{"id":215,"date":"2019-08-20T17:04:08","date_gmt":"2019-08-20T21:04:08","guid":{"rendered":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/back-matter\/proofs-identities-and-toolkit-functions\/"},"modified":"2019-08-20T17:04:08","modified_gmt":"2019-08-20T21:04:08","slug":"proofs-identities-and-toolkit-functions","status":"publish","type":"back-matter","link":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/back-matter\/proofs-identities-and-toolkit-functions\/","title":{"raw":"Proofs, Identities, and Toolkit Functions","rendered":"Proofs, Identities, and Toolkit Functions"},"content":{"raw":"
\n

Appendix<\/h3>\n
\n

Important Proofs and Derivations<\/h4>\n

Product Rule<\/strong><\/p>\n

[latex]{\\mathrm{log}}_{a}xy={\\mathrm{log}}_{a}x+{\\mathrm{log}}_{a}y[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

Let[latex]\\,m={\\mathrm{log}}_{a}x\\,[\/latex]and[latex]\\,n={\\mathrm{log}}_{a}y.[\/latex]<\/p>\n

Write in exponent form.<\/p>\n

[latex]x={a}^{m}\\,[\/latex]and[latex]\\,y={a}^{n}.[\/latex]<\/p>\n

Multiply.<\/p>\n

[latex]xy={a}^{m}{a}^{n}={a}^{m+n}[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {a}^{m+n}& =& xy\\hfill \\\\ \\hfill {\\mathrm{log}}_{a}\\left(xy\\right)& =& m+n\\hfill \\\\ & =& {\\mathrm{log}}_{a}x+{\\mathrm{log}}_{b}y\\hfill \\end{array}[\/latex]<\/p>\n

Change of Base Rule<\/strong><\/p>\n

[latex]\\begin{array}{l}\\hfill \\\\ {\\mathrm{log}}_{a}b=\\frac{{\\mathrm{log}}_{c}b}{{\\mathrm{log}}_{c}a}\\hfill \\\\ {\\mathrm{log}}_{a}b=\\frac{1}{{\\mathrm{log}}_{b}a}\\hfill \\end{array}[\/latex]<\/p>\n

where[latex]\\,x\\,[\/latex]and[latex]\\,y\\,[\/latex]are positive, and[latex]\\,a>0,a\\ne 1.[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

Let[latex]\\,x={\\mathrm{log}}_{a}b.[\/latex]<\/p>\n

Write in exponent form.<\/p>\n

[latex]{a}^{x}=b[\/latex]<\/p>\n

Take the[latex]\\,{\\mathrm{log}}_{c}\\,[\/latex]of both sides.<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {\\mathrm{log}}_{c}{a}^{x}& =& {\\mathrm{log}}_{c}b\\hfill \\\\ \\hfill x{\\mathrm{log}}_{c}a& =& {\\mathrm{log}}_{c}b\\hfill \\\\ \\hfill x& =& \\frac{{\\mathrm{log}}_{c}b}{{\\mathrm{log}}_{c}a}\\hfill \\\\ \\hfill {\\mathrm{log}}_{a}b& =& \\frac{{\\mathrm{log}}_{c}b}{{\\mathrm{log}}_{a}b}\\hfill \\end{array}[\/latex]<\/p>\n

When[latex]\\,c=b,[\/latex]<\/p>\n

[latex]{\\mathrm{log}}_{a}b=\\frac{{\\mathrm{log}}_{b}b}{{\\mathrm{log}}_{b}a}=\\frac{1}{{\\mathrm{log}}_{b}a}[\/latex]<\/p>\n

Heron\u2019s Formula<\/strong><\/p>\n

[latex]A=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex]<\/p>\n

where[latex]\\,s=\\frac{a+b+c}{2}[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

Let[latex]\\,a,[\/latex][latex]b,[\/latex]and[latex]\\,c\\,[\/latex]be the sides of a triangle, and[latex]\\,h\\,[\/latex]be the height.<\/p>\n\n

\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 1.<\/strong>[\/caption]\n\n<\/div>\n

So[latex]\\,s=\\frac{a+b+c}{2}[\/latex].<\/p>\n

We can further name the parts of the base in each triangle established by the height such that[latex]\\,p+q=c.[\/latex]<\/p>\n\n

\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Figure 2.<\/strong>[\/caption]\n\n<\/div>\n

Using the Pythagorean Theorem,[latex]\\,{h}^{2}+{p}^{2}={a}^{2}\\,[\/latex]and[latex]\\,{h}^{2}+{q}^{2}={b}^{2}.[\/latex]<\/p>\n

Since[latex]\\,q=c-p,[\/latex]then[latex]\\,{q}^{2}={\\left(c-p\\right)}^{2}.\\,[\/latex]Expanding, we find that[latex]\\,{q}^{2}={c}^{2}-2cp+{p}^{2}.[\/latex]<\/p>\n

We can then add[latex]\\,{h}^{2}\\,[\/latex]to each side of the equation to get[latex]\\,{h}^{2}+{q}^{2}={h}^{2}+{c}^{2}-2cp+{p}^{2}.[\/latex]<\/p>\n

Substitute this result into the equation[latex]\\,{h}^{2}+{q}^{2}={b}^{2}\\,[\/latex]yields[latex]\\,{b}^{2}={h}^{2}+{c}^{2}-2cp+{p}^{2}.[\/latex]<\/p>\n

Then replacing[latex]\\,{h}^{2}+{p}^{2}\\,[\/latex]with[latex]\\,{a}^{2}\\,[\/latex]gives[latex]\\,{b}^{2}={a}^{2}-2cp+{c}^{2}.[\/latex]<\/p>\n

Solve for[latex]\\,p\\,[\/latex]to get<\/p>\n

[latex]p=\\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2c}[\/latex]<\/p>\n

Since[latex]\\,{h}^{2}={a}^{2}-{p}^{2},[\/latex]we get an expression in terms of[latex]\\,a,[\/latex][latex]b,[\/latex]and [latex]\\,c.[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {h}^{2}& =& {a}^{2}-{p}^{2}\\hfill \\\\ & =& \\left(a+p\\right)\\left(a-p\\right)\\hfill \\\\ & =& \\left[a+\\frac{\\left({a}^{2}+{c}^{2}-{b}^{2}\\right)}{2c}\\right]\\left[a-\\frac{\\left({a}^{2}+{c}^{2}-{b}^{2}\\right)}{2c}\\right]\\hfill \\\\ & =& \\frac{\\left(2ac+{a}^{2}+{c}^{2}-{b}^{2}\\right)\\left(2ac-{a}^{2}-{c}^{2}+{b}^{2}\\right)}{4{c}^{2}}\\hfill \\\\ & =& \\frac{\\left({\\left(a+c\\right)}^{2}-{b}^{2}\\right)\\left({b}^{2}-{\\left(a-c\\right)}^{2}\\right)}{4{c}^{2}}\\hfill \\\\ & =& \\frac{\\left(a+b+c\\right)\\left(a+c-b\\right)\\left(b+a-c\\right)\\left(b-a+c\\right)}{4{c}^{2}}\\hfill \\\\ & =& \\frac{\\left(a+b+c\\right)\\left(-a+b+c\\right)\\left(a-b+c\\right)\\left(a+b-c\\right)}{4{c}^{2}}\\hfill \\\\ & =& \\frac{2s\\cdot \\left(2s-a\\right)\\cdot \\left(2s-b\\right)\\left(2s-c\\right)}{4{c}^{2}}\\hfill \\end{array}[\/latex]<\/p>\n

Therefore,<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {h}^{2}& =& \\frac{4s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}{{c}^{2}}\\hfill \\\\ \\hfill h& =& \\frac{2\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}}{c}\\hfill \\end{array}[\/latex]<\/p>\n

And since[latex]\\,A=\\frac{1}{2}ch,[\/latex]then<\/p>\n

[latex]\\begin{array}{ccc}\\hfill A& =& \\frac{1}{2}c\\frac{2\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}}{c}\\hfill \\\\ & =& \\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}\\hfill \\end{array}[\/latex]<\/p>\n

Properties of the Dot Product<\/strong><\/p>\n

[latex]u\u00b7v=v\u00b7u[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

[latex]\\begin{array}{cc}\\hfill u\u00b7v& =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{v}_{1},{v}_{2},...{v}_{n}\u232a\\hfill \\\\ & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+...+{u}_{n}{v}_{n}\\hfill \\\\ & ={v}_{1}{u}_{1}+{v}_{2}{u}_{2}+...+{v}_{n}{v}_{n}\\hfill \\\\ & =\u2329{v}_{1},{v}_{2},...{v}_{n}\u232a\u00b7\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\\hfill \\\\ & =v\u00b7u\\hfill \\end{array}[\/latex]<\/p>\n

[latex]u\u00b7\\left(v+w\\right)=u\u00b7v+u\u00b7w[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

[latex]\\begin{array}{cc}\\hfill u\u00b7\\left(v+w\\right)& =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\\left(\u2329{v}_{1},{v}_{2},...{v}_{n}\u232a+\u2329{w}_{1},{w}_{2},...{w}_{n}\u232a\\right)\\hfill \\\\ & =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{v}_{1}+{w}_{1},{v}_{2}+{w}_{2},...{v}_{n}+{w}_{n}\u232a\\hfill \\\\ & =\u2329{u}_{1}\\left({v}_{1}+{w}_{1}\\right),{u}_{2}\\left({v}_{2}+{w}_{2}\\right),...{u}_{n}\\left({v}_{n}+{w}_{n}\\right)\u232a\\hfill \\\\ & =\u2329{u}_{1}{v}_{1}+{u}_{1}{w}_{1},{u}_{2}{v}_{2}+{u}_{2}{w}_{2},...{u}_{n}{v}_{n}+{u}_{n}{w}_{n}\u232a\\hfill \\\\ & =\u2329{u}_{1}{v}_{1},{u}_{2}{v}_{2},...,{u}_{n}{v}_{n}\u232a+\u2329{u}_{1}{w}_{1},{u}_{2}{w}_{2},...,{u}_{n}{w}_{n}\u232a\\hfill \\\\ & =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{v}_{1},{v}_{2},...{v}_{n}\u232a+\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{w}_{1},{w}_{2},...{w}_{n}\u232a\\hfill \\\\ & =u\u00b7v+u\u00b7w\\hfill \\end{array}[\/latex]<\/p>\n

[latex]u\u00b7u={|u|}^{2}[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

[latex]\\begin{array}{cc}\\hfill u\u00b7u& =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\\hfill \\\\ & ={u}_{1}{u}_{1}+{u}_{2}{u}_{2}+...+{u}_{n}{u}_{n}\\hfill \\\\ & ={u}_{1}{}^{2}+{u}_{2}{}^{2}+...+{u}_{n}{}^{2}\\hfill \\\\ & =|\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a{|}^{2}\\hfill \\\\ & =v\u00b7u\\hfill \\end{array}[\/latex]<\/p>\n

Standard Form of the Ellipse centered at the Origin<\/strong><\/p>\n

[latex]1=\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}[\/latex]<\/p>\n

Derivation<\/strong><\/p>\n

An ellipse consists of all the points for which the sum of distances from two foci is constant:<\/p>\n

[latex]\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}+\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}=\\text{constant}[\/latex]<\/p>\n\n

\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"An Figure 3.<\/strong>[\/caption]\n\n<\/div>\n

Consider a vertex.<\/p>\n\n

\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"An Figure 4.<\/strong>[\/caption]\n\n<\/div>\n

Then,[latex]\\,\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}+\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}=2a[\/latex]<\/p>\n

Consider a covertex.<\/p>\n\n

\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"An Figure 5.<\/strong>[\/caption]\n\n<\/div>\n

Then[latex]\\,{b}^{2}+{c}^{2}={a}^{2}.[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill \\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}+\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}& =& 2a\\hfill \\\\ \\hfill \\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}& =& 2a-\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {\\left(x+c\\right)}^{2}+{y}^{2}& =& {\\left(2a-\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\right)}^{2}\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}-4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+{\\left(x-c\\right)}^{2}+{y}^{2}\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}-4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{y}^{2}\\hfill \\\\ \\hfill 2cx& =& 4{a}^{2}-4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}-2cx\\hfill \\\\ \\hfill 4cx-4{a}^{2}& =& 4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill -\\frac{1}{4a}\\left(4cx-4{a}^{2}\\right)& =& \\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill a-\\frac{c}{a}x& =& \\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {a}^{2}-2xc+\\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {\\left(x-c\\right)}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}-2xc+\\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}-2xc+{c}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}+\\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}+{c}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}+\\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}+{c}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}-{c}^{2}& =& {x}^{2}-\\frac{{c}^{2}}{{a}^{2}}{x}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}-{c}^{2}& =& {x}^{2}\\left(1-\\frac{{c}^{2}}{{a}^{2}}\\right)+{y}^{2}\\hfill \\end{array}[\/latex]<\/p>\n

Let[latex]\\,1=\\frac{{a}^{2}}{{a}^{2}}.[\/latex]<\/p>\n\n

[latex]\\begin{array}{ccc}\\hfill {a}^{2}-{c}^{2}& =& {x}^{2}\\left(\\frac{{a}^{2}-{c}^{2}}{{a}^{2}}\\right)+{y}^{2}\\hfill \\\\ \\hfill 1& =& \\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{a}^{2}-{c}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n

Because[latex]\\,{b}^{2}+{c}^{2}={a}^{2},[\/latex]then[latex]\\,{b}^{2}={a}^{2}-{c}^{2}.[\/latex]<\/p>\n\n

[latex]\\begin{array}{ccc}\\hfill 1& =& \\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{a}^{2}-{c}^{2}}\\hfill \\\\ \\hfill 1& =& \\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n

Standard Form of the Hyperbola<\/strong><\/p>\n

[latex]1=\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}[\/latex]<\/p>\n

Derivation<\/strong><\/p>\n

A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances between two fixed points is constant.<\/p>\n\n

\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"Side-by-side Figure 6.<\/strong>[\/caption]\n\n<\/div>\n

Diagram 1: The difference of the distances from Point P<\/em> to the foci is constant:<\/p>\n

[latex]\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}=\\text{constant}[\/latex]<\/p>\n

Diagram 2: When the point is a vertex, the difference is[latex]\\,2a.[\/latex]<\/p>\n

[latex]\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}=2a[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill \\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}& =& 2a\\hfill \\\\ \\hfill \\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}& =& 2a\\hfill \\\\ \\hfill \\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}& =& 2a+\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {\\left(x+c\\right)}^{2}+{y}^{2}& =& \\left(2a+\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\right)\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}}+{y}^{2}\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{y}^{2}\\hfill \\\\ \\hfill 2cx& =& 4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}-2cx\\hfill \\\\ \\hfill 4cx-4{a}^{2}& =& 4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill cx-{a}^{2}& =& a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {\\left(cx-{a}^{2}\\right)}^{2}& =& {a}^{2}\\left({\\left(x-c\\right)}^{2}+{y}^{2}\\right)\\hfill \\\\ \\hfill {c}^{2}{x}^{2}-2{a}^{2}{c}^{2}{x}^{2}+{a}^{4}& =& {a}^{2}{x}^{2}-2{a}^{2}{c}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\\hfill \\\\ \\hfill {c}^{2}{x}^{2}+{a}^{4}& =& {a}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\\hfill \\\\ \\hfill {a}^{4}-{a}^{2}{c}^{2}& =& {a}^{2}{x}^{2}-{c}^{2}{x}^{2}+{a}^{2}{y}^{2}\\hfill \\\\ \\hfill {a}^{2}\\left({a}^{2}-{c}^{2}\\right)& =& \\left({a}^{2}-{c}^{2}\\right){x}^{2}+{a}^{2}{y}^{2}\\hfill \\\\ \\hfill {a}^{2}\\left({a}^{2}-{c}^{2}\\right)& =& \\left({c}^{2}-{a}^{2}\\right){x}^{2}-{a}^{2}{y}^{2}\\hfill \\end{array}[\/latex]<\/p>\n

Define[latex]\\,b\\,[\/latex]as a positive number such that[latex]\\,{b}^{2}={c}^{2}-{a}^{2}.[\/latex]<\/p>\n\n

[latex]\\begin{array}{ccc}\\hfill {a}^{2}{b}^{2}& =& {b}^{2}{x}^{2}-{a}^{2}{y}^{2}\\hfill \\\\ \\hfill \\frac{{a}^{2}{b}^{2}}{{a}^{2}{b}^{2}}& =& \\frac{{b}^{2}{x}^{2}}{{a}^{2}{b}^{2}}-\\frac{{a}^{2}{y}^{2}}{{a}^{2}{b}^{2}}\\hfill \\\\ \\hfill 1& =& \\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
\n

Trigonometric Identities<\/h4>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Pythagorean Identity<\/td>\n[latex]\\begin{array}{l}{\\mathrm{cos}}^{2}t+{\\mathrm{sin}}^{2}t=1\\\\ 1+{\\mathrm{tan}}^{2}t={\\mathrm{sec}}^{2}t\\\\ 1+{\\mathrm{cot}}^{2}t={\\mathrm{csc}}^{2}t\\end{array}[\/latex]<\/td>\n<\/tr>\n
Even-Odd Identities<\/td>\n[latex]\\begin{array}{l}\\mathrm{cos}\\left(-t\\right)=cos\\,t\\hfill \\\\ \\mathrm{sec}\\left(-t\\right)=\\mathrm{sec}\\,t\\hfill \\\\ \\mathrm{sin}\\left(-t\\right)=-\\mathrm{sin}\\,t\\hfill \\\\ \\mathrm{tan}\\left(-t\\right)=-\\mathrm{tan}\\,t\\hfill \\\\ \\mathrm{csc}\\left(-t\\right)=-\\mathrm{csc}\\,t\\hfill \\\\ \\mathrm{cot}\\left(-t\\right)=-\\mathrm{cot}\\,t\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Cofunction Identities<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ \\mathrm{cos}\\,t=\\mathrm{sin}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{sin}\\,t=\\mathrm{cos}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{tan}\\,t=\\mathrm{cot}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{cot}\\,t=\\mathrm{tan}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{sec}\\,t=\\mathrm{csc}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{csc}\\,t=\\mathrm{sec}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Fundamental Identities<\/td>\n[latex]\\begin{array}{l}\\mathrm{tan}\\,t=\\frac{\\mathrm{sin}\\,t}{\\mathrm{cos}\\,t}\\hfill \\\\ \\mathrm{sec}\\,t=\\frac{1}{\\mathrm{cos}\\,t}\\hfill \\\\ \\mathrm{csc}\\,t=\\frac{1}{\\mathrm{sin}\\,t}\\hfill \\\\ cot\\,t=\\frac{1}{\\text{tan}\\,t}=\\frac{\\text{cos}\\,t}{\\text{sin}\\,t}\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Sum and Difference Identities<\/td>\n[latex]\\begin{array}{l}\\mathrm{cos}\\left(\\alpha +\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\mathrm{sin}\\left(\\alpha -\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\mathrm{tan}\\left(\\alpha +\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\\\ \\mathrm{tan}\\left(\\alpha -\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta }{1+\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Double-Angle Formulas<\/td>\n[latex]\\begin{array}{l}\\mathrm{sin}\\left(2\\theta \\right)=2\\mathrm{sin}\\,\\theta \\,\\mathrm{cos}\\,\\theta \\hfill \\\\ \\mathrm{cos}\\left(2\\theta \\right)={\\mathrm{cos}}^{2}\\theta -{\\mathrm{sin}}^{2}\\theta \\hfill \\\\ \\mathrm{cos}\\left(2\\theta \\right)=1-2{\\mathrm{sin}}^{2}\\theta \\hfill \\\\ \\mathrm{cos}\\left(2\\theta \\right)=2{\\mathrm{cos}}^{2}\\theta -1\\hfill \\\\ \\mathrm{tan}\\left(2\\theta \\right)=\\frac{2\\mathrm{tan}\\,\\theta }{1-{\\mathrm{tan}}^{2}\\theta }\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Half-Angle Formulas<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ \\mathrm{sin}\\frac{\\alpha }{2}=\u00b1\\sqrt{\\frac{1-\\mathrm{cos}\\,\\alpha }{2}}\\hfill \\\\ \\mathrm{cos}\\frac{\\alpha }{2}=\u00b1\\sqrt{\\frac{1+\\mathrm{cos}\\,\\alpha }{2}}\\hfill \\\\ \\mathrm{tan}\\frac{\\alpha }{2}=\u00b1\\sqrt{\\frac{1-\\mathrm{cos}\\,\\alpha }{1+\\mathrm{cos}\\,\\alpha }}\\hfill \\\\ \\mathrm{tan}\\frac{\\alpha }{2}=\\frac{\\mathrm{sin}\\,\\alpha }{1+\\mathrm{cos}\\,\\alpha }\\hfill \\\\ \\mathrm{tan}\\frac{\\alpha }{2}=\\frac{1-\\mathrm{cos}\\,\\alpha }{\\mathrm{sin}\\,\\alpha }\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Reduction Formulas<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ {\\mathrm{sin}}^{2}\\theta =\\frac{1-\\mathrm{cos}\\left(2\\theta \\right)}{2}\\hfill \\\\ {\\mathrm{cos}}^{2}\\theta =\\frac{1+\\mathrm{cos}\\left(2\\theta \\right)}{2}\\hfill \\\\ {\\mathrm{tan}}^{2}\\theta =\\frac{1-\\mathrm{cos}\\left(2\\theta \\right)}{1+\\mathrm{cos}\\left(2\\theta \\right)}\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Product-to-Sum Formulas<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ \\mathrm{cos}\\alpha \\mathrm{cos}\\beta =\\frac{1}{2}\\left[\\mathrm{cos}\\left(\\alpha -\\beta \\right)+\\mathrm{cos}\\left(\\alpha +\\beta \\right)\\right]\\hfill \\\\ \\mathrm{sin}\\alpha \\mathrm{cos}\\beta =\\frac{1}{2}\\left[\\mathrm{sin}\\left(\\alpha +\\beta \\right)+\\mathrm{sin}\\left(\\alpha -\\beta \\right)\\right]\\hfill \\\\ \\mathrm{sin}\\alpha \\mathrm{sin}\\beta =\\frac{1}{2}\\left[\\mathrm{cos}\\left(\\alpha -\\beta \\right)-\\mathrm{cos}\\left(\\alpha +\\beta \\right)\\right]\\hfill \\\\ \\mathrm{cos}\\alpha \\mathrm{sin}\\beta =\\frac{1}{2}\\left[\\mathrm{sin}\\left(\\alpha +\\beta \\right)-\\mathrm{sin}\\left(\\alpha -\\beta \\right)\\right]\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Sum-to-Product Formulas<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ \\mathrm{sin}\\alpha +\\mathrm{sin}\\beta =2\\mathrm{sin}\\left(\\frac{\\alpha +\\beta }{2}\\right)\\mathrm{cos}\\left(\\frac{\\alpha -\\beta }{2}\\right)\\hfill \\\\ \\mathrm{sin}\\alpha -\\mathrm{sin}\\beta =2\\mathrm{sin}\\left(\\frac{\\alpha -\\beta }{2}\\right)\\mathrm{cos}\\left(\\frac{\\alpha +\\beta }{2}\\right)\\hfill \\\\ \\mathrm{cos}\\alpha -\\mathrm{cos}\\beta =-2\\mathrm{sin}\\left(\\frac{\\alpha +\\beta }{2}\\right)\\mathrm{sin}\\left(\\frac{\\alpha -\\beta }{2}\\right)\\hfill \\\\ \\mathrm{cos}\\alpha +\\mathrm{cos}\\beta =2\\mathrm{cos}\\left(\\frac{\\alpha +\\beta }{2}\\right)\\mathrm{cos}\\left(\\frac{\\alpha -\\beta }{2}\\right)\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Law of Sines<\/td>\n[latex]\\begin{array}{l}\\frac{\\mathrm{sin}\\,\\alpha }{a}=\\frac{\\mathrm{sin}\\,\\beta }{b}=\\frac{\\mathrm{sin}\\,\\gamma }{c}\\hfill \\\\ \\frac{a}{\\mathrm{sin}\\,\\alpha }=\\frac{b}{\\mathrm{sin}\\,\\beta }=\\frac{c}{\\mathrm{sin}\\,\\gamma }\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Law of Cosines<\/td>\n[latex]\\begin{array}{c}{a}^{2}={b}^{2}+{c}^{2}-2bc\\,\\mathrm{cos}\\,\\alpha \\\\ {b}^{2}={a}^{2}+{c}^{2}-2ac\\,\\mathrm{cos}\\,\\beta \\\\ {c}^{2}={a}^{2}+{b}^{2}-2ab\\,\\text{cos}\\,\\gamma \\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
\n

ToolKit Functions<\/h4>\n
\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"Three Figure 7.<\/strong>[\/caption]\n\n<\/div>\n
\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"Three Figure 8.<\/strong>[\/caption]\n\n<\/div>\n
\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"Three Figure 9.<\/strong>[\/caption]\n\n<\/div>\n<\/div>\n
\n

Trigonometric Functions<\/h4>\n

Unit Circle<\/strong><\/p>\n\n

\n\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"Graph Figure 10.<\/strong>[\/caption]\n\n<\/div>\n\n\n\n\n\n\n\n\n\n\n
Angle<\/th>\n[latex]0[\/latex]<\/th>\n[latex]\\frac{\\pi }{6},\\text{or 30}\u00b0[\/latex]<\/th>\n[latex]\\frac{\\pi }{4},\\text{or 45}\u00b0[\/latex]<\/th>\n[latex]\\frac{\\pi }{3},\\text{or 60}\u00b0[\/latex]<\/th>\n[latex]\\frac{\\pi }{2},\\text{or 90}\u00b0[\/latex]<\/th>\n<\/tr>\n<\/thead>\n
Cosine<\/strong><\/td>\n1<\/td>\n[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n[latex]\\frac{1}{2}[\/latex]<\/td>\n0<\/td>\n<\/tr>\n
Sine<\/strong><\/td>\n0<\/td>\n[latex]\\frac{1}{2}[\/latex]<\/td>\n[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n1<\/td>\n<\/tr>\n
Tangent<\/strong><\/td>\n0<\/td>\n[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n1<\/td>\n[latex]\\sqrt{3}[\/latex]<\/td>\nUndefined<\/td>\n<\/tr>\n
Secant<\/strong><\/td>\n1<\/td>\n[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n[latex]\\sqrt{2}[\/latex]<\/td>\n2<\/td>\nUndefined<\/td>\n<\/tr>\n
Cosecant<\/strong><\/td>\nUndefined<\/td>\n2<\/td>\n[latex]\\sqrt{2}[\/latex]<\/td>\n[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n1<\/td>\n<\/tr>\n
Cotangent<\/strong><\/td>\nUndefined<\/td>\n[latex]\\sqrt{3}[\/latex]<\/td>\n1<\/td>\n[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>","rendered":"
\n

Appendix<\/h3>\n
\n

Important Proofs and Derivations<\/h4>\n

Product Rule<\/strong><\/p>\n

[latex]{\\mathrm{log}}_{a}xy={\\mathrm{log}}_{a}x+{\\mathrm{log}}_{a}y[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

Let[latex]\\,m={\\mathrm{log}}_{a}x\\,[\/latex]and[latex]\\,n={\\mathrm{log}}_{a}y.[\/latex]<\/p>\n

Write in exponent form.<\/p>\n

[latex]x={a}^{m}\\,[\/latex]and[latex]\\,y={a}^{n}.[\/latex]<\/p>\n

Multiply.<\/p>\n

[latex]xy={a}^{m}{a}^{n}={a}^{m+n}[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {a}^{m+n}& =& xy\\hfill \\\\ \\hfill {\\mathrm{log}}_{a}\\left(xy\\right)& =& m+n\\hfill \\\\ & =& {\\mathrm{log}}_{a}x+{\\mathrm{log}}_{b}y\\hfill \\end{array}[\/latex]<\/p>\n

Change of Base Rule<\/strong><\/p>\n

[latex]\\begin{array}{l}\\hfill \\\\ {\\mathrm{log}}_{a}b=\\frac{{\\mathrm{log}}_{c}b}{{\\mathrm{log}}_{c}a}\\hfill \\\\ {\\mathrm{log}}_{a}b=\\frac{1}{{\\mathrm{log}}_{b}a}\\hfill \\end{array}[\/latex]<\/p>\n

where[latex]\\,x\\,[\/latex]and[latex]\\,y\\,[\/latex]are positive, and[latex]\\,a>0,a\\ne 1.[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

Let[latex]\\,x={\\mathrm{log}}_{a}b.[\/latex]<\/p>\n

Write in exponent form.<\/p>\n

[latex]{a}^{x}=b[\/latex]<\/p>\n

Take the[latex]\\,{\\mathrm{log}}_{c}\\,[\/latex]of both sides.<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {\\mathrm{log}}_{c}{a}^{x}& =& {\\mathrm{log}}_{c}b\\hfill \\\\ \\hfill x{\\mathrm{log}}_{c}a& =& {\\mathrm{log}}_{c}b\\hfill \\\\ \\hfill x& =& \\frac{{\\mathrm{log}}_{c}b}{{\\mathrm{log}}_{c}a}\\hfill \\\\ \\hfill {\\mathrm{log}}_{a}b& =& \\frac{{\\mathrm{log}}_{c}b}{{\\mathrm{log}}_{a}b}\\hfill \\end{array}[\/latex]<\/p>\n

When[latex]\\,c=b,[\/latex]<\/p>\n

[latex]{\\mathrm{log}}_{a}b=\\frac{{\\mathrm{log}}_{b}b}{{\\mathrm{log}}_{b}a}=\\frac{1}{{\\mathrm{log}}_{b}a}[\/latex]<\/p>\n

Heron\u2019s Formula<\/strong><\/p>\n

[latex]A=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}[\/latex]<\/p>\n

where[latex]\\,s=\\frac{a+b+c}{2}[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

Let[latex]\\,a,[\/latex][latex]b,[\/latex]and[latex]\\,c\\,[\/latex]be the sides of a triangle, and[latex]\\,h\\,[\/latex]be the height.<\/p>\n

\n
\"A
Figure 1.<\/strong><\/figcaption><\/figure>\n<\/div>\n

So[latex]\\,s=\\frac{a+b+c}{2}[\/latex].<\/p>\n

We can further name the parts of the base in each triangle established by the height such that[latex]\\,p+q=c.[\/latex]<\/p>\n

\n
\"A
Figure 2.<\/strong><\/figcaption><\/figure>\n<\/div>\n

Using the Pythagorean Theorem,[latex]\\,{h}^{2}+{p}^{2}={a}^{2}\\,[\/latex]and[latex]\\,{h}^{2}+{q}^{2}={b}^{2}.[\/latex]<\/p>\n

Since[latex]\\,q=c-p,[\/latex]then[latex]\\,{q}^{2}={\\left(c-p\\right)}^{2}.\\,[\/latex]Expanding, we find that[latex]\\,{q}^{2}={c}^{2}-2cp+{p}^{2}.[\/latex]<\/p>\n

We can then add[latex]\\,{h}^{2}\\,[\/latex]to each side of the equation to get[latex]\\,{h}^{2}+{q}^{2}={h}^{2}+{c}^{2}-2cp+{p}^{2}.[\/latex]<\/p>\n

Substitute this result into the equation[latex]\\,{h}^{2}+{q}^{2}={b}^{2}\\,[\/latex]yields[latex]\\,{b}^{2}={h}^{2}+{c}^{2}-2cp+{p}^{2}.[\/latex]<\/p>\n

Then replacing[latex]\\,{h}^{2}+{p}^{2}\\,[\/latex]with[latex]\\,{a}^{2}\\,[\/latex]gives[latex]\\,{b}^{2}={a}^{2}-2cp+{c}^{2}.[\/latex]<\/p>\n

Solve for[latex]\\,p\\,[\/latex]to get<\/p>\n

[latex]p=\\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2c}[\/latex]<\/p>\n

Since[latex]\\,{h}^{2}={a}^{2}-{p}^{2},[\/latex]we get an expression in terms of[latex]\\,a,[\/latex][latex]b,[\/latex]and [latex]\\,c.[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {h}^{2}& =& {a}^{2}-{p}^{2}\\hfill \\\\ & =& \\left(a+p\\right)\\left(a-p\\right)\\hfill \\\\ & =& \\left[a+\\frac{\\left({a}^{2}+{c}^{2}-{b}^{2}\\right)}{2c}\\right]\\left[a-\\frac{\\left({a}^{2}+{c}^{2}-{b}^{2}\\right)}{2c}\\right]\\hfill \\\\ & =& \\frac{\\left(2ac+{a}^{2}+{c}^{2}-{b}^{2}\\right)\\left(2ac-{a}^{2}-{c}^{2}+{b}^{2}\\right)}{4{c}^{2}}\\hfill \\\\ & =& \\frac{\\left({\\left(a+c\\right)}^{2}-{b}^{2}\\right)\\left({b}^{2}-{\\left(a-c\\right)}^{2}\\right)}{4{c}^{2}}\\hfill \\\\ & =& \\frac{\\left(a+b+c\\right)\\left(a+c-b\\right)\\left(b+a-c\\right)\\left(b-a+c\\right)}{4{c}^{2}}\\hfill \\\\ & =& \\frac{\\left(a+b+c\\right)\\left(-a+b+c\\right)\\left(a-b+c\\right)\\left(a+b-c\\right)}{4{c}^{2}}\\hfill \\\\ & =& \\frac{2s\\cdot \\left(2s-a\\right)\\cdot \\left(2s-b\\right)\\left(2s-c\\right)}{4{c}^{2}}\\hfill \\end{array}[\/latex]<\/p>\n

Therefore,<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {h}^{2}& =& \\frac{4s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}{{c}^{2}}\\hfill \\\\ \\hfill h& =& \\frac{2\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}}{c}\\hfill \\end{array}[\/latex]<\/p>\n

And since[latex]\\,A=\\frac{1}{2}ch,[\/latex]then<\/p>\n

[latex]\\begin{array}{ccc}\\hfill A& =& \\frac{1}{2}c\\frac{2\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}}{c}\\hfill \\\\ & =& \\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}\\hfill \\end{array}[\/latex]<\/p>\n

Properties of the Dot Product<\/strong><\/p>\n

[latex]u\u00b7v=v\u00b7u[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

[latex]\\begin{array}{cc}\\hfill u\u00b7v& =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{v}_{1},{v}_{2},...{v}_{n}\u232a\\hfill \\\\ & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+...+{u}_{n}{v}_{n}\\hfill \\\\ & ={v}_{1}{u}_{1}+{v}_{2}{u}_{2}+...+{v}_{n}{v}_{n}\\hfill \\\\ & =\u2329{v}_{1},{v}_{2},...{v}_{n}\u232a\u00b7\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\\hfill \\\\ & =v\u00b7u\\hfill \\end{array}[\/latex]<\/p>\n

[latex]u\u00b7\\left(v+w\\right)=u\u00b7v+u\u00b7w[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

[latex]\\begin{array}{cc}\\hfill u\u00b7\\left(v+w\\right)& =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\\left(\u2329{v}_{1},{v}_{2},...{v}_{n}\u232a+\u2329{w}_{1},{w}_{2},...{w}_{n}\u232a\\right)\\hfill \\\\ & =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{v}_{1}+{w}_{1},{v}_{2}+{w}_{2},...{v}_{n}+{w}_{n}\u232a\\hfill \\\\ & =\u2329{u}_{1}\\left({v}_{1}+{w}_{1}\\right),{u}_{2}\\left({v}_{2}+{w}_{2}\\right),...{u}_{n}\\left({v}_{n}+{w}_{n}\\right)\u232a\\hfill \\\\ & =\u2329{u}_{1}{v}_{1}+{u}_{1}{w}_{1},{u}_{2}{v}_{2}+{u}_{2}{w}_{2},...{u}_{n}{v}_{n}+{u}_{n}{w}_{n}\u232a\\hfill \\\\ & =\u2329{u}_{1}{v}_{1},{u}_{2}{v}_{2},...,{u}_{n}{v}_{n}\u232a+\u2329{u}_{1}{w}_{1},{u}_{2}{w}_{2},...,{u}_{n}{w}_{n}\u232a\\hfill \\\\ & =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{v}_{1},{v}_{2},...{v}_{n}\u232a+\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{w}_{1},{w}_{2},...{w}_{n}\u232a\\hfill \\\\ & =u\u00b7v+u\u00b7w\\hfill \\end{array}[\/latex]<\/p>\n

[latex]u\u00b7u={|u|}^{2}[\/latex]<\/p>\n

Proof:<\/strong><\/p>\n

[latex]\\begin{array}{cc}\\hfill u\u00b7u& =\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\u00b7\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a\\hfill \\\\ & ={u}_{1}{u}_{1}+{u}_{2}{u}_{2}+...+{u}_{n}{u}_{n}\\hfill \\\\ & ={u}_{1}{}^{2}+{u}_{2}{}^{2}+...+{u}_{n}{}^{2}\\hfill \\\\ & =|\u2329{u}_{1},{u}_{2},...{u}_{n}\u232a{|}^{2}\\hfill \\\\ & =v\u00b7u\\hfill \\end{array}[\/latex]<\/p>\n

Standard Form of the Ellipse centered at the Origin<\/strong><\/p>\n

[latex]1=\\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}[\/latex]<\/p>\n

Derivation<\/strong><\/p>\n

An ellipse consists of all the points for which the sum of distances from two foci is constant:<\/p>\n

[latex]\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}+\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}=\\text{constant}[\/latex]<\/p>\n

\n
\"An
Figure 3.<\/strong><\/figcaption><\/figure>\n<\/div>\n

Consider a vertex.<\/p>\n

\n
\"An
Figure 4.<\/strong><\/figcaption><\/figure>\n<\/div>\n

Then,[latex]\\,\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}+\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}=2a[\/latex]<\/p>\n

Consider a covertex.<\/p>\n

\n
\"An
Figure 5.<\/strong><\/figcaption><\/figure>\n<\/div>\n

Then[latex]\\,{b}^{2}+{c}^{2}={a}^{2}.[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill \\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}+\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}& =& 2a\\hfill \\\\ \\hfill \\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}& =& 2a-\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {\\left(x+c\\right)}^{2}+{y}^{2}& =& {\\left(2a-\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\right)}^{2}\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}-4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+{\\left(x-c\\right)}^{2}+{y}^{2}\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}-4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{y}^{2}\\hfill \\\\ \\hfill 2cx& =& 4{a}^{2}-4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}-2cx\\hfill \\\\ \\hfill 4cx-4{a}^{2}& =& 4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill -\\frac{1}{4a}\\left(4cx-4{a}^{2}\\right)& =& \\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill a-\\frac{c}{a}x& =& \\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {a}^{2}-2xc+\\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {\\left(x-c\\right)}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}-2xc+\\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}-2xc+{c}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}+\\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}+{c}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}+\\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}+{c}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}-{c}^{2}& =& {x}^{2}-\\frac{{c}^{2}}{{a}^{2}}{x}^{2}+{y}^{2}\\hfill \\\\ \\hfill {a}^{2}-{c}^{2}& =& {x}^{2}\\left(1-\\frac{{c}^{2}}{{a}^{2}}\\right)+{y}^{2}\\hfill \\end{array}[\/latex]<\/p>\n

Let[latex]\\,1=\\frac{{a}^{2}}{{a}^{2}}.[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {a}^{2}-{c}^{2}& =& {x}^{2}\\left(\\frac{{a}^{2}-{c}^{2}}{{a}^{2}}\\right)+{y}^{2}\\hfill \\\\ \\hfill 1& =& \\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{a}^{2}-{c}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n

Because[latex]\\,{b}^{2}+{c}^{2}={a}^{2},[\/latex]then[latex]\\,{b}^{2}={a}^{2}-{c}^{2}.[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill 1& =& \\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{a}^{2}-{c}^{2}}\\hfill \\\\ \\hfill 1& =& \\frac{{x}^{2}}{{a}^{2}}+\\frac{{y}^{2}}{{b}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n

Standard Form of the Hyperbola<\/strong><\/p>\n

[latex]1=\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}[\/latex]<\/p>\n

Derivation<\/strong><\/p>\n

A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances between two fixed points is constant.<\/p>\n

\n
\"Side-by-side
Figure 6.<\/strong><\/figcaption><\/figure>\n<\/div>\n

Diagram 1: The difference of the distances from Point P<\/em> to the foci is constant:<\/p>\n

[latex]\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}=\\text{constant}[\/latex]<\/p>\n

Diagram 2: When the point is a vertex, the difference is[latex]\\,2a.[\/latex]<\/p>\n

[latex]\\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}=2a[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill \\sqrt{{\\left(x-\\left(-c\\right)\\right)}^{2}+{\\left(y-0\\right)}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{\\left(y-0\\right)}^{2}}& =& 2a\\hfill \\\\ \\hfill \\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}-\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}& =& 2a\\hfill \\\\ \\hfill \\sqrt{{\\left(x+c\\right)}^{2}+{y}^{2}}& =& 2a+\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {\\left(x+c\\right)}^{2}+{y}^{2}& =& \\left(2a+\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\right)\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}}+{y}^{2}\\hfill \\\\ \\hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{y}^{2}\\hfill \\\\ \\hfill 2cx& =& 4{a}^{2}+4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}-2cx\\hfill \\\\ \\hfill 4cx-4{a}^{2}& =& 4a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill cx-{a}^{2}& =& a\\sqrt{{\\left(x-c\\right)}^{2}+{y}^{2}}\\hfill \\\\ \\hfill {\\left(cx-{a}^{2}\\right)}^{2}& =& {a}^{2}\\left({\\left(x-c\\right)}^{2}+{y}^{2}\\right)\\hfill \\\\ \\hfill {c}^{2}{x}^{2}-2{a}^{2}{c}^{2}{x}^{2}+{a}^{4}& =& {a}^{2}{x}^{2}-2{a}^{2}{c}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\\hfill \\\\ \\hfill {c}^{2}{x}^{2}+{a}^{4}& =& {a}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\\hfill \\\\ \\hfill {a}^{4}-{a}^{2}{c}^{2}& =& {a}^{2}{x}^{2}-{c}^{2}{x}^{2}+{a}^{2}{y}^{2}\\hfill \\\\ \\hfill {a}^{2}\\left({a}^{2}-{c}^{2}\\right)& =& \\left({a}^{2}-{c}^{2}\\right){x}^{2}+{a}^{2}{y}^{2}\\hfill \\\\ \\hfill {a}^{2}\\left({a}^{2}-{c}^{2}\\right)& =& \\left({c}^{2}-{a}^{2}\\right){x}^{2}-{a}^{2}{y}^{2}\\hfill \\end{array}[\/latex]<\/p>\n

Define[latex]\\,b\\,[\/latex]as a positive number such that[latex]\\,{b}^{2}={c}^{2}-{a}^{2}.[\/latex]<\/p>\n

[latex]\\begin{array}{ccc}\\hfill {a}^{2}{b}^{2}& =& {b}^{2}{x}^{2}-{a}^{2}{y}^{2}\\hfill \\\\ \\hfill \\frac{{a}^{2}{b}^{2}}{{a}^{2}{b}^{2}}& =& \\frac{{b}^{2}{x}^{2}}{{a}^{2}{b}^{2}}-\\frac{{a}^{2}{y}^{2}}{{a}^{2}{b}^{2}}\\hfill \\\\ \\hfill 1& =& \\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
\n

Trigonometric Identities<\/h4>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
Pythagorean Identity<\/td>\n[latex]\\begin{array}{l}{\\mathrm{cos}}^{2}t+{\\mathrm{sin}}^{2}t=1\\\\ 1+{\\mathrm{tan}}^{2}t={\\mathrm{sec}}^{2}t\\\\ 1+{\\mathrm{cot}}^{2}t={\\mathrm{csc}}^{2}t\\end{array}[\/latex]<\/td>\n<\/tr>\n
Even-Odd Identities<\/td>\n[latex]\\begin{array}{l}\\mathrm{cos}\\left(-t\\right)=cos\\,t\\hfill \\\\ \\mathrm{sec}\\left(-t\\right)=\\mathrm{sec}\\,t\\hfill \\\\ \\mathrm{sin}\\left(-t\\right)=-\\mathrm{sin}\\,t\\hfill \\\\ \\mathrm{tan}\\left(-t\\right)=-\\mathrm{tan}\\,t\\hfill \\\\ \\mathrm{csc}\\left(-t\\right)=-\\mathrm{csc}\\,t\\hfill \\\\ \\mathrm{cot}\\left(-t\\right)=-\\mathrm{cot}\\,t\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Cofunction Identities<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ \\mathrm{cos}\\,t=\\mathrm{sin}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{sin}\\,t=\\mathrm{cos}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{tan}\\,t=\\mathrm{cot}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{cot}\\,t=\\mathrm{tan}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{sec}\\,t=\\mathrm{csc}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\\\ \\mathrm{csc}\\,t=\\mathrm{sec}\\left(\\frac{\\pi }{2}-t\\right)\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Fundamental Identities<\/td>\n[latex]\\begin{array}{l}\\mathrm{tan}\\,t=\\frac{\\mathrm{sin}\\,t}{\\mathrm{cos}\\,t}\\hfill \\\\ \\mathrm{sec}\\,t=\\frac{1}{\\mathrm{cos}\\,t}\\hfill \\\\ \\mathrm{csc}\\,t=\\frac{1}{\\mathrm{sin}\\,t}\\hfill \\\\ cot\\,t=\\frac{1}{\\text{tan}\\,t}=\\frac{\\text{cos}\\,t}{\\text{sin}\\,t}\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Sum and Difference Identities<\/td>\n[latex]\\begin{array}{l}\\mathrm{cos}\\left(\\alpha +\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\mathrm{cos}\\left(\\alpha -\\beta \\right)=\\mathrm{cos}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{sin}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\mathrm{sin}\\left(\\alpha +\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta +\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\mathrm{sin}\\left(\\alpha -\\beta \\right)=\\mathrm{sin}\\,\\alpha \\,\\mathrm{cos}\\,\\beta -\\mathrm{cos}\\,\\alpha \\,\\mathrm{sin}\\,\\beta \\hfill \\\\ \\mathrm{tan}\\left(\\alpha +\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha +\\mathrm{tan}\\,\\beta }{1-\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\\\ \\mathrm{tan}\\left(\\alpha -\\beta \\right)=\\frac{\\mathrm{tan}\\,\\alpha -\\mathrm{tan}\\,\\beta }{1+\\mathrm{tan}\\,\\alpha \\,\\mathrm{tan}\\,\\beta }\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Double-Angle Formulas<\/td>\n[latex]\\begin{array}{l}\\mathrm{sin}\\left(2\\theta \\right)=2\\mathrm{sin}\\,\\theta \\,\\mathrm{cos}\\,\\theta \\hfill \\\\ \\mathrm{cos}\\left(2\\theta \\right)={\\mathrm{cos}}^{2}\\theta -{\\mathrm{sin}}^{2}\\theta \\hfill \\\\ \\mathrm{cos}\\left(2\\theta \\right)=1-2{\\mathrm{sin}}^{2}\\theta \\hfill \\\\ \\mathrm{cos}\\left(2\\theta \\right)=2{\\mathrm{cos}}^{2}\\theta -1\\hfill \\\\ \\mathrm{tan}\\left(2\\theta \\right)=\\frac{2\\mathrm{tan}\\,\\theta }{1-{\\mathrm{tan}}^{2}\\theta }\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Half-Angle Formulas<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ \\mathrm{sin}\\frac{\\alpha }{2}=\u00b1\\sqrt{\\frac{1-\\mathrm{cos}\\,\\alpha }{2}}\\hfill \\\\ \\mathrm{cos}\\frac{\\alpha }{2}=\u00b1\\sqrt{\\frac{1+\\mathrm{cos}\\,\\alpha }{2}}\\hfill \\\\ \\mathrm{tan}\\frac{\\alpha }{2}=\u00b1\\sqrt{\\frac{1-\\mathrm{cos}\\,\\alpha }{1+\\mathrm{cos}\\,\\alpha }}\\hfill \\\\ \\mathrm{tan}\\frac{\\alpha }{2}=\\frac{\\mathrm{sin}\\,\\alpha }{1+\\mathrm{cos}\\,\\alpha }\\hfill \\\\ \\mathrm{tan}\\frac{\\alpha }{2}=\\frac{1-\\mathrm{cos}\\,\\alpha }{\\mathrm{sin}\\,\\alpha }\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Reduction Formulas<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ {\\mathrm{sin}}^{2}\\theta =\\frac{1-\\mathrm{cos}\\left(2\\theta \\right)}{2}\\hfill \\\\ {\\mathrm{cos}}^{2}\\theta =\\frac{1+\\mathrm{cos}\\left(2\\theta \\right)}{2}\\hfill \\\\ {\\mathrm{tan}}^{2}\\theta =\\frac{1-\\mathrm{cos}\\left(2\\theta \\right)}{1+\\mathrm{cos}\\left(2\\theta \\right)}\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Product-to-Sum Formulas<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ \\mathrm{cos}\\alpha \\mathrm{cos}\\beta =\\frac{1}{2}\\left[\\mathrm{cos}\\left(\\alpha -\\beta \\right)+\\mathrm{cos}\\left(\\alpha +\\beta \\right)\\right]\\hfill \\\\ \\mathrm{sin}\\alpha \\mathrm{cos}\\beta =\\frac{1}{2}\\left[\\mathrm{sin}\\left(\\alpha +\\beta \\right)+\\mathrm{sin}\\left(\\alpha -\\beta \\right)\\right]\\hfill \\\\ \\mathrm{sin}\\alpha \\mathrm{sin}\\beta =\\frac{1}{2}\\left[\\mathrm{cos}\\left(\\alpha -\\beta \\right)-\\mathrm{cos}\\left(\\alpha +\\beta \\right)\\right]\\hfill \\\\ \\mathrm{cos}\\alpha \\mathrm{sin}\\beta =\\frac{1}{2}\\left[\\mathrm{sin}\\left(\\alpha +\\beta \\right)-\\mathrm{sin}\\left(\\alpha -\\beta \\right)\\right]\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Sum-to-Product Formulas<\/td>\n[latex]\\begin{array}{l}\\hfill \\\\ \\mathrm{sin}\\alpha +\\mathrm{sin}\\beta =2\\mathrm{sin}\\left(\\frac{\\alpha +\\beta }{2}\\right)\\mathrm{cos}\\left(\\frac{\\alpha -\\beta }{2}\\right)\\hfill \\\\ \\mathrm{sin}\\alpha -\\mathrm{sin}\\beta =2\\mathrm{sin}\\left(\\frac{\\alpha -\\beta }{2}\\right)\\mathrm{cos}\\left(\\frac{\\alpha +\\beta }{2}\\right)\\hfill \\\\ \\mathrm{cos}\\alpha -\\mathrm{cos}\\beta =-2\\mathrm{sin}\\left(\\frac{\\alpha +\\beta }{2}\\right)\\mathrm{sin}\\left(\\frac{\\alpha -\\beta }{2}\\right)\\hfill \\\\ \\mathrm{cos}\\alpha +\\mathrm{cos}\\beta =2\\mathrm{cos}\\left(\\frac{\\alpha +\\beta }{2}\\right)\\mathrm{cos}\\left(\\frac{\\alpha -\\beta }{2}\\right)\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Law of Sines<\/td>\n[latex]\\begin{array}{l}\\frac{\\mathrm{sin}\\,\\alpha }{a}=\\frac{\\mathrm{sin}\\,\\beta }{b}=\\frac{\\mathrm{sin}\\,\\gamma }{c}\\hfill \\\\ \\frac{a}{\\mathrm{sin}\\,\\alpha }=\\frac{b}{\\mathrm{sin}\\,\\beta }=\\frac{c}{\\mathrm{sin}\\,\\gamma }\\hfill \\end{array}[\/latex]<\/td>\n<\/tr>\n
Law of Cosines<\/td>\n[latex]\\begin{array}{c}{a}^{2}={b}^{2}+{c}^{2}-2bc\\,\\mathrm{cos}\\,\\alpha \\\\ {b}^{2}={a}^{2}+{c}^{2}-2ac\\,\\mathrm{cos}\\,\\beta \\\\ {c}^{2}={a}^{2}+{b}^{2}-2ab\\,\\text{cos}\\,\\gamma \\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n
\n

ToolKit Functions<\/h4>\n
\n
\"Three
Figure 7.<\/strong><\/figcaption><\/figure>\n<\/div>\n
\n
\"Three
Figure 8.<\/strong><\/figcaption><\/figure>\n<\/div>\n
\n
\"Three
Figure 9.<\/strong><\/figcaption><\/figure>\n<\/div>\n<\/div>\n
\n

Trigonometric Functions<\/h4>\n

Unit Circle<\/strong><\/p>\n

\n
\"Graph
Figure 10.<\/strong><\/figcaption><\/figure>\n<\/div>\n\n\n\n\n\n\n\n\n\n\n
Angle<\/th>\n[latex]0[\/latex]<\/th>\n[latex]\\frac{\\pi }{6},\\text{or 30}\u00b0[\/latex]<\/th>\n[latex]\\frac{\\pi }{4},\\text{or 45}\u00b0[\/latex]<\/th>\n[latex]\\frac{\\pi }{3},\\text{or 60}\u00b0[\/latex]<\/th>\n[latex]\\frac{\\pi }{2},\\text{or 90}\u00b0[\/latex]<\/th>\n<\/tr>\n<\/thead>\n
Cosine<\/strong><\/td>\n1<\/td>\n[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n[latex]\\frac{1}{2}[\/latex]<\/td>\n0<\/td>\n<\/tr>\n
Sine<\/strong><\/td>\n0<\/td>\n[latex]\\frac{1}{2}[\/latex]<\/td>\n[latex]\\frac{\\sqrt{2}}{2}[\/latex]<\/td>\n[latex]\\frac{\\sqrt{3}}{2}[\/latex]<\/td>\n1<\/td>\n<\/tr>\n
Tangent<\/strong><\/td>\n0<\/td>\n[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n1<\/td>\n[latex]\\sqrt{3}[\/latex]<\/td>\nUndefined<\/td>\n<\/tr>\n
Secant<\/strong><\/td>\n1<\/td>\n[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n[latex]\\sqrt{2}[\/latex]<\/td>\n2<\/td>\nUndefined<\/td>\n<\/tr>\n
Cosecant<\/strong><\/td>\nUndefined<\/td>\n2<\/td>\n[latex]\\sqrt{2}[\/latex]<\/td>\n[latex]\\frac{2\\sqrt{3}}{3}[\/latex]<\/td>\n1<\/td>\n<\/tr>\n
Cotangent<\/strong><\/td>\nUndefined<\/td>\n[latex]\\sqrt{3}[\/latex]<\/td>\n1<\/td>\n[latex]\\frac{\\sqrt{3}}{3}[\/latex]<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n","protected":false},"author":291,"menu_order":1,"template":"","meta":{"pb_show_title":null,"pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"back-matter-type":[],"contributor":[],"license":[],"class_list":["post-215","back-matter","type-back-matter","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/back-matter\/215","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/back-matter"}],"about":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/types\/back-matter"}],"author":[{"embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/users\/291"}],"version-history":[{"count":0,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/back-matter\/215\/revisions"}],"metadata":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/back-matter\/215\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/media?parent=215"}],"wp:term":[{"taxonomy":"back-matter-type","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/pressbooks\/v2\/back-matter-type?post=215"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/contributor?post=215"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/integrations.pressbooks.network\/testinternalcloneforcomparison\/wp-json\/wp\/v2\/license?post=215"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}