# 0.1 Proofs, identities, and toolkit functions

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## 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

root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was  1350  bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after  3  hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
how do I attempted a trig number as a starter
cos 18 ____ sin 72 evaluate