# 0.1 Proofs, identities, and toolkit functions

 Page 1 / 1

## Important proofs and derivations

Product Rule

${\mathrm{log}}_{a}xy={\mathrm{log}}_{a}x+{\mathrm{log}}_{a}y$

Proof:

Let $\text{\hspace{0.17em}}m={\mathrm{log}}_{a}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n={\mathrm{log}}_{a}y.$

Write in exponent form.

$x={a}^{m}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y={a}^{n}.$

Multiply.

$xy={a}^{m}{a}^{n}={a}^{m+n}$

$\begin{array}{ccc}\hfill {a}^{m+n}& =& xy\hfill \\ \hfill {\mathrm{log}}_{a}\left(xy\right)& =& m+n\hfill \\ & =& {\mathrm{log}}_{a}x+{\mathrm{log}}_{b}y\hfill \end{array}$

Change of Base Rule

$\begin{array}{l}\hfill \\ {\mathrm{log}}_{a}b=\frac{{\mathrm{log}}_{c}b}{{\mathrm{log}}_{c}a}\hfill \\ {\mathrm{log}}_{a}b=\frac{1}{{\mathrm{log}}_{b}a}\hfill \end{array}$

where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are positive, and $\text{\hspace{0.17em}}a>0,a\ne 1.$

Proof:

Let $\text{\hspace{0.17em}}x={\mathrm{log}}_{a}b.$

Write in exponent form.

${a}^{x}=b$

Take the $\text{\hspace{0.17em}}{\mathrm{log}}_{c}\text{\hspace{0.17em}}$ of both sides.

$\begin{array}{ccc}\hfill {\mathrm{log}}_{c}{a}^{x}& =& {\mathrm{log}}_{c}b\hfill \\ \hfill x{\mathrm{log}}_{c}a& =& {\mathrm{log}}_{c}b\hfill \\ \hfill x& =& \frac{{\mathrm{log}}_{c}b}{{\mathrm{log}}_{c}a}\hfill \\ \hfill {\mathrm{log}}_{a}b& =& \frac{{\mathrm{log}}_{c}b}{{\mathrm{log}}_{a}b}\hfill \end{array}$

When $\text{\hspace{0.17em}}c=b,$

${\mathrm{log}}_{a}b=\frac{{\mathrm{log}}_{b}b}{{\mathrm{log}}_{b}a}=\frac{1}{{\mathrm{log}}_{b}a}$

Heron’s Formula

$A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$

where $\text{\hspace{0.17em}}s=\frac{a+b+c}{2}$

Proof:

Let $\text{\hspace{0.17em}}a,$ $b,$ and $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ be the sides of a triangle, and $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ be the height.

So $\text{\hspace{0.17em}}s=\frac{a+b+c}{2}$ .

We can further name the parts of the base in each triangle established by the height such that $\text{\hspace{0.17em}}p+q=c.$

Using the Pythagorean Theorem, $\text{\hspace{0.17em}}{h}^{2}+{p}^{2}={a}^{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{h}^{2}+{q}^{2}={b}^{2}.$

Since $\text{\hspace{0.17em}}q=c-p,$ then $\text{\hspace{0.17em}}{q}^{2}={\left(c-p\right)}^{2}.\text{\hspace{0.17em}}$ Expanding, we find that $\text{\hspace{0.17em}}{q}^{2}={c}^{2}-2cp+{p}^{2}.$

We can then add $\text{\hspace{0.17em}}{h}^{2}\text{\hspace{0.17em}}$ to each side of the equation to get $\text{\hspace{0.17em}}{h}^{2}+{q}^{2}={h}^{2}+{c}^{2}-2cp+{p}^{2}.$

Substitute this result into the equation $\text{\hspace{0.17em}}{h}^{2}+{q}^{2}={b}^{2}\text{\hspace{0.17em}}$ yields $\text{\hspace{0.17em}}{b}^{2}={h}^{2}+{c}^{2}-2cp+{p}^{2}.$

Then replacing $\text{\hspace{0.17em}}{h}^{2}+{p}^{2}\text{\hspace{0.17em}}$ with $\text{\hspace{0.17em}}{a}^{2}\text{\hspace{0.17em}}$ gives $\text{\hspace{0.17em}}{b}^{2}={a}^{2}-2cp+{c}^{2}.$

Solve for $\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ to get

$p=\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2c}$

Since $\text{\hspace{0.17em}}{h}^{2}={a}^{2}-{p}^{2},$ we get an expression in terms of $\text{\hspace{0.17em}}a,$ $b,$ and $\text{\hspace{0.17em}}c.$

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

Therefore,

$\begin{array}{ccc}\hfill {h}^{2}& =& \frac{4s\left(s-a\right)\left(s-b\right)\left(s-c\right)}{{c}^{2}}\hfill \\ \hfill h& =& \frac{2\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}{c}\hfill \end{array}$

And since $\text{\hspace{0.17em}}A=\frac{1}{2}ch,$ then

$\begin{array}{ccc}\hfill A& =& \frac{1}{2}c\frac{2\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}{c}\hfill \\ & =& \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\hfill \end{array}$

Properties of the Dot Product

$u·v=v·u$

Proof:

$\begin{array}{cc}\hfill u·v& =⟨{u}_{1},{u}_{2},...{u}_{n}⟩·⟨{v}_{1},{v}_{2},...{v}_{n}⟩\hfill \\ & ={u}_{1}{v}_{1}+{u}_{2}{v}_{2}+...+{u}_{n}{v}_{n}\hfill \\ & ={v}_{1}{u}_{1}+{v}_{2}{u}_{2}+...+{v}_{n}{v}_{n}\hfill \\ & =⟨{v}_{1},{v}_{2},...{v}_{n}⟩·⟨{u}_{1},{u}_{2},...{u}_{n}⟩\hfill \\ & =v·u\hfill \end{array}$

$u·\left(v+w\right)=u·v+u·w$

Proof:

$\begin{array}{cc}\hfill u·\left(v+w\right)& =⟨{u}_{1},{u}_{2},...{u}_{n}⟩·\left(⟨{v}_{1},{v}_{2},...{v}_{n}⟩+⟨{w}_{1},{w}_{2},...{w}_{n}⟩\right)\hfill \\ & =⟨{u}_{1},{u}_{2},...{u}_{n}⟩·⟨{v}_{1}+{w}_{1},{v}_{2}+{w}_{2},...{v}_{n}+{w}_{n}⟩\hfill \\ & =⟨{u}_{1}\left({v}_{1}+{w}_{1}\right),{u}_{2}\left({v}_{2}+{w}_{2}\right),...{u}_{n}\left({v}_{n}+{w}_{n}\right)⟩\hfill \\ & =⟨{u}_{1}{v}_{1}+{u}_{1}{w}_{1},{u}_{2}{v}_{2}+{u}_{2}{w}_{2},...{u}_{n}{v}_{n}+{u}_{n}{w}_{n}⟩\hfill \\ & =⟨{u}_{1}{v}_{1},{u}_{2}{v}_{2},...,{u}_{n}{v}_{n}⟩+⟨{u}_{1}{w}_{1},{u}_{2}{w}_{2},...,{u}_{n}{w}_{n}⟩\hfill \\ & =⟨{u}_{1},{u}_{2},...{u}_{n}⟩·⟨{v}_{1},{v}_{2},...{v}_{n}⟩+⟨{u}_{1},{u}_{2},...{u}_{n}⟩·⟨{w}_{1},{w}_{2},...{w}_{n}⟩\hfill \\ & =u·v+u·w\hfill \end{array}$

$u·u={|u|}^{2}$

Proof:

$\begin{array}{cc}\hfill u·u& =⟨{u}_{1},{u}_{2},...{u}_{n}⟩·⟨{u}_{1},{u}_{2},...{u}_{n}⟩\hfill \\ & ={u}_{1}{u}_{1}+{u}_{2}{u}_{2}+...+{u}_{n}{u}_{n}\hfill \\ & ={u}_{1}{}^{2}+{u}_{2}{}^{2}+...+{u}_{n}{}^{2}\hfill \\ & =|⟨{u}_{1},{u}_{2},...{u}_{n}⟩{|}^{2}\hfill \\ & =v·u\hfill \end{array}$

Standard Form of the Ellipse centered at the Origin

$1=\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}$

Derivation

An ellipse consists of all the points for which the sum of distances from two foci is constant:

$\sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y-0\right)}^{2}}+\sqrt{{\left(x-c\right)}^{2}+{\left(y-0\right)}^{2}}=\text{constant}$

Consider a vertex.

Then, $\text{\hspace{0.17em}}\sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y-0\right)}^{2}}+\sqrt{{\left(x-c\right)}^{2}+{\left(y-0\right)}^{2}}=2a$

Consider a covertex.

Then $\text{\hspace{0.17em}}{b}^{2}+{c}^{2}={a}^{2}.$

$\begin{array}{ccc}\hfill \sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y-0\right)}^{2}}+\sqrt{{\left(x-c\right)}^{2}+{\left(y-0\right)}^{2}}& =& 2a\hfill \\ \hfill \sqrt{{\left(x+c\right)}^{2}+{y}^{2}}& =& 2a-\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill \\ \hfill {\left(x+c\right)}^{2}+{y}^{2}& =& {\left(2a-\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\right)}^{2}\hfill \\ \hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{\left(x-c\right)}^{2}+{y}^{2}\hfill \\ \hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{y}^{2}\hfill \\ \hfill 2cx& =& 4{a}^{2}-4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}-2cx\hfill \\ \hfill 4cx-4{a}^{2}& =& 4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill \\ \hfill -\frac{1}{4a}\left(4cx-4{a}^{2}\right)& =& \sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill \\ \hfill a-\frac{c}{a}x& =& \sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill \\ \hfill {a}^{2}-2xc+\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {\left(x-c\right)}^{2}+{y}^{2}\hfill \\ \hfill {a}^{2}-2xc+\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}-2xc+{c}^{2}+{y}^{2}\hfill \\ \hfill {a}^{2}+\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}+{c}^{2}+{y}^{2}\hfill \\ \hfill {a}^{2}+\frac{{c}^{2}}{{a}^{2}}{x}^{2}& =& {x}^{2}+{c}^{2}+{y}^{2}\hfill \\ \hfill {a}^{2}-{c}^{2}& =& {x}^{2}-\frac{{c}^{2}}{{a}^{2}}{x}^{2}+{y}^{2}\hfill \\ \hfill {a}^{2}-{c}^{2}& =& {x}^{2}\left(1-\frac{{c}^{2}}{{a}^{2}}\right)+{y}^{2}\hfill \end{array}$

Let $\text{\hspace{0.17em}}1=\frac{{a}^{2}}{{a}^{2}}.$

$\begin{array}{ccc}\hfill {a}^{2}-{c}^{2}& =& {x}^{2}\left(\frac{{a}^{2}-{c}^{2}}{{a}^{2}}\right)+{y}^{2}\hfill \\ \hfill 1& =& \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}-{c}^{2}}\hfill \end{array}$

Because $\text{\hspace{0.17em}}{b}^{2}+{c}^{2}={a}^{2},$ then $\text{\hspace{0.17em}}{b}^{2}={a}^{2}-{c}^{2}.$

$\begin{array}{ccc}\hfill 1& =& \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{a}^{2}-{c}^{2}}\hfill \\ \hfill 1& =& \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}\hfill \end{array}$

Standard Form of the Hyperbola

$1=\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}$

Derivation

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.

Diagram 1: The difference of the distances from Point P to the foci is constant:

$\sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y-0\right)}^{2}}-\sqrt{{\left(x-c\right)}^{2}+{\left(y-0\right)}^{2}}=\text{constant}$

Diagram 2: When the point is a vertex, the difference is $\text{\hspace{0.17em}}2a.$

$\sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y-0\right)}^{2}}-\sqrt{{\left(x-c\right)}^{2}+{\left(y-0\right)}^{2}}=2a$

$\begin{array}{ccc}\hfill \sqrt{{\left(x-\left(-c\right)\right)}^{2}+{\left(y-0\right)}^{2}}-\sqrt{{\left(x-c\right)}^{2}+{\left(y-0\right)}^{2}}& =& 2a\hfill \\ \hfill \sqrt{{\left(x+c\right)}^{2}+{y}^{2}}-\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}& =& 2a\hfill \\ \hfill \sqrt{{\left(x+c\right)}^{2}+{y}^{2}}& =& 2a+\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill \\ \hfill {\left(x+c\right)}^{2}+{y}^{2}& =& \left(2a+\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\right)\hfill \\ \hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}}+{y}^{2}\hfill \\ \hfill {x}^{2}+2cx+{c}^{2}+{y}^{2}& =& 4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}+{x}^{2}-2cx+{y}^{2}\hfill \\ \hfill 2cx& =& 4{a}^{2}+4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}-2cx\hfill \\ \hfill 4cx-4{a}^{2}& =& 4a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill \\ \hfill cx-{a}^{2}& =& a\sqrt{{\left(x-c\right)}^{2}+{y}^{2}}\hfill \\ \hfill {\left(cx-{a}^{2}\right)}^{2}& =& {a}^{2}\left({\left(x-c\right)}^{2}+{y}^{2}\right)\hfill \\ \hfill {c}^{2}{x}^{2}-2{a}^{2}{c}^{2}{x}^{2}+{a}^{4}& =& {a}^{2}{x}^{2}-2{a}^{2}{c}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\hfill \\ \hfill {c}^{2}{x}^{2}+{a}^{4}& =& {a}^{2}{x}^{2}+{a}^{2}{c}^{2}+{a}^{2}{y}^{2}\hfill \\ \hfill {a}^{4}-{a}^{2}{c}^{2}& =& {a}^{2}{x}^{2}-{c}^{2}{x}^{2}+{a}^{2}{y}^{2}\hfill \\ \hfill {a}^{2}\left({a}^{2}-{c}^{2}\right)& =& \left({a}^{2}-{c}^{2}\right){x}^{2}+{a}^{2}{y}^{2}\hfill \\ \hfill {a}^{2}\left({a}^{2}-{c}^{2}\right)& =& \left({c}^{2}-{a}^{2}\right){x}^{2}-{a}^{2}{y}^{2}\hfill \end{array}$

Define $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ as a positive number such that $\text{\hspace{0.17em}}{b}^{2}={c}^{2}-{a}^{2}.$

$\begin{array}{ccc}\hfill {a}^{2}{b}^{2}& =& {b}^{2}{x}^{2}-{a}^{2}{y}^{2}\hfill \\ \hfill \frac{{a}^{2}{b}^{2}}{{a}^{2}{b}^{2}}& =& \frac{{b}^{2}{x}^{2}}{{a}^{2}{b}^{2}}-\frac{{a}^{2}{y}^{2}}{{a}^{2}{b}^{2}}\hfill \\ \hfill 1& =& \frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}\hfill \end{array}$

## Trigonometric identities

 Pythagorean Identity $\begin{array}{l}{\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1\\ 1+{\mathrm{tan}}^{2}t={\mathrm{sec}}^{2}t\\ 1+{\mathrm{cot}}^{2}t={\mathrm{csc}}^{2}t\end{array}$ Even-Odd Identities $\begin{array}{l}\mathrm{cos}\left(-t\right)=cos\text{\hspace{0.17em}}t\hfill \\ \mathrm{sec}\left(-t\right)=\mathrm{sec}\text{\hspace{0.17em}}t\hfill \\ \mathrm{sin}\left(-t\right)=-\mathrm{sin}\text{\hspace{0.17em}}t\hfill \\ \mathrm{tan}\left(-t\right)=-\mathrm{tan}\text{\hspace{0.17em}}t\hfill \\ \mathrm{csc}\left(-t\right)=-\mathrm{csc}\text{\hspace{0.17em}}t\hfill \\ \mathrm{cot}\left(-t\right)=-\mathrm{cot}\text{\hspace{0.17em}}t\hfill \end{array}$ Cofunction Identities $\begin{array}{l}\hfill \\ \mathrm{cos}\text{\hspace{0.17em}}t=\mathrm{sin}\left(\frac{\pi }{2}-t\right)\hfill \\ \mathrm{sin}\text{\hspace{0.17em}}t=\mathrm{cos}\left(\frac{\pi }{2}-t\right)\hfill \\ \mathrm{tan}\text{\hspace{0.17em}}t=\mathrm{cot}\left(\frac{\pi }{2}-t\right)\hfill \\ \mathrm{cot}\text{\hspace{0.17em}}t=\mathrm{tan}\left(\frac{\pi }{2}-t\right)\hfill \\ \mathrm{sec}\text{\hspace{0.17em}}t=\mathrm{csc}\left(\frac{\pi }{2}-t\right)\hfill \\ \mathrm{csc}\text{\hspace{0.17em}}t=\mathrm{sec}\left(\frac{\pi }{2}-t\right)\hfill \end{array}$ Fundamental Identities $\begin{array}{l}\mathrm{tan}\text{\hspace{0.17em}}t=\frac{\mathrm{sin}\text{\hspace{0.17em}}t}{\mathrm{cos}\text{\hspace{0.17em}}t}\hfill \\ \mathrm{sec}\text{\hspace{0.17em}}t=\frac{1}{\mathrm{cos}\text{\hspace{0.17em}}t}\hfill \\ \mathrm{csc}\text{\hspace{0.17em}}t=\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}t}\hfill \\ cot\text{\hspace{0.17em}}t=\frac{1}{\text{tan}\text{\hspace{0.17em}}t}=\frac{\text{cos}\text{\hspace{0.17em}}t}{\text{sin}\text{\hspace{0.17em}}t}\hfill \end{array}$ Sum and Difference Identities $\begin{array}{l}\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ \mathrm{cos}\left(\alpha -\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ \mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ \mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta \hfill \\ \mathrm{tan}\left(\alpha +\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\hfill \\ \mathrm{tan}\left(\alpha -\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha -\mathrm{tan}\text{\hspace{0.17em}}\beta }{1+\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\hfill \end{array}$ Double-Angle Formulas $\begin{array}{l}\mathrm{sin}\left(2\theta \right)=2\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \\ \mathrm{cos}\left(2\theta \right)={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \hfill \\ \mathrm{cos}\left(2\theta \right)=1-2{\mathrm{sin}}^{2}\theta \hfill \\ \mathrm{cos}\left(2\theta \right)=2{\mathrm{cos}}^{2}\theta -1\hfill \\ \mathrm{tan}\left(2\theta \right)=\frac{2\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }\hfill \end{array}$ Half-Angle Formulas $\begin{array}{l}\hfill \\ \mathrm{sin}\frac{\alpha }{2}=±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ \mathrm{cos}\frac{\alpha }{2}=±\sqrt{\frac{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }{2}}\hfill \\ \mathrm{tan}\frac{\alpha }{2}=±\sqrt{\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }}\hfill \\ \mathrm{tan}\frac{\alpha }{2}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{1+\mathrm{cos}\text{\hspace{0.17em}}\alpha }\hfill \\ \mathrm{tan}\frac{\alpha }{2}=\frac{1-\mathrm{cos}\text{\hspace{0.17em}}\alpha }{\mathrm{sin}\text{\hspace{0.17em}}\alpha }\hfill \end{array}$ Reduction Formulas $\begin{array}{l}\hfill \\ {\mathrm{sin}}^{2}\theta =\frac{1-\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ {\mathrm{cos}}^{2}\theta =\frac{1+\mathrm{cos}\left(2\theta \right)}{2}\hfill \\ {\mathrm{tan}}^{2}\theta =\frac{1-\mathrm{cos}\left(2\theta \right)}{1+\mathrm{cos}\left(2\theta \right)}\hfill \end{array}$ Product-to-Sum Formulas $\begin{array}{l}\hfill \\ \mathrm{cos}\alpha \mathrm{cos}\beta =\frac{1}{2}\left[\mathrm{cos}\left(\alpha -\beta \right)+\mathrm{cos}\left(\alpha +\beta \right)\right]\hfill \\ \mathrm{sin}\alpha \mathrm{cos}\beta =\frac{1}{2}\left[\mathrm{sin}\left(\alpha +\beta \right)+\mathrm{sin}\left(\alpha -\beta \right)\right]\hfill \\ \mathrm{sin}\alpha \mathrm{sin}\beta =\frac{1}{2}\left[\mathrm{cos}\left(\alpha -\beta \right)-\mathrm{cos}\left(\alpha +\beta \right)\right]\hfill \\ \mathrm{cos}\alpha \mathrm{sin}\beta =\frac{1}{2}\left[\mathrm{sin}\left(\alpha +\beta \right)-\mathrm{sin}\left(\alpha -\beta \right)\right]\hfill \end{array}$ Sum-to-Product Formulas $\begin{array}{l}\hfill \\ \mathrm{sin}\alpha +\mathrm{sin}\beta =2\mathrm{sin}\left(\frac{\alpha +\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta }{2}\right)\hfill \\ \mathrm{sin}\alpha -\mathrm{sin}\beta =2\mathrm{sin}\left(\frac{\alpha -\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha +\beta }{2}\right)\hfill \\ \mathrm{cos}\alpha -\mathrm{cos}\beta =-2\mathrm{sin}\left(\frac{\alpha +\beta }{2}\right)\mathrm{sin}\left(\frac{\alpha -\beta }{2}\right)\hfill \\ \mathrm{cos}\alpha +\mathrm{cos}\beta =2\mathrm{cos}\left(\frac{\alpha +\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta }{2}\right)\hfill \end{array}$ Law of Sines $\begin{array}{l}\frac{\mathrm{sin}\text{\hspace{0.17em}}\alpha }{a}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\beta }{b}=\frac{\mathrm{sin}\text{\hspace{0.17em}}\gamma }{c}\hfill \\ \frac{a}{\mathrm{sin}\text{\hspace{0.17em}}\alpha }=\frac{b}{\mathrm{sin}\text{\hspace{0.17em}}\beta }=\frac{c}{\mathrm{sin}\text{\hspace{0.17em}}\gamma }\hfill \end{array}$ Law of Cosines $\begin{array}{c}{a}^{2}={b}^{2}+{c}^{2}-2bc\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha \\ {b}^{2}={a}^{2}+{c}^{2}-2ac\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta \\ {c}^{2}={a}^{2}+{b}^{2}-2ab\text{\hspace{0.17em}}\text{cos}\text{\hspace{0.17em}}\gamma \end{array}$

## Trigonometric functions

Unit Circle

Angle $0$
Cosine 1 $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{1}{2}$ 0
Sine 0 $\frac{1}{2}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{3}}{2}$ 1
Tangent 0 $\frac{\sqrt{3}}{3}$ 1 $\sqrt{3}$ Undefined
Secant 1 $\frac{2\sqrt{3}}{3}$ $\sqrt{2}$ 2 Undefined
Cosecant Undefined 2 $\sqrt{2}$ $\frac{2\sqrt{3}}{3}$ 1
Cotangent Undefined $\sqrt{3}$ 1 $\frac{\sqrt{3}}{3}$ 0

x exposant 4 + 4 x exposant 3 + 8 exposant 2 + 4 x + 1 = 0
x exposent4+4x exposent3+8x exposent2+4x+1=0
HERVE
How can I solve for a domain and a codomains in a given function?
ranges
EDWIN
Thank you I mean range sir.
Oliver
proof for set theory
don't you know?
Inkoom
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5)  and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has