for Cosine<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/sumdifsin\">Sum and Difference Identities for Sine<\/a><\/li>\n<li><a href=\"http:\/\/openstaxcollege.org\/l\/sumdiftan\">Sum and Difference Identities for Tangent<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id2332636\" class=\"key-equations\">\n<h3>Key Equations<\/h3>\n<table id=\"fs-id2332643\" summary=\"..\">\n<tbody>\n<tr>\n<td>Sum Formula for Cosine<\/td>\n<td>[latex]\\mathrm{cos}\\left(\\alpha +\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\mathrm{sin}\\,\\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Difference Formula for Cosine<\/td>\n<td>[latex]\\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Sum Formula for Sine<\/td>\n<td>[latex]\\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Difference Formula for Sine<\/td>\n<td>[latex]\\mathrm{sin}\\left(\\alpha -\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Sum Formula for Tangent<\/td>\n<td>[latex]\\mathrm{tan}\\left(\\alpha +\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Difference Formula for Tangent<\/td>\n<td>[latex]\\mathrm{tan}\\left(\\alpha -\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta }{1+\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Cofunction identities<\/td>\n<td>[latex]\\begin{array}{ccc}\\hfill \\mathrm{sin}\\,\\theta & =& \\mathrm{cos}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{cos}\\,\\theta & =& \\mathrm{sin}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{tan}\\,\\theta & =& \\mathrm{cot}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{cot}\\,\\theta & =& \\mathrm{tan}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{sec}\\,\\theta & =& \\mathrm{csc}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\\\ \\hfill \\mathrm{csc}\\,\\theta & =& \\mathrm{sec}\\left(\\frac{\\pi }{2}-\\theta \\right)\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"fs-id2224028\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id2224035\">\n<li>The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles.<\/li>\n<li>The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See <a class=\"autogenerated-content\" href=\"#Example_07_02_01\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_07_02_02\">(Figure)<\/a>.<\/li>\n<li>The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See <a class=\"autogenerated-content\" href=\"#Example_07_02_03\">(Figure)<\/a>.<\/li>\n<li>The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See <a class=\"autogenerated-content\" href=\"#Example_07_02_04\">(Figure)<\/a>.<\/li>\n<li>The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles. See <a class=\"autogenerated-content\" href=\"#Example_07_02_05\">(Figure)<\/a>.<\/li>\n<li>The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See <a class=\"autogenerated-content\" href=\"#Example_07_02_06\">(Figure)<\/a>.<\/li>\n<li>The cofunction identities apply to complementary angles and pairs of reciprocal functions. See <a class=\"autogenerated-content\" href=\"#Example_07_02_07\">(Figure)<\/a>.<\/li>\n<li>Sum and difference formulas are useful in verifying identities. See <a class=\"autogenerated-content\" href=\"#Example_07_02_08\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_07_02_09\">(Figure)<\/a>.<\/li>\n<li>Application problems are often easier to solve by using sum and difference formulas. See <a class=\"autogenerated-content\" href=\"#Example_07_02_10\">(Figure)<\/a> and <a class=\"autogenerated-content\" href=\"#Example_07_02_10\">(Figure)<\/a>.<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id2633717\" class=\"textbox exercises\">\n<h3>Section Exercises<\/h3>\n<div id=\"fs-id2633720\" class=\"bc-section section\">\n<h4>Verbal<\/h4>\n<div id=\"fs-id2633726\">\n<div id=\"fs-id2633727\">\n<p id=\"fs-id2633728\">Explain the basis for the cofunction identities and when they apply.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1879715\">The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures[latex]\\,x,[\/latex]the second angle measures[latex]\\,\\frac{\\pi }{2}-x.\\,[\/latex]Then[latex]\\,\\mathrm{sin}x=\\mathrm{cos}\\left(\\frac{\\pi }{2}-x\\right).\\,[\/latex]The same holds for the other cofunction identities. The key is that the angles are complementary.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1784023\">\n<div id=\"fs-id1784024\">\n<p id=\"fs-id1784025\">Is there only one way to evaluate[latex]\\,\\mathrm{cos}\\left(\\frac{5\\pi }{4}\\right)?\\,[\/latex]Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1943345\">\n<div id=\"fs-id1943346\">\n<p id=\"fs-id1943347\">Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for[latex]\\,f\\left(x\\right)=\\mathrm{sin}\\left(x\\right)\\,[\/latex]and[latex]\\,g\\left(x\\right)=\\mathrm{cos}\\left(x\\right).\\,[\/latex](Hint:[latex]\\,0-x=-x[\/latex])<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2169010\">[latex]\\mathrm{sin}\\left(-x\\right)=-\\mathrm{sin}x,[\/latex]so[latex]\\,\\mathrm{sin}x\\,[\/latex] is odd.[latex]\\,\\mathrm{cos}\\left(-x\\right)=\\mathrm{cos}\\left(0-x\\right)=\\mathrm{cos}x,[\/latex]so[latex]\\,\\mathrm{cos}x\\,[\/latex]is even.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2030146\" class=\"bc-section section\">\n<h4>Algebraic<\/h4>\n<p id=\"fs-id2030152\">For the following exercises, find the exact value.<\/p>\n<div id=\"fs-id1535795\">\n<div id=\"fs-id1535796\">\n<p id=\"fs-id1535797\">[latex]\\mathrm{cos}\\left(\\frac{7\\pi }{12}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2686776\">\n<div id=\"fs-id2686777\">\n<p id=\"fs-id2686778\">[latex]\\mathrm{cos}\\left(\\frac{\\pi }{12}\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1546201\">[latex]\\frac{\\sqrt{2}+\\sqrt{6}}{4}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1343104\">\n<div id=\"fs-id1343105\">\n<p id=\"fs-id1343106\">[latex]\\mathrm{sin}\\left(\\frac{5\\pi }{12}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1662806\">\n<div id=\"fs-id1662807\">\n<p id=\"fs-id1662808\">[latex]\\mathrm{sin}\\left(\\frac{11\\pi }{12}\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1617567\">[latex]\\frac{\\sqrt{6}-\\sqrt{2}}{4}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2327970\">\n<div id=\"fs-id2327971\">\n<p id=\"fs-id2327972\">[latex]\\mathrm{tan}\\left(-\\frac{\\pi }{12}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2641720\">\n<div id=\"fs-id2641721\">\n<p id=\"fs-id2641722\">[latex]\\mathrm{tan}\\left(\\frac{19\\pi }{12}\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2340651\">[latex]-2-\\sqrt{3}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<p id=\"fs-id1872453\">For the following exercises, rewrite in terms of[latex]\\,\\mathrm{sin}\\,x\\,[\/latex]and[latex]\\,\\mathrm{cos}\\,x.[\/latex]<\/p>\n<div id=\"fs-id1944688\">\n<div id=\"fs-id1944689\">\n<p id=\"fs-id1944690\">[latex]\\mathrm{sin}\\left(x+\\frac{11\\pi }{6}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1924818\">\n<div id=\"fs-id1924819\">\n<p id=\"fs-id1924820\">[latex]\\mathrm{sin}\\left(x-\\frac{3\\pi }{4}\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2907495\">[latex]-\\frac{\\sqrt{2}}{2}\\mathrm{sin}x-\\frac{\\sqrt{2}}{2}\\mathrm{cos}x[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2289260\">\n<div id=\"fs-id2289261\">\n<p id=\"fs-id2289262\">[latex]\\mathrm{cos}\\left(x-\\frac{5\\pi }{6}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1923485\">\n<div id=\"fs-id1923486\">\n<p id=\"fs-id1923487\">[latex]\\mathrm{cos}\\left(x+\\frac{2\\pi }{3}\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1784096\">[latex]-\\frac{1}{2}\\mathrm{cos}x-\\frac{\\sqrt{3}}{2}\\mathrm{sin}x[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<p id=\"fs-id1186242\">For the following exercises, simplify the given expression.<\/p>\n<div id=\"fs-id1587348\">\n<div id=\"fs-id1587349\">\n<p id=\"fs-id1587350\">[latex]\\mathrm{csc}\\left(\\frac{\\pi }{2}-t\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1931167\">\n<div id=\"fs-id1931168\">\n<p id=\"fs-id1931169\">[latex]\\mathrm{sec}\\left(\\frac{\\pi }{2}-\\theta \\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1973314\">[latex]\\mathrm{csc}\\theta[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1405117\">\n<div id=\"fs-id1405118\">\n<p id=\"fs-id1405119\">[latex]\\mathrm{cot}\\left(\\frac{\\pi }{2}-x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1223240\">\n<div id=\"fs-id1223241\">\n<p id=\"fs-id2168811\">[latex]\\mathrm{tan}\\left(\\frac{\\pi }{2}-x\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2780587\">[latex]\\mathrm{cot}x[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2306521\">\n<div id=\"fs-id2306522\">\n<p id=\"fs-id2306523\">[latex]\\mathrm{sin}\\left(2x\\right)\\,\\mathrm{cos}\\left(5x\\right)-\\mathrm{sin}\\left(5x\\right)\\,\\mathrm{cos}\\left(2x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1359485\">\n<div id=\"fs-id2164270\">\n<p id=\"fs-id2164271\">[latex]\\frac{\\mathrm{tan}\\left(\\frac{3}{2}x\\right)-\\mathrm{tan}\\left(\\frac{7}{5}x\\right)}{1+\\mathrm{tan}\\left(\\frac{3}{2}x\\right)\\mathrm{tan}\\left(\\frac{7}{5}x\\right)}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2148686\">[latex]\\mathrm{tan}\\left(\\frac{x}{10}\\right)[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<p id=\"fs-id1912452\">For the following exercises, find the requested information.<\/p>\n<div id=\"fs-id1912455\">\n<div id=\"fs-id1912456\">\n<p id=\"fs-id1912457\">Given that[latex]\\,\\mathrm{sin}\\,a=\\frac{2}{3}\\,[\/latex]and[latex]\\,\\mathrm{cos}\\,b=-\\frac{1}{4},[\/latex]with[latex]\\,a\\,[\/latex]and[latex]\\,b\\,[\/latex]both in the interval[latex]\\,\\left[\\frac{\\pi }{2},\\pi \\right),[\/latex]find[latex]\\,\\mathrm{sin}\\left(a+b\\right)\\,[\/latex]and[latex]\\,\\mathrm{cos}\\left(a-b\\right).[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1256165\">\n<div id=\"fs-id1256166\">\n<p id=\"fs-id1256167\">Given that[latex]\\,\\mathrm{sin}\\,a=\\frac{4}{5},[\/latex]and[latex]\\,\\mathrm{cos}\\,b=\\frac{1}{3},[\/latex]with[latex]\\,a\\,[\/latex]and[latex]\\,b\\,[\/latex]both in the interval[latex]\\,\\left[0,\\frac{\\pi }{2}\\right),[\/latex]find[latex]\\,\\mathrm{sin}\\left(a-b\\right)\\,[\/latex]and[latex]\\,\\mathrm{cos}\\left(a+b\\right).[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2011929\">[latex]\\begin{array}{ccccc}\\hfill \\mathrm{sin}\\left(a-b\\right)& =& \\left(\\frac{4}{5}\\right)\\left(\\frac{1}{3}\\right)-\\left(\\frac{3}{5}\\right)\\left(\\frac{2\\sqrt{2}}{3}\\right)\\hfill & =& \\frac{4-6\\sqrt{2}}{15}\\hfill \\\\ \\hfill \\mathrm{cos}\\left(a+b\\right)& =& \\left(\\frac{3}{5}\\right)\\left(\\frac{1}{3}\\right)-\\left(\\frac{4}{5}\\right)\\left(\\frac{2\\sqrt{2}}{3}\\right)\\hfill & =& \\frac{3-8\\sqrt{2}}{15}\\hfill \\end{array}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<p id=\"fs-id2478807\">For the following exercises, find the exact value of each expression.<\/p>\n<div id=\"fs-id2478810\">\n<div id=\"fs-id2478811\">\n<p id=\"fs-id2478812\">[latex]\\mathrm{sin}\\left({\\mathrm{cos}}^{-1}\\left(0\\right)-{\\mathrm{cos}}^{-1}\\left(\\frac{1}{2}\\right)\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2126676\">\n<div id=\"fs-id2126677\">\n<p id=\"fs-id2126678\">[latex]\\mathrm{cos}\\left({\\mathrm{cos}}^{-1}\\left(\\frac{\\sqrt{2}}{2}\\right)+{\\mathrm{sin}}^{-1}\\left(\\frac{\\sqrt{3}}{2}\\right)\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2229931\">[latex]\\frac{\\sqrt{2}-\\sqrt{6}}{4}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2337567\">\n<div id=\"fs-id2336035\">\n<p id=\"fs-id2336036\">[latex]\\mathrm{tan}\\left({\\mathrm{sin}}^{-1}\\left(\\frac{1}{2}\\right)-{\\mathrm{cos}}^{-1}\\left(\\frac{1}{2}\\right)\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1408229\" class=\"bc-section section\">\n<h4>Graphical<\/h4>\n<p id=\"fs-id2259918\">For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. Confirm your answer using a graphing calculator.<\/p>\n<div id=\"fs-id2259922\">\n<div id=\"fs-id2259924\">\n<p id=\"fs-id2259925\">[latex]\\mathrm{cos}\\left(\\frac{\\pi }{2}-x\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1411055\">[latex]\\mathrm{sin}x[\/latex]<\/p>\n<p><span id=\"fs-id1615462\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19143955\/CNX_Precalc_Figure_07_02_201.jpg\" alt=\"Graph of y=sin(x) from -2pi to 2pi.\" \/><\/span><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1555893\">\n<div id=\"fs-id1555894\">\n<p id=\"fs-id1555895\">[latex]\\mathrm{sin}\\left(\\pi -x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2960991\">\n<div id=\"fs-id2960992\">\n<p id=\"fs-id1430117\">[latex]\\mathrm{tan}\\left(\\frac{\\pi }{3}+x\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2113190\">[latex]\\mathrm{cot}\\left(\\frac{\\pi }{6}-x\\right)[\/latex]<\/p>\n<p><span id=\"fs-id2233832\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19144001\/CNX_Precalc_Figure_07_02_203.jpg\" alt=\"Graph of y=cot(pi\/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi\/6.\" \/><\/span><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1502696\">\n<div id=\"fs-id1502697\">\n<p id=\"fs-id1502698\">[latex]\\mathrm{sin}\\left(\\frac{\\pi }{3}+x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1724458\">\n<div id=\"fs-id1724459\">\n<p id=\"fs-id1724460\">[latex]\\mathrm{tan}\\left(\\frac{\\pi }{4}-x\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2307484\">[latex]\\mathrm{cot}\\left(\\frac{\\pi }{4}+x\\right)[\/latex]<\/p>\n<p><span id=\"fs-id1877975\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19144008\/CNX_Precalc_Figure_07_02_205.jpg\" alt=\"Graph of y=cot(pi\/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi\/4.\" \/><\/span><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1252765\">\n<div id=\"fs-id1252766\">\n<p id=\"fs-id1252768\">[latex]\\mathrm{cos}\\left(\\frac{7\\pi }{6}+x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1681120\">\n<div id=\"fs-id2255437\">\n<p id=\"fs-id2255438\">[latex]\\mathrm{sin}\\left(\\frac{\\pi }{4}+x\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1752329\">[latex]\\frac{\\mathrm{sin}x}{\\sqrt{2}}+\\frac{\\mathrm{cos}x}{\\sqrt{2}}[\/latex]<\/p>\n<p><span id=\"fs-id2199923\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3252\/2018\/07\/19144020\/CNX_Precalc_Figure_07_02_207.jpg\" alt=\"Graph of y = sin(x) \/ rad2 + cos(x) \/ rad2 - it looks like the sin curve shifted by pi\/4.\" \/><\/span><\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2199934\">\n<div id=\"fs-id2199935\">\n<p id=\"fs-id1672664\">[latex]\\mathrm{cos}\\left(\\frac{5\\pi }{4}+x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id2710551\">For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think[latex]\\,2x=x+x.[\/latex]<br \/>\n)<\/p>\n<div id=\"fs-id1493462\">\n<div id=\"fs-id2308858\">\n<p id=\"fs-id2308859\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(4x\\right)-\\mathrm{sin}\\left(3x\\right)\\mathrm{cos}\\,x,g\\left(x\\right)=\\mathrm{sin}\\,x\\,\\mathrm{cos}\\left(3x\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2253381\">They are the same.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1086612\">\n<div id=\"fs-id1086613\">\n<p id=\"fs-id1086614\">[latex]f\\left(x\\right)=\\mathrm{cos}\\left(4x\\right)+\\mathrm{sin}\\,x\\,\\mathrm{sin}\\left(3x\\right),g\\left(x\\right)=-\\mathrm{cos}\\,x\\,\\mathrm{cos}\\left(3x\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2102136\">\n<div id=\"fs-id2102137\">\n<p id=\"fs-id2102138\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(3x\\right)\\mathrm{cos}\\left(6x\\right),g\\left(x\\right)=-\\mathrm{sin}\\left(3x\\right)\\mathrm{cos}\\left(6x\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2190228\">They are the different, try[latex]\\,g\\left(x\\right)=\\mathrm{sin}\\left(9x\\right)-\\mathrm{cos}\\left(3x\\right)\\mathrm{sin}\\left(6x\\right).[\/latex]<\/details>\n<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1367081\">\n<div id=\"fs-id1367082\">\n<p id=\"fs-id1367083\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(4x\\right),g\\left(x\\right)=\\mathrm{sin}\\left(5x\\right)\\mathrm{cos}\\,x-\\mathrm{cos}\\left(5x\\right)\\mathrm{sin}\\,x[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1573766\">\n<div id=\"fs-id1573767\">\n<p id=\"fs-id1573768\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(2x\\right),g\\left(x\\right)=2\\,\\mathrm{sin}\\,x\\,\\mathrm{cos}\\,x[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1544808\">They are the same.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1544812\">\n<div id=\"fs-id1544813\">\n<p id=\"fs-id1544814\">[latex]f\\left(\\theta \\right)=\\mathrm{cos}\\left(2\\theta \\right),g\\left(\\theta \\right)={\\mathrm{cos}}^{2}\\theta -{\\mathrm{sin}}^{2}\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1687448\">\n<div id=\"fs-id1687449\">\n<p id=\"fs-id1687450\">[latex]f\\left(\\theta \\right)=\\mathrm{tan}\\left(2\\theta \\right),g\\left(\\theta \\right)=\\frac{\\mathrm{tan}\\,\\theta }{1+{\\mathrm{tan}}^{2}\\theta }[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1186012\">They are the different, try[latex]\\,g\\left(\\theta \\right)=\\frac{2\\,\\mathrm{tan}\\theta }{1-{\\mathrm{tan}}^{2}\\theta }.[\/latex]<\/details>\n<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2888541\">\n<div id=\"fs-id2888542\">\n<p id=\"fs-id2888544\">[latex]f\\left(x\\right)=\\mathrm{sin}\\left(3x\\right)\\mathrm{sin}\\,x,g\\left(x\\right)={\\mathrm{sin}}^{2}\\left(2x\\right){\\mathrm{cos}}^{2}x-{\\mathrm{cos}}^{2}\\left(2x\\right){\\mathrm{sin}}^{2}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1664714\">\n<div id=\"fs-id1664715\">\n<p id=\"fs-id1664716\">[latex]f\\left(x\\right)=\\mathrm{tan}\\left(-x\\right),g\\left(x\\right)=\\frac{\\mathrm{tan}\\,x-\\mathrm{tan}\\left(2x\\right)}{1-\\mathrm{tan}\\,x\\,\\mathrm{tan}\\left(2x\\right)}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1547118\">They are different, try[latex]\\,g\\left(x\\right)=\\frac{\\mathrm{tan}x-\\mathrm{tan}\\left(2x\\right)}{1+\\mathrm{tan}x\\mathrm{tan}\\left(2x\\right)}.[\/latex]<\/details>\n<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\">\n<h4>Technology<\/h4>\n<p id=\"fs-id1429674\">For the following exercises, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point.<\/p>\n<div id=\"fs-id1429678\">\n<div id=\"fs-id1429679\">\n<p id=\"fs-id1429680\">[latex]\\mathrm{sin}\\left(75\u00b0\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1604725\">\n<div id=\"fs-id1604726\">\n<p id=\"fs-id1604727\">[latex]\\mathrm{sin}\\left(195\u00b0\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1874102\">[latex]-\\frac{\\sqrt{3}-1}{2\\sqrt{2}},\\text{or }-0.2588[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2160300\">\n<div id=\"fs-id2160301\">\n<p id=\"fs-id2160302\">[latex]\\mathrm{cos}\\left(165\u00b0\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2613308\">\n<div id=\"fs-id2613309\">\n<p id=\"fs-id2613310\">[latex]\\mathrm{cos}\\left(345\u00b0\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1676814\">[latex]\\frac{1+\\sqrt{3}}{2\\sqrt{2}},[\/latex]or 0.9659<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2291639\">\n<div id=\"fs-id2291640\">\n<p id=\"fs-id2291641\">[latex]\\mathrm{tan}\\left(-15\u00b0\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2244374\" class=\"bc-section section\">\n<h4>Extensions<\/h4>\n<p id=\"fs-id2244379\">For the following exercises, prove the identities provided.<\/p>\n<div id=\"fs-id2244382\">\n<div id=\"fs-id2244383\">\n<p id=\"fs-id2244384\">[latex]\\mathrm{tan}\\left(x+\\frac{\\pi }{4}\\right)=\\frac{\\mathrm{tan}\\,x+1}{1-\\mathrm{tan}\\,x}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2884049\">[latex]\\begin{array}{ccc}\\hfill \\mathrm{tan}\\left(x+\\frac{\\pi }{4}\\right)& =& \\\\ \\hfill \\frac{\\mathrm{tan}x+\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)}{1-\\mathrm{tan}x\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)}& =& \\\\ \\hfill \\frac{\\mathrm{tan}x+1}{1-\\mathrm{tan}x\\left(1\\right)}& =& \\frac{\\mathrm{tan}x+1}{1-\\mathrm{tan}x}\\hfill \\end{array}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1743113\">\n<div id=\"fs-id1743114\">\n<p id=\"fs-id1743115\">[latex]\\frac{\\mathrm{tan}\\left(a+b\\right)}{\\mathrm{tan}\\left(a-b\\right)}=\\frac{\\mathrm{sin}\\,a\\,\\mathrm{cos}\\,a+\\mathrm{sin}\\,b\\,\\mathrm{cos}\\,b}{\\mathrm{sin}\\,a\\,\\mathrm{cos}\\,a-\\mathrm{sin}\\,b\\,\\mathrm{cos}\\,b}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2024685\">\n<div id=\"fs-id2024686\">\n<p id=\"fs-id2024687\">[latex]\\frac{\\mathrm{cos}\\left(a+b\\right)}{\\mathrm{cos}\\,a\\,\\mathrm{cos}\\,b}=1-\\mathrm{tan}\\,a\\,\\mathrm{tan}\\,b[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2810960\">[latex]\\begin{array}{ccc}\\hfill \\frac{\\mathrm{cos}\\left(a+b\\right)}{\\mathrm{cos}a\\mathrm{cos}b}& =& \\\\ \\hfill \\frac{\\mathrm{cos}a\\mathrm{cos}b}{\\mathrm{cos}a\\mathrm{cos}b}-\\frac{\\mathrm{sin}a\\mathrm{sin}b}{\\mathrm{cos}a\\mathrm{cos}b}& =& 1-\\mathrm{tan}a\\mathrm{tan}b\\hfill \\end{array}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2467728\">\n<div id=\"fs-id2467729\">\n<p id=\"fs-id2467730\">[latex]\\mathrm{cos}\\left(x+y\\right)\\mathrm{cos}\\left(x-y\\right)={\\mathrm{cos}}^{2}x-{\\mathrm{sin}}^{2}y[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1831266\">\n<div id=\"fs-id1831267\">\n<p id=\"fs-id1831268\">[latex]\\frac{\\mathrm{cos}\\left(x+h\\right)-\\mathrm{cos}\\,x}{h}=\\mathrm{cos}\\,x\\frac{\\mathrm{cos}\\,h-1}{h}-\\mathrm{sin}\\,x\\frac{\\mathrm{sin}\\,h}{h}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2167170\">[latex]\\begin{array}{ccc}\\hfill \\frac{\\mathrm{cos}\\left(x+h\\right)-\\mathrm{cos}x}{h}& =& \\\\ \\hfill \\frac{\\mathrm{cos}x\\mathrm{cosh}-\\mathrm{sin}x\\mathrm{sinh}-\\mathrm{cos}x}{h}& =& \\\\ \\hfill \\frac{\\mathrm{cos}x\\left(\\mathrm{cosh}-1\\right)-\\mathrm{sin}x\\mathrm{sinh}}{h}& =& \\mathrm{cos}x\\frac{\\mathrm{cos}h-1}{h}-\\mathrm{sin}x\\frac{\\mathrm{sin}h}{h}\\hfill \\end{array}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n<p id=\"fs-id1647857\">For the following exercises, prove or disprove the statements.<\/p>\n<div id=\"fs-id1647860\">\n<div id=\"fs-id1647861\">\n<p id=\"fs-id1647862\">[latex]\\mathrm{tan}\\left(u+v\\right)=\\frac{\\mathrm{tan}\\,u+\\mathrm{tan}\\,v}{1-\\mathrm{tan}\\,u\\,\\mathrm{tan}\\,v}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id2575246\">\n<div id=\"fs-id2575247\">\n<p id=\"fs-id2575248\">[latex]\\mathrm{tan}\\left(u-v\\right)=\\frac{\\mathrm{tan}\\,u-\\mathrm{tan}\\,v}{1+\\mathrm{tan}\\,u\\,\\mathrm{tan}\\,v}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id1369518\">True<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id1369522\">\n<div id=\"fs-id1369523\">\n<p id=\"fs-id1369524\">[latex]\\frac{\\mathrm{tan}\\left(x+y\\right)}{1+\\mathrm{tan}\\,x\\,\\mathrm{tan}\\,x}=\\frac{\\mathrm{tan}\\,x+\\mathrm{tan}\\,y}{1-{\\mathrm{tan}}^{2}x\\,{\\mathrm{tan}}^{2}y}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1859738\">\n<div id=\"fs-id1859739\">\n<p id=\"fs-id1859740\">If[latex]\\,\\alpha ,\\beta ,[\/latex]and[latex]\\,\\gamma \\,[\/latex]are angles in the same triangle, then prove or disprove[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\,\\gamma .[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<details>\n<summary>Show Solution<\/summary>\n<p id=\"fs-id2290243\">True. Note that[latex]\\,\\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\left(\\pi -\\gamma \\right)\\,[\/latex]and expand the right hand side.<\/p>\n<\/details>\n<\/div>\n<\/div>\n<div id=\"fs-id2640056\">\n<div id=\"fs-id2640057\">\n<p id=\"fs-id2640058\">If[latex]\\,\\alpha ,\\beta ,[\/latex]and[latex]\\,y\\,[\/latex]are angles in the same triangle, then prove or disprove[latex]\\,\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta +\\mathrm{tan}\\,\\gamma =\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta \\,\\mathrm{tan}\\,\\gamma[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":291,"menu_order":3,"template":"","meta":{"pb_show_title":null,"pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-139","chapter","type-chapter","status-publish","hentry"],"part":134,"_links":{"self":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/139","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/users\/291"}],"version-history":[{"count":1,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/139\/revisions"}],"predecessor-version":[{"id":140,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/139\/revisions\/140"}],"part":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/parts\/134"}],"metadata":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapters\/139\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/media?parent=139"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/chapter-type?post=139"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/contributor?post=139"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/license?post=139"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}