A | Proofs, Identities, and Toolkit Functions

Important Proofs and Derivations

Product Rule

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

Proof:

Let $m={\log }_{a}x$ and $n={\log }_{a}y.$

Write in exponent form.

$x=a^m$ and $y=a^n.$

Multiply.

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

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

Change of Base Rule

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

where $x$ and $y$ are positive, and $a>0,a\ne 1.$

Proof:

Let $x={\log }_{a}b.$

Write in exponent form.

$a^x=b$

Take the ${\log }_c$ of both sides.

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

When $c=b,$

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

Heron’s Formula

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

where $s=\frac{a+b+c}{2}$

Proof:

Let $a,$ $b,$ and $c$ be the sides of a triangle, and $h$ be the height.

So $s=\frac{a+b+c}{2}$.

We can further name the parts of the base in each triangle established by the height such that $p+q=c.$

Using the Pythagorean Theorem, $h^2+p^2=a^2$ and $h^2+q^2=b^2.$

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

We can then add $h^2$ to each side of the equation to get $h^2+q^2=h^2+c^2-2cp+p^2.$

Substitute this result into the equation $h^2+q^2=b^2$ yields $b^2=h^2+c^2-2cp+p^2.$

Then replacing $h^2+p^2$ with $a^2$ gives $b^2=a^2-2cp+c^2.$

Solve for $p$ to get

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

Since $h^2=a^2-p^2,$ we get an expression in terms of $a,$ $b,$ and $c.$

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

Therefore,

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

And since $A=\frac{1}{2}ch,$ then

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

Properties of the Dot Product

$u·v=v·u$

Proof:

$\begin{array}{cc}\hfill u·v& =\langle u_1,u_2,\mathrm{...}{u}_n\rangle ·\langle v_1,v_2,\mathrm{...}{v}_n\rangle \hfill \\ & =u_1{v}_1+u_2{v}_2+\mathrm{...}+u_n{v}_n\hfill \\ & =v_1{u}_1+v_2{u}_2+\mathrm{...}+v_n{v}_n\hfill \\ & =\langle v_1,v_2,\mathrm{...}{v}_n\rangle ·\langle u_1,u_2,\mathrm{...}{u}_n\rangle \hfill \\ & =v·u\hfill \end{array}$

$u·\left(v+w\right)=u·v+u·w$

Proof:

$\begin{array}{cc}\hfill u·(v+w)& =\langle u_1,u_2,\mathrm{...}{u}_n\rangle ·\left(\langle v_1,v_2,\mathrm{...}{v}_n\rangle +\langle w_1,w_2,\mathrm{...}{w}_n\rangle \right)\hfill \\ & =\langle u_1,u_2,\mathrm{...}{u}_n\rangle ·\langle v_1+w_1,v_2+w_2,\mathrm{...}{v}_n+w_n\rangle \hfill \\ & =\langle u_1(v_1+w_1),u_2(v_2+w_2),\mathrm{...}{u}_n(v_n+w_n)\rangle \hfill \\ & =\langle u_1{v}_1+u_1{w}_1,u_2{v}_2+u_2{w}_2,\mathrm{...}{u}_n{v}_n+u_n{w}_n\rangle \hfill \\ & =\langle u_1{v}_1,u_2{v}_2,\mathrm{...},u_n{v}_n\rangle +\langle u_1{w}_1,u_2{w}_2,\mathrm{...},u_n{w}_n\rangle \hfill \\ & =\langle u_1,u_2,\mathrm{...}{u}_n\rangle ·\langle v_1,v_2,\mathrm{...}{v}_n\rangle +\langle u_1,u_2,\mathrm{...}{u}_n\rangle ·\langle w_1,w_2,\mathrm{...}{w}_n\rangle \hfill \\ & =u·v+u·w\hfill \end{array}$

$u·u={\left|u\right|}^2$

Proof:

$\begin{array}{cc}\hfill u·u& =\langle u_1,u_2,\mathrm{...}{u}_n\rangle ·\langle u_1,u_2,\mathrm{...}{u}_n\rangle \hfill \\ & =u_1{u}_1+u_2{u}_2+\mathrm{...}+u_n{u}_n\hfill \\ & =u_1{}^2+u_2{}^2+\mathrm{...}+u_n{}^2\hfill \\ & =|\langle u_1,u_2,\mathrm{...}{u}_n\rangle {|}^2\hfill \\ & =v·u\hfill \end{array}$

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

Consider a covertex.

Then $b^2+c^2=a^2.$

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

Let $1=\frac{a^2}{a^2}.$

Because $b^2+c^2=a^2,$ then $b^2=a^2-c^2.$

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:

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

Diagram 2: When the point is a vertex, the difference is $2a.$

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

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

Define $b$ as a positive number such that $b^2=c^2-a^2.$

Trigonometric Identities