Math Identities Proofs


"Usually mathematical proofs lead to some form of rehab."
– Professor James Gilbert

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Please Excuse My Dear Aunt Sally

Parentheses, Exponents, Multiplication, Division, Addition & Subtraction
If two quantities are equal with a third, then they are equal with each other. $$ \text{If } a=c \text{ and } b=c \text{, then } a=b. $$
Additive Multiplicative
Commutativity $a+b=b+a$ $a·b=b·a$
Associativity $(a+b)+c=a+(b+c)$ $(a·b)·c=a·(b·c)$
Identity $a+0=a=0+a$ $a·1=a=1·a$
Inverse $a+(-a)=0$ $a·a^{-1}=1$
Distributive property $a·(b+c)=a·b+a·c$
Distribution of one negative $\frac{-1}{\phantom{-}1}=\frac{\phantom{-}1}{-1}=-\frac{1}{1}=-1$
Distribution of two negatives $\frac{-1}{-1}=-\big(-\frac{1}{1}\big)=1$
For expressions $a=\frac{1}{1/a}$
For evaluating fractions $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}·\frac{d}{c}$
If $±$ is used on one side of an equation, then the equation has two solutions. $$x=a±b→ \begin{cases} x=a+b \\ x=a-b \end{cases}$$ If $±$ is used on two sides of an equation, then when it is one sign on one side, it is the same on the other. $$±x=a±b→ \begin{cases} +x=a+b \\ -x=a-b \end{cases}$$ If $±$ and $∓$ are used, then when one is positive, the other is negative. $$ ±x=a∓b→ \begin{cases} +x=a-b \\ -x=a+b \end{cases}$$ If $±$ is used in an exponent, it is used to indicate multiplication or division and the same rules above apply. $$x·y^{±1}=a±b→ \begin{cases} x·y=a+b \\ x/y=a-b \end{cases}$$
Symbol Meaning Example Translation
$||$ absolute value $|x|$ The positive value or magnitude of $x$
$!$ factorial $5!=1·2·3·4·5$ The product of all integers to the specified value
$\therefore$ therefore $x^2=4 \therefore x= \pm 2$ One is true, therefore the other is true.
$\bmod$ modulus $4 \bmod 3 = 1$ 4/3 has a remainder of 1.
$\forall$ for all $\forall x≥0$ for every non-negative value
$\isin$ element of $\forall x \isin \Z$ $x$ is an integer.
$\land$ and $x>0 \land y>0$ Both of these statements are true.
$\lor$ or $x>0 \lor y>0$ One or both of these statements are true.
Real $$\R$$ All common numbers and multiples of one
Integer $$\Z$$ Real whole numbers that can be expressed without using a fraction $$\lbrace …,-2,-1,0,1,2,… \rbrace$$
Natural $$\N$$ Positive real integers (argued whether or not zero is included)
Rational Any number that can be represented as a fraction inclusive of only natural numbers, and always either repeats or terminates as a decimal $$ \begin{array}{cc} 1/2 & 2/1 & 1.2 & 2.\overline{1} \end{array} $$
Irrational Any number that cannot be represented as a fraction, and never repeats or terminates as a decimal $$ \begin{array}{cc} \sqrt{2} & e & \pi & \phi \end{array} $$
Imaginary All common numbers that are multiples of $i$
Complex $$\Complex$$ Numbers comprised of both real and imaginary numbers
Series of Nine
Repeating values can always be represented as a fraction with the repeated value over an equal number of digits of 9’s.

$$.\overline{2}=\frac{2}{9}$$ $$.\overline{137}=\frac{137}{999}$$ $$.\overline{142857}=\frac{142857}{999999}=\frac{1}{7}$$

Orders of Ten
Where a number starts repeating is a matter of orders of 10, which can be isolated.

$$433.\overline{3}=400+\frac{3}{9}·10^2=\frac{1300}{3}$$ $$.35\overline{7}=\frac{35}{100}+\frac{7}{9}·10^{-2}=\frac{161}{450}$$ $$3.8\overline{3}=\frac{1}{2}+\frac{30}{9}=\frac{23}{6}$$

Proof Nine Repeating Equals One
Let $.\bar{9}$ equal a variable $$x=.\overline{9}$$ Multiply by $10$ $$10·x=9.\overline{9}$$ Separate $9.\bar{9}$ into $9$ and $.\bar{9}$ $$10·x=9+.\overline{9}$$ Substitute $.\bar{9}$ for $x$ $$10·x=9+x$$ Subtract $x$ $$9·x=9$$ Divide by $9$
A full rotation in one angular direction is 360° or two radians. This page will always use radians. $$180°=\pi$$
Undefined forms with no number value, therefore cannot be operated on algebraically.

$$\frac{0}{0}$$ $$\frac{∞}{∞}$$ $$0·∞$$ $$∞-∞$$ $$0^0$$ $$1^∞$$ $$∞^0$$
$17+5=22$ sum
$17-5=12$ difference
$17·5=22$ product
$\frac{17}{5}=3+\frac{2}{5}$ $17$: dividend
$5$: divisor
$17/5$: simplified fraction
$3$: quotient
$2$: remainder
$3+2/5$: proper fraction
$\frac{x}{x_∘}+\frac{y}{y_∘}=1$ $x$: abscissa
$x_∘$: horizontal axis intercept
$+$: operator
$y$: ordinate
$y$: vertical axis intercept
$1$: value
$y=a·x^2+b·x+c$ $y$: dependent variable
$a$: leading coefficient
$x$: independent variable
$2$: order
$b$: coefficient
$c$: constant
$(a+b·i)(a-b·i)=a^2+b^2$ ( expression = expression ) ← equation
factored form = expanded form
$a$: real component
$b·i$: imaginary component
(a±b∙i): roots
$\sqrt[n]{x}$ $n$: $n$th root
$x$: radicand
$x/x_∘$
$y/y_∘$
$a·x^2$
$b·x$
$(a±b·i)$
$a^2$
$b^2$
$\sqrt[n]{x}$
terms
General in mathematics means that all possible situations are represented or that something applies to multiple scenarios. A general equation reflects this and usually has all the terms on one side for easy manipulation.
Example 1
Using the quadratic formula to solve for:
Zeros of $x$ only (specific) $$x=\frac{-b \pm \sqrt{b^2-4·a·y_∘}}{2·a}$$ All values of $x$ (general) $$x=\frac{-b \pm \sqrt{b^2+4·a·(y-y_∘)}}{2·a}$$

Example 2
The conic sections general formula describes ellipses, circles, parabolas, hyperbolas, and lines. $$A·x^2+B·x·y+C·y^2+D·x+E·y+F=0$$
Even - symmetric about $y$-axis
$$f(-x)=f(x)$$
Odd - symmetric about origin
$$f(-x)=-f(x)$$
Inverse - reflected about $y=x$
$$f(g(x))=g(f(x))$$
Absolute value - magnitude; non-negative
$$ |x| = \begin{cases} a &x ≥ 0 \\ c &x \lt 0 \end{cases} $$
Periodic - repeated segment ($P$)
$$f(x+P)=f(x)$$
All images in this section were taken from Wolfram Alpha and edited
$$a·f(b·x+c)+d$$
  • If $|a| \gt 1$, $f$ stretches vertically.
  • If $|a| \lt 1$, $f$ compresses vertically.
  • If $a \lt 0$, $f$ flips vertically.
  • If $|b| \gt 1$, $f$ compresses horizontally.
  • If $|b| \lt 1$, $f$ stretches horizontally.
  • If $b \lt 0$, $f$ flips vertically.
  • If $c \gt 0$, $f$ shifts to the left.
  • If $c \lt 0$, $f$ shifts to the right.
  • If $d \gt 0$, $f$ shifts upwards.
  • If $d \lt 0$, $f$ shifts downwards.
Exponentiation is repeated multiplication. An exponent is the number of times to "times" a number.
  • Exponents do not commute: $x^a≠a^x$
  • Exponents do not associate: $x^{(a^b)}≠{(x^a)}^b$
  • Factoring is the condensing of terms: $x^3$
  • The reverse of factoring is expansion: $x·x·x$
$$x^{-a}=\frac{1}{x^a}$$
Deductive Logic
Multiplication is reversible with division, therefore exponential reduction yields inverses once reaching negatives.
$$x^3=1·x·x·x$$ $$3^3=27$$
$$x^2=1·x·x$$ $$3^2=9$$
$$x=1·x$$ $$3^1=3$$
$$x^0=1$$ $$3^0=1$$
$$x^{-1}=1/x$$ $$3^{-1}=1/3$$
$$x^{-2}=1/(x·x)$$ $$3^{-2}=1/9$$
$$x^{-3}=1/(x·x·x)$$ $$3^{-3}=1/27$$
$$x^a·x^{\pm b}=x^{a \pm b}$$
Deductive Logic
With the same base, $x^a$ is $x$ multiplied $a$ times, $x^b$ is $x$ multiplied $b$ times, and they are multiplied to each other, so then the exponents are added $$2^3·2^{-5}=2^{3-5}=2^{-2}=1/4$$
Evaluate starting from the highest exponent first, working downward, minding parenthesized terms along the way. $$x^{a^{b^{c^d}}}=x^{\Big(a^{\big(b^{(c^d)}\big)}\Big)}$$
$${(x^a·y^{\pm b})}^c=x^{a·c}·y^{\pm b·c}$$
Deductive Logic
Distribute the exponential $c$ by expanding into ${(x^a)}^c·{(y^{±b})}^c$. The terms are multiplied $c$ times

Example
Given $3·x–5·y=2$, evaluate $8^x/32^y$ $$\frac{8^x}{32^y}=\frac{{(2^3)}^x}{{(2^5)}^y}=\frac{2^{3·x}}{2^{5·y}}$$ Apply the power rule, then substitute $3·x–5·y=2$ $$\frac{2^{3·x}}{2^{5·y}}=2^{3·x-5·y}=2^2$$
Denominators denote a root, such as a square root if it is 2 $$x^{p/q}={(\sqrt[q]{x})}^p=\sqrt[q]{x^p}, \forall x≥0$$
$$\sqrt[n]{x·y^{\pm 1}}=\sqrt[n]{x}·\sqrt[n]{y^{\pm 1}}, x>0 \lor y>0$$
$$\sqrt[n]{x^n}=|x|, \{ 2·n | n \isin \Z \}$$
Scientific notation displays all numbers as multiples of ten to a power in accordance with the leading digit $$299792458=2.99792458·10^8$$ $$0.0820574=8.20574·10^{-2}$$
Square roots of negative numbers fail to yield real results. The imaginary unit satisfies the equation $i^2=–1$ $$i^2=-1 \therefore i=\sqrt{-1}$$
Powers of $i$ always reduce using a modulus of $4$; $$i^x=i^{(x \bmod 4)}$$ Even powers simplify into real numbers.
$$i^{-2}=i^2=i^6=-1$$ $$i^{-4}=i^0=i^4=1$$
Odd powers simplify into imaginary numbers.
$$i^{-3}=i^1=i^5=i$$ $$i^{-1}=i^3=i^7=-i$$
It follows that imaginary numbers with real integer exponents are a periodic series of $i, –1, –i, 1$.
Complex conjugation functions the same as real conjugation with the sign reversal on the imaginary part.
$$z=x+y·i$$ $$z^*=x-y·i$$
$$|z|^2=z·z^*=x^2+y^2$$
Division uses multiplication with the complex conjugate of the divisor in both the divisor and dividend. $$\frac{a+b·i}{c+d·i}=\frac{a+b·i}{c+d·i}·\frac{c-d·i}{c-d·i}=\frac{a·c+b·d+(b·c-a·d)·i}{c^2+d^2}$$
In a factored function, like terms in both a numerator and denominator as multipliers reduce to 1 $$\frac{a(x)·f(x)}{a(x)·g(x)}=\frac{f(x)}{g(x)}$$
Examples
$$\frac{(x-4)·(x+2)}{(x-4)·(x-2)}=\frac{(x+2)}{(x-2)}$$ $$\frac{(x-4)·(x+2)}{(4-x)·(x-2)}=-\frac{(x+2)}{(x-2)}$$
Second degree trinomial equations in the form $a·x^2+b·x+c=0$. The following is a more useful general form. $$y=a·x^2+b·x+y_∘$$
$$a·c·x^2 \pm (a·d+b·c)·x+b·d=(a·x \pm b)·(c·x \pm d)$$
The leading coefficient must be 1, otherwise the equation must be multiplied for it to be 1. Using the general form $$x^2 \pm b·x+y_∘=y$$ Add ${(b/2)}^2$ and subtract $y_∘$ $$x^2 \pm b·x+ {\Big(\frac{b}{2} \Big)}^2=y-y_∘+{\Big(\frac{b}{2} \Big)}^2$$ Factor $$\frac{{(2·x \pm b)}^2}{4}=y-y_∘+\frac{b^2}{4}$$
Jump to Circles
The general solutions of $y=a·x^2+b·x+y_∘$ are $$x=\frac{-b \pm \sqrt{b^2+4·a·(y-y_∘)}}{2·a}, a≠0$$
Jump to Parabolas
Proof
Given the general formula, subtract $y_∘$ from both sides $$y-y_∘=a·x^2+b·x$$ Multiply by $4·a$ $$4·a·(y-y_∘)=4·a^2·x^2+4·a·b·x$$ Add $b^2$, then factor the right, similar to completing the square $$b^2+4·a·(y-y_∘)={(2·a·x+b)}^2$$ Take the square root $$\pm \sqrt{b^2+4·a·(y-y_∘)}=2·a·x+b$$ Subtract $b$, and divide by $2·a$
$$Δ=b^2-4·a·y_∘$$
  • If $Δ>0$, the equation has two real factorable roots
  • If $Δ=0$, the equation has one factorable root squared
  • If $Δ \lt 0$, the equation has two complex factorable roots
Sum of Powers for Real Roots (Odd Exponents Only)
$$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}·b+a^{n-3}·b^2-...+a^2·b^{n-3}-a·b^{n-2}+b^{n-1}), \forall 2·n-1 \isin \N$$

Difference of Powers for Real Roots
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}·b+a^{n-3}·b^2+...+a^2·b^{n-3}+a·b^{n-2}+b^{n-1}), \forall 2·n \isin \N$$

Sum & Difference of Squares & Cubes
$$a^2+b^2=(a+i·b)(a-i·b)$$ $$a^2-b^2=(a+b)(a-b)$$ $$a^3 \pm b^3=(a \pm b)(a^2 \mp a·b+b^2)$$

Other Sums & Differences of Interest
$$a^4-b^4=(a+i·b)(a-i·b)(a+b)(a-b)$$ $$a^6+b^6=(a^2+b^2)(a^4-a^2·b^2+b^4)$$ $$a^{10}+b^{10}=(a^2+b^2)(a^8-a^6·b^2+a^4·b^4-a^2·b^6+b^8)$$
Each number in the given 'triangle' below the first is the sum of the numbers diagonally above it

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
Jump to Binomials
Binomial Coefficient
$$ \binom{n}{k}=\frac{n!}{(n-k)!·k!}=\frac{n·(n-1)(n-2)...(n-k+1)}{k!}$$

Binomial Theorem
The coefficients in the expanded form match the line $n+1$ down from the top on Pascal’s triangle $${(x+y)}^n=\binom{n}{0}·x^n+\binom{n}{1}·x^{n-1}·y+\binom{n}{2}·x^{n-2}·y^2+...+\binom{n}{n}·y^n, \forall n \isin \Z$$

Binomials Squared & Cubed
$${(a \pm b)}^2=a^2 \pm 2·a·b+b^2$$
$${(a \pm b)}^3=a^3 \pm 3·a^2·b+3·a·b^2 \pm b^3$$
In polynomial equations with matching factors, orders, and results, variable coefficients can be equated for solving $$a·x+b·y=13 \land 4·x+3·y=13$$ $$a=4 \qquad b=3$$
$$a·x+b·y=m \land (n-1)·x+\frac{p}{2}·y=m$$ $$a=(n-1) \qquad b=p/2$$
Jump to Circles
Jump to Ellipses
Jump to Parabolas
Given the rational function $N(x)/D(x)$, the division results in the equality $N(x)=D(x)·Q(x)+R(x)$, or $$\begin{array}{r} Q(x)+R(x)\\ D(x){\overline{\smash{\big)}\,N(x)}}\phantom{+R(x)}\\ \end{array}$$
Solving by Example
Given $(x^3+2·x^2+12)/(x–2)$, express in the following form with the missing (zero coefficient) terms included $$\begin{array}{r} x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \end{array}$$ Multiply the divisor with the coefficient and variable order necessary to cancel the first term with subtraction, meaning the sign must match. In this case $x^2$ $$\begin{array}{r} x^2\phantom{+0·x+12}\\ x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \phantom{0}-x^3+2·x^2\phantom{+0·x+12}\\ {\overline{\phantom{-x^3+}4·x^2+0·x+12}}\\ \end{array}$$ Repeat the process $$\begin{array}{r} x^2+4·x\phantom{+12}\\ x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \phantom{0}-x^3+2·x^2\phantom{+0·x+12}\\ \overline{\phantom{-x^3+}4·x^2+0·x+12}\\ \phantom{0}-4·x^2+8·x\phantom{+12}\\ \overline{\phantom{-4·x^2+}8·x+12}\\ \end{array}$$ Keep repeating until the first term no longer applies $$\begin{array}{r} Q=x^2+4·x+\phantom{0}8\\ x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \phantom{0}-x^3+2·x^2\phantom{+0·x+12}\\ \overline{\phantom{-x^3+}4·x^2+0·x+12}\\ \phantom{0}-4·x^2+8·x\phantom{+12}\\ {\overline{\phantom{-4·x^2+}8·x+12}}\\ -8·x+16\\ R=\overline{\phantom{-8·x}+28}\\ \end{array}$$ Substitute the values for $N(x)=D(x)·Q(x)+R(x)$ $$x^3+2·x^2+12=(x-2)(x^2+4·x+8)+28$$
Partial Fraction Decomposition
Given the rational function $N(x)/D(x)$, expand by the smallest possible roots of the denominator

1) Linear roots are treated as follows: $$\frac{N(x)}{(x+a)(x+b)}=\frac{A}{x+a}+\frac{B}{x+b}$$ 2) Quadratic roots are treated as follows $$\frac{N(x)}{a·x^2+b·x+c}=\frac{A·x+B}{a·x^2+b·x+c}$$ 3) Repeated roots are treated as follows $$\frac{N(x)}{{D(x)}^P}=\frac{A}{D(x)}+\frac{B}{{D(x)}^2}+...+\frac{C}{{D(x)}^P}$$ Example $$\frac{5·x}{(x-1)(x^2+2){(x+7)}^2}=\frac{A}{x-1}+\frac{B·x+C}{x^2+2}+\frac{D}{x+7}+\frac{E}{{(x+7)}^2}$$

Solving by Example
$$\frac{12·x}{x^2-10·x+16}$$ Factor $$\frac{12·x}{(x-8)(x-2)}$$ Decompose $$\frac{12·x}{(x-8)(x-2)}=\frac{A}{x-8}+\frac{B}{x-2}$$ Multiply by the factored denominator $$12·x=(x-2)·A+(x-8)·B$$ Enter values for $x$ to cancel terms $$x=2 \therefore 12·2=0·A-6·B \therefore B=-4$$ $$x=8 \therefore 12·8=6·A+0·B \therefore A=16$$ Substitute $A$ and $B$ to solve $$\frac{12·x}{(x-8)(x-2)}=\frac{16}{x-8}-\frac{4}{x-2}$$
Logarithms are inverse functions of exponents $$b^p=x \therefore \log_b x=p$$
Example
$$2^3=8 \therefore \log_2 8=3$$
$$\log_b b^p=p$$
Proof
$$\log_b x = p \land x=b^p \therefore \log_b b^p=p$$

Example
$$\log_2 8=3 \land 8=2^3 \therefore log_2 2^3=3$$
Given an equality, an equation may be raised as exponential powers with the same base value $$a=b↔x^a=x^b$$
Proof
Given $x^a=x^b$, take the logarithm of base $x$ $$\log_x x^a=\log_x x^b$$ Apply the base inverse property on each side
$$ b^{\log_b x^p}=x^p, \forall x \isin \R$$
Deductive Logic
Logarithms are the inverse functions of exponents, therefore operating a logarithm in an exponent with the same base cancels both

Example
$$\log_2 2^3=2^{log_2 3}=3$$
$$\log_b x^p=p·\log_b x$$
Proof
Let $a=\log_bx$ so that $x=b^a$, then exponentiate to $p$ $$x^p={(b^a)}^p$$ Distribute the power $$x^p=b^{a·p}$$ Take the logarithm using base $b$ $$\log_b x^p=\log_b b^{a·p}$$ Apply the base inverse property $$\log_b x^p=a·p$$ Substitute $a$ for its original form

Example 1
$$\log_2 64=\log_2 2^6=6·\log_2 2=6$$

Example 2
$$\log_2 x^{-1}=-\log_2 x$$
$$\log_b (M·N^{\pm 1})=\log_b M \pm \log_b N$$
Proof
Let $$ x=\log_b M \land ±y=\log_b N^{±1}$$ So that $$M=b^x \land N^{±1}=b^{±y}$$ Multiply the two equalities $$M·N^{±1}=b^x·b^{±y}$$ Apply the exponent power rule $$M·N^{±1}=b^{x±y}$$ Take the logarithm of base $b$ $$\log_b (M·N^{±1})=\log_b b^{x±y}$$ Apply the base inverse property $$\log_b (M·N^{±1})=x±y$$ Substitute $x$ and $±y$ for their original terms
$$\log_b (M \pm N)=\log_b M+\log_b \Big(1 \pm \frac{N}{M} \Big)$$
Proof
Given $log_b(M±N)$, multiply $N$ by $M/M$ $$\log_b \Big(M \pm \frac{M}{M}·N \Big)$$ Factor $$\log_b \Big(M· \Big( 1 \pm \frac{N}{M} \Big) \Big)$$ Apply the product rule
$$\log_b x = \frac{\log_a x}{\log_a b}, x>0 \land a>0 \land a≠1$$
Properties from the Change of Base Rule
$$\log_b x = \frac{1}{log_x b}$$ $$\log_{c^n} x=\frac{\log_c x}{n}$$ $$a^{\log_b x}=x^{\log_b a}$$

Proof
Let $c=\log_bx$ so that $b^c=x$, then take the logarithm of base $a$ $$\log_a b^c=\log_a x$$ Apply the power rule $$c·\log_a b=\log_a x$$ Solve for $c$ and sutstitute it for its original form

Proof of Properties
1) Using the change of base rule, substitute $a$ for $x$

2) Using the change of base rule, substitute $c^n$ for $b$ and $c$ for $a$

3) Take the logarithm $b$ of the given formula and apply the power rule
$$\log_{1/b} x=-\log_b x=log_b x^{-1}$$
Proof
Let $a=log_{1/b} x$ so that $x=b^{–a}$, then take the logarithm of base $b$ $$\log_b x= \log_b b^{-a}$$ Apply the base inverse property, then negate $$-\log_b x=a$$ The equality $–log_bx=log_bx^{–1}$ can be proven simply with both the power and quotient rules
Slope of a line $m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$
Midpoint formula $\big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\big)$
Horizontal line $y=c, \medspace m=0$
Vertical line $x=c, \medspace m=±∞$
General equation $A·x+B·y=C, \medspace x_∘=\frac{C}{A} \land y_∘=\frac{C}{B}$
Slope-intercept equation $y(x)=m·x+y_∘$
Point-slope equation $y-y_1=m·(x-x_1)$
Two point equation $y-y_1=\frac{y_2-y_1}{x_2-x_1}·(x-x_1)$
Two-intercept equation $\frac{x}{x_∘}+\frac{y}{y_∘}=1, \medspace x_∘≠0 \land y_∘≠0$

Proof of Two-Intercept Equation
Use the coordinates for the intercepts in the slope of a line equation, which are interchangeable $$m=\frac{0-y_∘}{x_∘-0}=\frac{y_∘-0}{0-x_∘}=-\frac{y_∘}{x_∘}$$ Substitute it into the slope-intercept equation $$y=y_∘-\frac{y_∘}{x_∘}·x$$ Divide by $y_∘$ and isolate $1$
Two Line Relations
Two equations that intersect will have the same $(x,y)$ coordinates and different slopes

Parallel lines $m_1=m_2$
Intercepting lines $m_1 \ne m_2$
Perpendicular lines $m_1=-1/m_2$

Intersection Solves by Addition
Two equations may be added to each other. Like variables must be on the same side of the equation. $$\begin{array}{r} -2·x+9·y=5\\ \phantom{-}2·x-5·y=3\\ \overline{\phantom{-3·x+}4·y=8}\\ \end{array}$$ When one variable is solved, its value can be plugged into either of the equations to solve for the other variable. If the solution appears as $0=n$, then the lines are parallel.

Intersection Solves by Isolation
Given two equations with different slopes $$y=m_1·x+y_1\qquad y=m_2·x+y_2$$ To solve for $x$, set the equations equal to each other $$m_1·x+y_1=m_2·x+y_2$$ Subtract $y_1$ and $m_2·x$ $$m_1·x-m_2·x=y_2-y_1$$ Factor $$(m_1-m_2)·x=y_2-y_1$$ Isolate $x$ $$x=\frac{y_2-y_1}{m_1-m_2}$$ The value of $x$ can be plugged into either of the equations to solve for $y$. If the solution appears as $n/0$, then the lines are parallel.
A matrix can be used to display the coefficients of linear expressions $$ \begin{array}{c} 3·x+2·y \\ 4·x-1·y \end{array} ↔ \begin{bmatrix} 3 & \phantom{0}2 \\ 4 & -1 \end{bmatrix} $$ A joined matrix can be used to display the coefficients of linear eqautions in the general form, with the solutions being a vector $$ \begin{array}{c} 3·x+2·y=25 \\ 4·x-1·y=\phantom{0}4 \end{array} ↔ \begin{bmatrix} \begin{array}{cc|c} 3 & \phantom{-}2 & 25 \\ 4 & -1 & \phantom{0}4 \end{array} \end{bmatrix} $$
Identity Matrix
The matrix form of the multiplicative identity. The diagonal from top left to bottom right is called the trace. $$ I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

Row Operations
A joined matrix representing linear equations allows for certain arithmetical operations on individual rows

1. Multiplication of a row $$ \begin{array}{cc} \\ 2 & · \end{array} \begin{bmatrix} \begin{array}{cc|c} 3 & \phantom{-}2 & 25 \\ 4 & -1 & \phantom{0}4 \end{array} \end{bmatrix} = \begin{bmatrix} \begin{array}{cc|c} 3 & \phantom{-}2 & 25 \\ 8 & -2 & \phantom{0}8 \end{array} \end{bmatrix} $$ 2. Addition of one row to another $$ \begin{array}{c} \text{R1} \\ \text{R2} \end{array} \begin{bmatrix} \begin{array}{cc|c} 3 & \phantom{-}2 & 25 \\ 8 & -2 & \phantom{0}8 \end{array} \end{bmatrix} $$ $$ \begin{array}{c} \text{R1\phantom{+R2}} \\ \text{R1+R2} \end{array} \begin{bmatrix} \begin{array}{cc|c} \phantom{0}3 & 2 & 25 \\ 11 & 0 & 33 \end{array} \end{bmatrix} $$ 3. Switching rows $$ \begin{bmatrix} \begin{array}{cc|c} 11 & 0 & 33 \\ \phantom{0}3 & 2 & 25 \end{array} \end{bmatrix} $$ The aim is to create an identity matrix to display the values for $(1·x,1·y)$ in the vector. If an identity matrix is unattainable, then the lines are parallel.

Example
To continue solving for the above, divide row 1 by 11 $$ \begin{bmatrix} \begin{array}{cc|c} 1 & 0 & \phantom{0}3 \\ 3 & 2 & 25 \end{array} \end{bmatrix} $$ Add row 1 multiplied by –3 to row 2 $$ \begin{bmatrix} \begin{array}{cc|c} 1 & 0 & \phantom{0}3 \\ 0 & 2 & 16 \end{array} \end{bmatrix} $$ Divide row 2 by 2 $$ \begin{bmatrix} \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 8 \end{array} \end{bmatrix} $$ The two equations intersect at $(x,y)=(3,8)$.
Closed planar objects with three straight sides, and a sum of lengths of any two sides greater than the third
Perimeter $a+b+c$
Area (non-obtuse) $b·h/2$
Sum of Angles $\pi$
Right (all others are oblique)
Obtuse
Acute
Equilateral
Isosceles
Scalene
Congruent triangles
Centroid (G)

Median intersection point

Line segments extend from vertices (corners) to the sides at the midpoint
Orthocenter (H)

Altitude intersection point

Line segments extend from vertices to the sides at which right angles are formed
Circumcenter (O)

Perpendicular bisector intersection point

Line segments extend at right angles from the midpoint of each side
Incenter (I)

Angle bisector intersection point

Line segments extend from vertices dividing the angles of the vertices in half
Nine-Point Circle (N=center)

Circle passing though the midpoint of each side, the point of altitude on each side, and the midpoint of lines between each vertex and the orthocenter
Image taken from Wolfram Alpha and edited

Euler Line

Line passing through G, H, N, and O of all non-equilateral triangles (In equilateral triangles, G, H, N, O and I all overlap in a single point)
Image taken from Wolfram Alpha and edited
In an equilateral triangle, the sum of distances from any inner point extending orthogonally to the sides is equal to the height of the triangle
Proof
Given side length $L$ and height $h$, the area of the triangle equals the sum of the areas of the inner triangles with altitudes $a$, $b$, and $c$ $$\frac{L·h}{2}=\frac{L·a}{2}+\frac{L·b}{2}+\frac{L·c}{2}$$ Multiply by $2/L$
$$h^2=x^2+y^2$$
Proof
The square $c^2$ has area equal to all the triangle areas subtracted from the largest square $$c·c=(a+b)(a+b)-4·\Big(\frac{1}{2}·a·b\Big)$$ Expand $$c^2=a^2+2·a·b+b^2-2·a·b$$ Cancel like terms
$$|d|=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$$
Proof
Apply the right angle theorem to $(x_1,y_1)$ and $(x_2,y_2)$

Determinant
$$ det \begin{bmatrix} a & b \\ c & d \end{bmatrix} =a·d-b·c $$

Cramer's Rule
Given a system of equations $$ \begin{bmatrix} \begin{array}{cc|c} x_1 & y_1 & c_1 \\ x_2 & y_2 & c_2 \end{array} \end{bmatrix} $$ Intersections can be solved using the determinant with determinants where the vector becomes the transformation $$ D= \begin{bmatrix} \begin{array}{cc} x_1 & y_1 \\ x_2 & y_2 \end{array} \end{bmatrix} $$ $$ Dx= \begin{bmatrix} \begin{array}{cc} c_1 & y_1 \\ c_2 & y_2 \end{array} \end{bmatrix} \qquad x=\frac{Dx}{D}$$ $$ Dy= \begin{bmatrix} \begin{array}{cc} x_1 & c_1 \\ x_2 & c_2 \end{array} \end{bmatrix} \qquad y=\frac{Dy}{D}$$

Matrix Inverse
link to multiplicative inverse
$$ \begin{bmatrix} \begin{array}{ccc} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{array} \end{bmatrix} $$
Jump to Regular Polygons
Jump to Circles
$$\sin(\theta)=\frac{\text{opp}}{\text{hyp}}$$ $$\cos(\theta)=\frac{\text{adj}}{\text{hyp}}$$
$$\csc(\theta)=\frac{\text{hyp}}{\text{opp}}$$ $$\sec(\theta)=\frac{\text{hyp}}{\text{adj}}$$
$$\tan(\theta)=\frac{\text{opp}}{\text{adj}}$$ $$\cot(\theta)=\frac{\text{adj}}{\text{opp}}$$

Simple Trig Angles
Two simple triangles help identify certain values
Function $$0$$ $$\pi/6$$ $$\pi/4$$ $$\pi/3$$ $$\pi/2$$
$\sin(\theta)$ $$0$$ $$1/2$$ $$1/\sqrt{2}$$ $$\sqrt{3}/2$$ $$1$$
$\cos(\theta)$ $$1$$ $$\sqrt{3}/2$$ $$1/\sqrt{2}$$ $$1/2$$ $$0$$
$\csc(\theta)$ $$∞$$ $$2$$ $$\sqrt{2}$$ $$2/\sqrt{3}$$ $$1$$
$\sec(\theta)$ $$1$$ $$2/\sqrt{3}$$ $$\sqrt{2}$$ $$2$$ $$∞$$
$\tan(\theta)$ $$0$$ $$1/\sqrt{3}$$ $$1$$ $$\sqrt{3}$$ $$∞$$
$\cot(\theta)$ $$∞$$ $$\sqrt{3}$$ $$1$$ $$1/\sqrt{3}$$ $$0$$
Closed planar objects with at least three straight sides and equal number of vertices (corners)
Perimeter Sum of side lengths
Area Sum of areas of triangles divided within shape
Sum of Angles $(\text{number of sides}-2)·\pi$
Equilateral
Equiangular
Regular
Irregular
Convex & Concave
  • $A=$ area
  • $n=$ number of sides
  • $s=$ length of sides
  • $h=$ apothem
  • $r=$ circumradius
Jump to Circles
Perimeter $n·s$
Area $n·s·h/2$
Sum of angles $(n-2)·\pi$
Angle (h,r) $\pi/n$
Circumradius $h·\sec \Big(\frac{\pi}{n}\Big) \land \frac{s}{2}·\csc \Big(\frac{\pi}{n}\Big)$
Area using apothem $A=n·h^2·\tan \Big(\frac{\pi}{n}\Big)$
Area using side $A=\frac{n·s^2}{4}·\cot \Big(\frac{\pi}{n}\Big)$
Area using circumradius $A=n·r^2·\sin \Big(\frac{\pi}{n}\Big)·\cos \Big(\frac{\pi}{n}\Big)$

Proof of Circumradius
With respect to the center angle between the apothem and circumradius, use the corresponding trig definitions for $h$ = adjacent, $b/2$ = opposite, and $r$ = hypotenuse, and solve for $r$ in each case
$$\sec \Big(\frac{\pi}{n}\Big)=\frac{r}{h}$$ $$\csc \Big(\frac{\pi}{n}\Big)=\frac{r}{s/2}$$

Proof of Area using Apothem
Set the two circumradius formulae equal to each other $$h·\sec \Big(\frac{\pi}{n}\Big)=\frac{s}{2}·\csc \Big(\frac{\pi}{n}\Big)$$ Multiply by $h$, then divide by the cosecant function $$h^2·\frac{\sec(\pi/n)}{\csc(\pi/n)}=\frac{s·h}{2}$$ Use the trig definitions of the secant and cosecant functions to simplify into the tangent function $$\frac{\sec(\pi/n)}{\csc(\pi/n)}=\frac{r/h}{r/s}=\frac{s}{h}=\tan \Big(\frac{\pi}{n}\Big)$$ Substitute $$h^2·\tan \Big(\frac{\pi}{n}\Big)=\frac{s·h}{2}$$ Multiply by the number of sides $$n·h^2·\tan \Big(\frac{\pi}{n}\Big)=n·\frac{s·h}{2}$$

Proof of Area using Circumradius
Using the first circumradius equality, isolate $h$ $$h=\frac{r}{\sec(\pi/n)}$$ Substitute for $h$ in the equation for the area using the apothem $$A=n·r^2·\frac{\tan(\pi/n)}{\sec^2(\pi/n)}$$ Use the trig definitions of the tangent and secant functions to simplify to sine and cosine $$\frac{\tan(\pi/n)}{\sec^2(\pi/n)}=\frac{s/h}{r^2/h^2}=\frac{s}{r}·\frac{h}{r}=\sin\Big(\frac{\pi}{n}\Big)·\cos\Big(\frac{\pi}{n}\Big)$$ Substitute $$A=n·r^2·\sin\Big(\frac{\pi}{n}\Big)·\cos\Big(\frac{\pi}{n}\Big)$$

Proof of Area using Side Length
Substitute the second circumradius equality into the area using the circumradius equation, then expand $$A=\frac{n·s^2}{4}·\csc^2\Big(\frac{\pi}{n}\Big)·\sin\Big(\frac{\pi}{n}\Big)·\cos\Big(\frac{\pi}{n}\Big)$$ Use the trig definitions to simplify the trig functions into cotangent, $$\csc^2\Big(\frac{\pi}{n}\Big)·\sin\Big(\frac{\pi}{n}\Big)·\cos\Big(\frac{\pi}{n}\Big)=\frac{r^2}{s^2}·\frac{s}{r}·\frac{h}{r}=\frac{h}{s}=\cot\Big(\frac{\pi}{n}\Big)$$ Substitute $$A=\frac{n·s^2}{4}·\cot\Big(\frac{\pi}{n}\Big)$$
Two-dimensional subsets of a 3D (double) cone surface in which the shapes are determined by the intersection of a 2D plane
Image taken from CK-12 Foundation and edited
Jump to Circles
Jump to Ellipses
Jump to Parabolas
Jump to Hyperbolas
General Equation
$$A⋅x^2+B⋅x⋅y+C⋅y^2+D⋅x+E⋅y+F=0$$ All conic sections will be represented without rotation until the section on rotations. In other words, $B=0$.

Discriminant & Related Properties
$$\Delta=B^2-4⋅a⋅c$$ Assuming no other values:
  • If $B=0$, the function is even either vertically or horizontally (not rotated)
  • If $Δ \lt 0$, the function is an ellipse
  • If $Δ \lt 0$, $B=0$, and $A \lt C$, the function is an ellipse with a horizontal major axis
  • If $Δ \lt 0$, $B=0$, and $A \gt C$, the function is an ellipse with a vertical major axis
  • If $Δ \lt 0$, $B=0$, and $A=C$, the function is a circle
  • If $Δ \lt 0$, $A=C$, $B=D=E=0$, the function is a circle centered at the origin
  • If $Δ=0$, the function is a parabola
  • If $C=D=F=0$, the function is a vertical parabola with the vertex at the origin
  • If $A=E=F=0$, the function is a horizontal parabola with the vertex at the origin
  • If $D=E=0$, the function is either an ellipse or hyperbola centered at the origin
  • If $Δ \gt 0$, the function is a hyperbola
  • If $Δ \gt 0$ and $A+C=0$, the function is a rectangular hyperbola, meaning the asymptotes are perpendicular

Directrix

Eccentricity
Eccentricity is a function's deviation from being circular, given by $e=c/a$; $$\text{eccentricity}=\frac{\text{distance from any point to the focus}}{\text{distance from any point to the directrix}}$$
  • Two conic sections are congruent if the eccentricity of each are equal
  • Circle eccentricities are always 0
  • Ellipse eccentricities are always between 0 and 1
  • Parabola eccentricities are always 1
  • Hyperbola eccentricities are always greater than 1
  • Line eccentricities are infinite

Degenerate Conics
Conics when either a 2D plane intercepts the vertex of a double cone, or the result of the general equation yields a non-function by real algebraic definition
No result $A⋅x^2+A⋅y^2=-1$
Point $A⋅x^2+A⋅y^2=0$
Line $D⋅x+E⋅y+F=0$
Intersecting lines $(x-a)⋅(y+a)=0$
Parallel lines $(x-a)⋅(x-b)=0$
Universal Properties
Closed curves with all points equidistant to an internal point
Circumference $2·\pi·r$
Area $\pi·r^2$
Arc length $\theta·r$
Sector area $\theta·r^2/2$
Chord length (k) $2·r·\sin\Big(\frac{\theta}{2}\Big)$
Segment area $\frac{r^2}{2}·\Big(\theta-2·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)\Big)$
Conic general equation $A·(x^2+y^2)+D·x+E·y+F=0,A≠0$
Standard equation ${(x-x_∘)}^2+{(y-y_∘)}^2=r^2$
Conic-standard conversions $x_∘=-\frac{D}{2·A}$
$y_∘=-\frac{E}{2·A}$
$r^2=\frac{D^2+E^2-4·A·F}{4·A^2}$
Focus Coordinates $(x_∘,y_∘)$
Eccentricity $0$
Directrix None

Deductive Logic for Area
Apply the area function of regular polygons, using the circumference as the perimeter and the radius as the apothem $$A=(2·\pi·r)·\frac{r}{2}$$

Proof of Arc Length
The arc length is a fraction of the circumference, therefore can be found by the ratio to it and its angle $$\frac{a}{2·\pi·r}=\frac{\theta}{2·\pi}$$

Proof of Sector Area
The sector area is a fraction of the circle area, therefore can be found by the ratio to it and its angle $$\frac{A_S}{\pi·r^2}=\frac{\theta}{2·\pi}$$

Proof of Chord Length
Use the radius and half the chord length to form a right triangle Use the sine function for the angle $$\sin\Big(\frac{\theta}{2}\Big)=\frac{k}{2·r}$$ Multiply by $2·r$

Proof of Segment Area
The segment area is the triangular area between the center and chord subtracted from the sector area $$A_S=\frac{\theta·r^2}{2}-A_t$$ For the triangular area, use the chord length equation for the base and trig definition with respect to the radius to find the height
$$k=2·r·\sin\Big(\frac{\theta}{2}\Big)$$ $$h=r·\cos\Big(\frac{\theta}{2}\Big)$$
Substitute the triangular area using these values for $(k·h)/2$ $$A_S=\frac{\theta·r^2}{2}-\frac{r^2}{2}·2·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)$$ Factor $$A_S=\frac{r^2}{2}·\Big(\theta-2·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)\Big)$$

Conic-Standard Conversion
Given the conic general equation with the properties for a circle, group the $x$ terms and $y$ terms, and isolate the constant $$A·x^2+D·x+A·y^2+E·y=-F$$ Divide by $A$ $$x^2+\frac{D}{A}·x+y^2+\frac{E}{A}·y=-\frac{F}{A}$$ Complete the square for the $x$ and $y$ terms $$x^2+\frac{D}{A}·x+\frac{D^2}{4·A^2}+y^2+\frac{E}{A}·y+\frac{E^2}{4·A^2}=-\frac{F}{A}+\frac{D^2}{4·A^2}+\frac{E^2}{4·A^2}$$ Factor $${\Big(x+\frac{D}{2·A}\Big)}^2+{\Big(y+\frac{E}{2·A}\Big)}^2=\frac{D^2+E^2-4·A·F}{4·A^2}$$ Equate the coefficients to the standard equation

Annulus
A ring formed by two concentric circles in which all features of circles apply with respect to differences involving two radii
Perimeter $2·\pi·(R+r)$
Area $\pi·(R^2-r^2)$
Sector Area $\theta·(R^2-r^2)/2$
Universal Properties
Closed ovular curves whose points are the result of a constant sum between two internal points
Jump to Hyperbolas
Perimeter $4·a\lt 2·\pi·a, a\gt b$
Area $\pi·a·b$
Conic General Equation $A·x^2+C·y^2+D·x+E·y+F=0,A·C>0$
Standard Equation ${(x-x_∘)}^2/a^2+{(y-y_∘)}^2/b^2=1$
Conic-Standard Conversions $A=b^2$
$C=a^2$
$D=-2·b^2·x_∘$
$E=-2·a^2·y_∘$
$F=b^2·{x_∘}^2+a^2·{y_∘}^2-a^2·b^2$
$a^2=C$
$b^2=A$
$x_∘=-\frac{D}{2·A}$
$y_∘=-\frac{E}{2·C}$
Orientation horizontal if $C>A \land a>b$
vertical if $A>C \land b>a$
Center Coordinates $(x_∘,y_∘)$
Foci Coordinates $f=\big(x_∘ \pm \sqrt{a^2-b^2}, y_∘\big),a>b$ $f=\left(x_∘,y_∘ \pm \sqrt{b^2-a^2}\right),b>a$
Eccentricity $e=f/a=\sqrt{1-b^2/a^2},a>b$
$e=f/b=\sqrt{1-a^2/b^2},b>a$
Directrix $x=\pm a/e,a>b$
$y=\pm b/e,b>a$

Conic-Standard Conversion
Given the standard equation $$\frac{{(x-x_∘)}^2}{a^2}+\frac{{(y-y_∘)}^2}{b^2}=1$$ Multiply by $a^2·b^2$ $$b^2·{(x-x_∘)}^2+a^2·{(y-y_∘)}^2=a^2·b^2$$ Expand $$b^2·x^2-2·b^2·x·x_∘+b^2·x_∘^2+a^2·y^2-2·a^2·y·y_∘+a^2·y_∘^2=a^2·b^2$$ Rearrange to appear as the conic general equation $$b^2·x^2+a^2·y^2-2·b^2·x_∘·x-2·a^2·y_∘·y+b^2·x_∘^2+a^2·y_∘^2-a^2·b^2=0$$ Equate the coefficients to the conic equation
Universal Properties
Open mirrored curves whose points are the same distance between a common internal point and an exterenal line.
Image taken from Varsity Tutors and edited
Conic General Equations vertical: $A·x^2+D·x+E·y+F=0$
horizontal: $C·y^2+D·x+E·y+F=0$
The following are in vertical form
Standard Equation $y=a·x^2+b·x+y_∘$
Vertex Equation $y=a·(x-x_∘)^2+y_∘$
Intercept Equation $y=a·(x-x_1)(x-x_2)$
Discriminant $\Delta=b^2-4·a·y_∘$ If $Δ \lt 0$, two $x$-intercepts
If $Δ=0$, one $x$-intercept
If $Δ>0$, no $x$-intercepts
Conic-Standard Conversion $a=-A/E$
$b=-D/E$
$y_∘=-F/E$
Vertex Coordinates $(x_∘,y_∘)=\big(-\frac{b}{2·a},y_∘-\frac{b^2}{4·a}\big)$
$x$-Intercepts $\lbrace (x_1,0),(x_2,0) \rbrace = \big(\frac{-b \pm \sqrt{b^2-4·a·y_∘}}{2·a},0\big)$
Fucus Length from Vertex $f=\frac{1}{4·a}$
Focus Coordinates $\big(-\frac{b}{2·a},y_∘-\frac{b^2+1}{4·a}\big)$
Eccentricity $1$
Drectrix $y=-f$

Conic-Standard Conversion
Rearrange the conic general equation to isolate the $y$ term $$E·y=-A·x^2-D·x-F$$ Divide by $E$ $$y=-\frac{A}{E}·x^2-\frac{D}{E}·x-\frac{F}{E}$$ Equate the coefficients to the standard equation

Standard-Vertex Conversion
Given the standard equation, isolate the $x$ terms $$y-y_∘=a·x^2+b·x$$ Divide by $a$ $$\frac{y-y_∘}{a}=x^2+\frac{b·x}{a}$$ Complete the square $$\frac{y-y_∘}{a}+\frac{b^2}{4·a^2}=x^2+\frac{b·x}{a}+\frac{b^2}{4·a^2}$$ Factor $$\frac{y-y_∘}{a}+\frac{b^2}{4·a^2}={\bigg(x+\frac{b}{2·a}\bigg)}^2$$ Multiply by $a$ $$y-y_∘+\frac{b^2}{4·a}=a·{\bigg(x+\frac{b}{2·a}\bigg)}^2$$ Isolate $y$ $$y=a·{\bigg(x+\frac{b}{2·a}\bigg)}^2-\frac{b^2}{4·a}+y_∘$$ Equate the coefficients to the vertex equation

Standard-Intercept Conversion
Given the quadratic formula with $y=0$, find the zeros of $x$
Universal Properties
A mirrored set of open mirrored curvers whose points are the difference between two common internal points.
Image taken from BYJU's and edited
Conic General Equation $A·x^2+C·y^2+D·x+E·y+F=0, A·C \lt 0$
Standard Equation $(x-x_∘)^2/a^2+(y-y_∘)^2/b^2=1, a^2·b^2 \lt 0$
Standard Equation (Real Terms Only) $\pm (x-x_∘)^2/a^2 \mp (y-y_∘)^2/b^2=1$
Conic-Standard Conversion $A=b^2$
$C=a^2$ $D=-2·b^2·x_∘$
$E=-2·a^2·y_∘$
$F=b^2·{x_∘}^2+a^2·{y_∘}^2-a^2·b^2$
$a^2=C$
$b^2=A$
$x_∘=-\frac{D}{2·A}$
$y_∘=-\frac{E}{2·C}$
Orientation horizontal if $C \lt 0; a \isin ℂ$
vertical if $A \lt 0; b \isin ℂ$
Vertices $(x_∘ \pm a,y_∘) \lor (x_∘,y_∘ \pm b)$
Focus Coordinates $(x_∘ \pm \sqrt{a^2+b^2},y_∘) \lor (x_∘,y_∘ \pm \sqrt{a^2+b^2})$
Eccentricity $e=f/a=\sqrt{1+b^2/a^2}$
Directrix $x=\pm a^2/f \lor y=\pm b^2/f$
Asymptotes $y= \pm b·(x-x_∘)/a+y_∘$

Conic-Standard Conversion
The proof is the same for ellipses, however since by definition $A·C \lt 0$, either $a$ or $b$ must be imaginary. For the real-numbers-only representation, one of the terms must be negative for $a$ and $b$ to both be positive.
Conic section rotations occur exclusively when in the general equation $B \neq 0$ $$A·x^2+B·x·y+C·y^2+D·x+E·y+F=0$$ The angle of rotation is used to identify a new axis

PEEKSURE
$$x=x'·\cos(\theta)-y'·\sin(\theta)$$ $$y=x'·\sin(\theta)+y'·\cos(\theta)$$ Substitute the values into the general equation $$A·\big( \big)^2+B·\big( \big)·\big( \big)+C·\big( \big)^2+D·\big( \big)+E·\big( \big)+F=0$$
The trig functions are the functions of rotation on a circumference. The triangular functions (href) are a specific case. $$\sin(\theta)=y/r\qquad\cos(\theta)=x/r$$ $$\csc(\theta)=r/y\qquad\sec(\theta)=r/x$$ $$\tan(\theta)=y/x\qquad\cot(\theta)=x/y$$ $$\theta=\sin^{-1}(y/r)\qquad\theta=\cos^{-1}(x/r)$$ $$\theta=\csc^{-1}(r/y)\qquad\theta=\sec^{-1}(r/x)$$ $$\theta=\tan^{-1}(y/x)\qquad\theta=\cot^{-1}(x/y)$$
Jump to Sum & Difference Identities
Sources reference inverses like $\arcsin(θ)$ to avoid confusion. Here they are listed as above to save space, while reciprocal functions are as $$\sin(\theta)^{-1}=\frac{1}{\sin(\theta)}=\csc(\theta)≠\sin^{-1}(\theta)$$ However, powers other than $-1$ always mean the same in either of the following $$\sin(\theta)^n=\sin^n(\theta),\medspace\forall n≠-1$$
$$\sin(-\theta)=-\sin(\theta)\qquad\cos(-\theta)=\cos(\theta)$$ $$\csc(-\theta)=-\csc(\theta)\qquad\sec(-\theta)=\sec(\theta)$$ $$\tan(-\theta)=-\tan(\theta)\qquad\cot(-\theta)=-\cot(\theta)$$ The yielded sign value depends on which quadrant are the values of $(x,y)$

$\text{II: $\pi/2\lt 0\le\pi$}$ $\text{I: $0\le\theta\le\pi/2$}$
$$ \begin{array}{cc} \sin(\theta):+ & \csc(\theta):+ \\ \cos(\theta):- & \sec(\theta):- \\ \tan(\theta):- & \cot(\theta):- \end{array} $$ $$ \begin{array}{cc} \sin(\theta):+ & \csc(\theta):+ \\ \cos(\theta):+ & \sec(\theta):+ \\ \tan(\theta):+ & \cot(\theta):+ \end{array} $$
$\text{III: $\pi/2\lt\theta\lt3⋅\pi/2$}$ $\text{IV: $3⋅\pi/2\le\theta\lt 2⋅\pi$}$
$$ \begin{array}{cc} \sin(\theta):- & \csc(\theta):- \\ \cos(\theta):- & \sec(\theta):- \\ \tan(\theta):+ & \cot(\theta):+ \end{array} $$ $$ \begin{array}{cc} \sin(\theta):- & \csc(\theta):- \\ \cos(\theta):+ & \sec(\theta):+ \\ \tan(\theta):- & \cot(\theta):- \end{array} $$
Jump to Sum & Difference Identities
Cofunctions
$$\sin(\theta)=\frac{1}{\csc(\theta)} \qquad \cos(\theta)=\frac{1}{\sec{\theta}}$$ $$\tan(\theta)=\frac{\sin{\theta}}{\cos{\theta}}=\frac{1}{\cot{\theta}}$$

Complementary Angles
$$\sin\bigg(\frac{\pi}{2}-\theta\bigg)=\cos(\theta) \qquad\cos\bigg(\frac{\pi}{2}-\theta\bigg)=\sin(\theta)$$ $$\csc\bigg(\frac{\pi}{2}-\theta\bigg)=\sec(\theta) \qquad\sec\bigg(\frac{\pi}{2}-\theta\bigg)=\csc(\theta)$$ $$\tan\bigg(\frac{\pi}{2}-\theta\bigg)=\cot(\theta) \qquad\cot\bigg(\frac{\pi}{2}-\theta\bigg)=\tan(\theta)$$
Jump to Sum & Difference Identities
Inverse Cofunctions
$$\sin^{-1}(1/\theta)=\csc^{-1}(\theta),\medspace |\theta| \ge 1$$ $$\cos^{-1}(1/\theta)=\sec^{-1}(\theta),\medspace |\theta| \ge 1$$ $$\tan^{-1}(1/\theta)=\cot^{-1}(\theta),\medspace \theta \gt 0\phantom{0}$$

Inverse Complementary Angles
$$\sin^{-1}(\theta)+\cos^{-1}(\theta)=\pi/2,\medspace |\theta| \le 1$$ $$\csc^{-1}(\theta)+\sec^{-1}(\theta)=\pi/2 \phantom{,\medspace |\theta| \le 1}$$ $$\tan^{-1}(\theta)+\cot^{-1}(\theta)=\pi/2,\medspace |\theta| \ge 1$$
Function Period Domain Range Graph
$$\sin(\theta)$$ $$2·\pi$$ $(-∞,∞)$ $[-1,1]$
$$\cos(\theta)$$ $$2·\pi$$ $(-∞,∞)$ $[-1,1]$
$$\tan(\theta)$$ $$\pi$$ $(-∞,∞),\medspace x\ne(2·n-1)·\frac{\pi}{2}\medspace \forall n \isin \Z$ $(-∞,∞)$
$$\cot(\theta)$$ $$\pi$$ $(-∞,∞),\medspace x\ne(2·n-1)·\frac{\pi}{2}\medspace \forall n \isin \Z$ $(-∞,∞)$
$$\sec(\theta)$$ $$2·\pi$$ $(-∞,∞),\medspace x\ne(2·n-1)·\frac{\pi}{2}\medspace \forall n \isin \Z$ $|y|\ge 1$
$$\csc(\theta)$$ $$2·\pi$$ $(-∞,∞),\medspace x\ne 2·n·\pi\medspace \forall n \isin \Z$ $|y|\ge 1$
$$\sin^{-1}(\theta)$$ $$-$$ $|x|\le 1$ $|y|\le \pi/2$
$$\cos^{-1}(\theta)$$ $$-$$ $|x|\le 1$ $0\le y\le\pi$
$$\tan^{-1}(\theta)$$ $$-$$ $(-∞,∞)$ $|y|\le \pi/2$
$$\cot^{-1}(\theta)$$ $$-$$ $(-∞,∞)$ $0\lt y\lt\pi$
$$\sec^{-1}(\theta)$$ $$-$$ $|x|\ge 1$ $0\le y\le \pi, y\ne\pi/2$
$$\csc^{-1}(\theta)$$ $$-$$ $|x|\ge 1$ $|y|\le \pi/2,y\ne 0$
  • The complex plain is a two-dimensional number set whose exponential nature is rotational
  • Rational powers of $i$ expand into complex numbers with multipliers of $1$ and $i^1$
  • Powers of $i^x$ function the same as radians on the trig unit circle with $x=2$ equivalent to one radian
  • It follows that $i^x$ is a two component number on the unit circle equal to $\cos(x·\pi/2)+i·\sin(x·\pi/2)$

Examples
$$\sqrt{i}=\cos\bigg(\frac{\pi}{4}\bigg)+i·\sin\bigg(\frac{\pi}{4}\bigg)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}·i$$ $$i^{5/3}=\cos\bigg(\frac{5·\pi}{6}\bigg)+i·\sin\bigg(\frac{5·\pi}{6}\bigg)=-\frac{\sqrt{3}}{2}+\frac{1}{2}·i$$
Jump to Complex Number System
$$\sin^2(\theta)+\cos^2(\theta)=1$$ $$\tan^2(\theta)+1=\sec^2(\theta)$$ $$\cot^2(\theta)+1=\csc^2(\theta)$$
Jump to Sum & Difference Identities
Jump to Half Angle Identities
Proofs
In each case, given the right angle theorem, divide by the hypotenuse, $x$ component, and $y$ component respectively $$\frac{h^2=x^2+y^2}{h^2}\medspace→\medspace 1=\frac{x^2}{h^2}+\frac{y^2}{h^2}$$ $$\frac{h^2=x^2+y^2}{x^2}\medspace→\medspace \frac{h^2}{x^2}=1+\frac{y^2}{x^2}$$ $$\frac{h^2=x^2+y^2}{y^2}\medspace→\medspace \frac{h^2}{y^2}=\frac{x^2}{y^2}+1$$ In each case, substitute the right angle definitions
$$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}$$
Proof
Given the triangle area, substitute the height of each base with its right angle definition for sine $$\frac{1}{2}·b·c·\sin(A)=\frac{1}{2}·a·c·\sin(B)=\frac{1}{2}·a·b·\sin(C)$$ Multiply by $2/(a·b·c)$
$$c^2=a^2+b^2-2·a·b·\cos(C)$$ $$C=\cos^{-1}\bigg(\frac{a^2+b^2-c^2}{2·a·b}\bigg)$$
Proof
Insert the perpendicular line to c to represent the height of the triangle, dividing $c$ as $c_1$ and $c_2$ Find $c_1$ and $c_2$ by taking the cosine of $a$ and $b$ $$\cos(A)=\frac{c_2}{b}\qquad\cos(B)=\frac{c_1}{a}$$ Solve for $c_1$ and $c_2$ $$c_2=b·\cos(A)\qquad c_1=a·cos(B)$$ Add $c_1$ to $c_2$ for $c$ $$c_2+c_1=c=b·\cos(A)+a·cos(B)$$ Multiply by $c$ $$c^2=b·c·\cos(A)+a·c·cos(B)$$ Repeat the process for the other two sides $$a^2=c·a·\cos(B)+b·a·cos(C)$$ $$b^2=a·b·\cos(C)+c·b·cos(A)$$ Add the last two equations and subtract the first $$a^2+b^2-c^2=c·a·\cos(B)+b·a·cos(C)+a·b·\cos(C)+c·b·cos(A)-b·c·\cos(A)-a·c·cos(B)$$ Simplify $$a^2+b^2-c^2=2·a·b·\cos(C)$$ Isolate $c^2$
$$\sin(a\pm b)=\sin(a)·\cos(b)\pm\cos(a)·\sin(b)$$ $$\cos(a\pm b)=\cos(a)·\cos(b)\mp\sin(a)·\sin(b)$$ $$\tan(a\pm b)=\frac{\tan(a)\pm\tan(b)}{1\mp\tan(a)·\tan(b)}$$ $$\cot(a\pm b)=\frac{\cot(a)·\cot(b)\mp1}{\cot(b)\pm \cot(a)}$$
Jump to Double Angle Identities
Proof of Sine & Cosine
On the unit circle, plot points for $(1,0)$, and for the angles $b$, $a-b$, and $a$ respectively in a positive rotation Determine the coordinates of each point given their angles using the standard trig definitions The distances displayed are equal to each other. Use the distance formula for each set equal to each other. $$\sqrt{\big(1-\cos(a-b)^2\big)^2+\big(0-\sin(a-b)^2\big)^2}=\sqrt{\big(\cos(b)-\cos(a)\big)^2+\big(\sin(b)-\sin(a)\big)^2}$$ Square $$\big(1-\cos(a-b)^2\big)^2+\big(0-\sin(a-b)^2\big)^2=\big(\cos(b)-\cos(a)\big)^2+\big(\sin(b)-\sin(a)\big)^2$$ Expand the left $$1-2·\cos(a-b)+\cos^2(a-b)+\sin^2(a-b)=\big(\cos(b)-\cos(a)\big)^2+\big(\sin(b)-\sin(a)\big)^2$$ Use the right angle identity to simplify to one $$1-2·\cos(a-b)+1=\big(\cos(b)-\cos(a)\big)^2+\big(\sin(b)-\sin(a)\big)^2$$ Expand the right $$2-2·\cos(a-b)=\cos^2(b)-2·\cos(a)·\cos(b)+\cos^2(a)+\sin^2(b)-2·\sin(a)·\sin(b)+\sin^2(a)$$ Use the right angle identitiy to simplify to 1 in two instances $$2-2·\cos(a-b)=-2·\cos(a)·\cos(b)-2·\sin(a)·\sin(b)+2$$ Subtract 2 $$-2·\cos(a-b)=-2·\cos(a)·\cos(b)-2·\sin(a)·\sin(b)$$ Divide by $-2$ $$\cos(a-b)=\cos(a)·\cos(b)+\sin(a)·\sin(b)$$ Substitute $-b$ for $b$ $$\cos\big(a-(-b)\big)=\cos(a)·\cos(-b)+\sin(a)·\sin(-b)$$ Apply the even/odd identities $$\cos(a+b)=\cos(a)·\cos(b)-\sin(a)·\sin(b)$$ Substitute $\pi/2-(a+b)$ for $a+b$ and $\pi/2-a$ for $a$. Note that $b$ in the left term is a separate addend from the shift. $$\cos\bigg(\frac{\pi}{2}-a-b\bigg)=\cos\bigg(\frac{\pi}{2}-a\bigg)·\cos(b)-\sin\bigg(\frac{\pi}{2}-a\bigg)·\sin(b)$$ Apply the complimentary angle identities $$\sin(a-b)=\sin(a)·\cos(b)-\cos(a)·\sin(b)$$ Substitute $-b$ for $b$ $$\sin\big(a-(-b)\big)=\sin(a)·\cos(-b)-\cos(a)·\sin(-b)$$ Apply the even/odd identities $$\sin(a+b)=\sin(a)·\cos(b)+\cos(a)·\sin(b)$$

Proof of Tangent
Substitute tangent for its sine/cosine cofunction $$\tan(a\pm b)=\frac{\sin(a\pm b)}{\cos(a\pm b)}$$ Substitute the sum & difference identities $$\tan(a\pm b)=\frac{\sin(a)·\cos(b)\pm\cos(a)·\sin(b)}{\cos(a)·\cos(b)\mp\sin(a)·\sin(b)}$$ Divide the numerator and denominator by $\cos(a)·\cos(b)$ $$\tan(a\pm b)= \frac{\big(\sin(a)·\cos(b)\pm\cos(a)·\sin(b)\big)/\big(\cos(a)·\cos(b)\big)} {\big(\cos(a)·\cos(b)\mp\sin(a)·\sin(b)\big)/\big(\cos(a)·\cos(b)\big)}$$ Cancel like terms $$\frac{\sin(a)/\cos(a)\pm\sin(b)/\cos(b)}{1\mp\big(\sin(a)·\sin(b)\big)/\big(\cos(a)·\cos(b)\big)}$$ SubSubstitute the tangent function for the sines over cosines

Proof of Cotangent
Substitute cotangent for its cosine/sine cofunction $$\cot(a\pm b)=\frac{\cos(a\pm b)}{\sin(a\pm b)}$$ Substitute the sum & difference identities $$\cot(a\pm b)=\frac{\cos(a)·\cos(b)\mp\sin(a)·\sin(b)}{\sin(a)·\cos(b)\pm\cos(a)·\sin(b)}$$ Divide the numerator and denominator by $\sin(a)·\sin(b)$ $$\tan(a\pm b)= \frac{\big(\cos(a)·\cos(b)\mp\sin(a)·\sin(b)\big)/\big(\sin(a)·\sin(b)\big)} {\big(\sin(a)·\cos(b)\pm\cos(a)·\sin(b)\big)/\big(\sin(a)·\sin(b)\big)}$$ Cancel like terms $$\frac{\big(\cos(a)·\cos(b)\big)/\big(\sin(a)·\sin(b)\big)\mp 1}{\cos(b)/\sin(b)\pm\cos(a)/\sin(a)}$$
$$2·\sin(a)·\sin(b)=\cos(a-b)-\cos(a+b)$$ $$2·\cos(a)·\cos(b)=\cos(a-b)+\cos(a+b)$$ $$2·\sin(a)·\cos(b)=\sin(a+b)-\sin(a-b)$$ $$2·\cos(a)·\sin(b)=\sin(a+b)-\sin(a-b)$$
Proofs
Substitute the sum and difference identities for the terms on the right, cancel like terms, and factor
$$\sin(a) \pm \sin(b)=2·\sin\bigg(\frac{a \pm b}{2} \bigg)·\cos\bigg(\frac{a \mp b}{2} \bigg)$$ $$\cos(a)+\cos(b)=2·\cos\bigg(\frac{a+b}{2} \bigg)·\cos\bigg(\frac{a-b}{2} \bigg)$$ $$\cos(a)-\cos(b)=-2·\sin\bigg(\frac{a+b}{2} \bigg)·\sin\bigg(\frac{a-b}{2} \bigg)$$
Proof
$$\sin(2·\theta)=2·\sin(\theta)·\cos(\theta)$$ $$\cos(2·\theta)=$$
$$\csc(2·\theta)=\frac{1}{2}·\sec(\theta)·\csc(\theta)$$ $$\sec(2·\theta)=$$
$$\tan(2·\theta)=$$ $$\cot(2·\theta)=$$

Proofs
Use the same angle twice in the sum identities for each function
Example
Circle segment area
$$\bigg|\sin\bigg(\frac{\theta}{2}\bigg)\bigg|=$$ $$\bigg|\cos\bigg(\frac{\theta}{2}\bigg)\bigg|=$$ $$\bigg|\tan\bigg(\frac{\theta}{2}\bigg)\bigg|=$$ $$\bigg|\cot\bigg(\frac{\theta}{2}\bigg)\bigg|=$$
Proof of Sine
Proof of Cosine
Proof of Tangent
Proof of Cotangent
$$\sin^2(\theta)=\frac{1-\cos(2·\theta)}{2}$$ $$\cos^2(\theta)=\frac{1+\cos(2·\theta)}{2}$$ $$\tan^2(\theta)=\frac{1-\cos(2·\theta)}{1+\cos(2·\theta)}$$ $$\cot^2(\theta)=\frac{1+\cos(2·\theta)}{1-\cos(2·\theta)}$$
Proofs
$$(r,\theta)=\bigg(\sqrt{x^2+y^2},\tan^{-1}\bigg(\frac{y}{x}\bigg)\bigg),\medspace x\ne 0$$ $$(x,y)=\big(r·\sin(\theta),r·\cos(\theta)\big)$$
Limits are determinations of specific values in a function where values might not exist. $$\lim_{x→2}\frac{x^2-4}{x-2}=\frac{0}{0}$$ $$\lim_{x→2}\frac{x^2-4}{x-2}=\frac{(x+2)·(x-2)}{x-2}=x+2=4$$
$$\lim\limits_{x→a^+}f(x)$$ The limit as a function approaches a point from the right
$$\lim\limits_{x→a^-}f(x)$$ The limit as a function approaches a point from the left
$$\lim\limits_{x→a}f(x)$$ The limit as a function approaches a point from the both sides
A function $f(x)$ is continuous at point $a$ if the limit exists and $f(a)$ is defined;

  • The limit exists when $\lim_{{x→a}^-}f(x)=\lim_{{x→a}^+}f(x)=\lim_{{x→a}}f(x)$
  • There is no change in functionality, meaning there are no corners at $f(a)$
  • $\lim_{{x→a}}f(x)=f(a)$
The limit of a constant function is equal to the constant $$\lim\limits_{x→a} c=c$$ The limit of a constant multiple of a function equals the product of the constant with the limit of the function $$\lim\limits_{x→a} c·f(x)=c·\lim\limits_{x→a} f(x)$$ The limit of sums is the sum of the limits $$\lim\limits_{x→a} \big(f(x)\pm g(x)\big)=\lim\limits_{x→a} f(x)\pm \lim\limits_{x→a} g(x)$$ The limit of products is the product of the limits $$\lim\limits_{x→a} \big(f(x)·g(x)^{\pm 1}\big)=\lim\limits_{x→a} f(x)· \lim\limits_{x→a} g(x)^{\pm 1}$$
Equation of a line $\lim_{x→a}f(x)=f(a)=m·a+b$
Horizontal asymptote The line $x=c$ is defined as $\lim_{x→c}f(x)=\pm ∞$
Vertical asymptote The line $y=c$ is defined as $\lim_{x→\pm ∞}f(x)=c$
Given the transitive property, and three functions related as $f(x)≤g(x)≤h(x)$ converging to a point $$\text{If }\lim_{x→a}f(x)=\lim_{x→a}h(x)=L\text{ then }\lim_{x→a}g(x)=L$$
$$\lim\limits_{x→0}\frac{\sin(x)}{x}=1\qquad\lim\limits_{x→0}\frac{\cos(x)-1}{x}=0$$
Proof of Sine Limit
Using the incircle of a regular octagon with an apothem of magnitude 1 to define points in the sector for $0 \le \theta \le \pi/4$ Add to the top right corner a triangular region, forming an overall right triangle The following determination can be made inside of triangle $OAD$ $$\text{arc} AC \lt |AB|+|BC| \lt \big(|AB|+|BD|=|AD|\big)$$ $$\land \medspace |AD|=|OA|·\tan(\theta)=\tan(\theta)$$ $$\therefore \text{arc} AC \lt\tan(\theta)$$ The apothem is 1 and the arc length $AC$ is the angle $\theta$. Substitute tangent for its cofunctions, then isolate $\cos(\theta)$ $$\theta \lt \frac{\sin(\theta)}{\cos(\theta)} \medspace → \medspace \cos(\theta) \lt \frac{\sin(\theta)}{\theta}$$ Connect point $C$ perpendicularly to $|OA|$ at point $E$, and connect a line segment to $|AC|$ Note that $|CE|=|OC|·\sin(\theta)=\sin(\theta)$ and $|CE|\lt |AC|\lt arc AC$. It follows that $$\sin(\theta)\lt\theta\medspace\therefore\medspace\frac{\sin(\theta)}{\theta}\lt 1$$ Combine the two inequalities involving $\sin(\theta)/\theta$ $$\cos(\theta)\lt\frac{\sin(\theta)}{\theta}\lt 1$$ Apply the squeeze theorem for the limit on the right $$\lim_{\theta→0^+}\cos(\theta)=1\medspace\land\medspace\lim_{\theta→0^+}1=1\medspace\therefore\medspace\lim_{\theta→0^+}\frac{\sin(\theta)}{\theta}=1$$ Sine is an odd function, so its nature tending to zero may be determined with sign reversal $$\frac{\sin(-\theta)}{-\theta}=\frac{-\sin(\theta)}{-\theta}=\frac{\sin(\theta)}{\theta}$$ Since the function is the same from the reverse direction, the limit exists $$\lim_{\theta→0^-}\frac{\sin(\theta)}{\theta}=\lim_{\theta→0^+}\frac{\sin(\theta)}{\theta}=\lim_{\theta→0}\frac{\sin(\theta)}{\theta}=1$$

Proof of Cosine Limit
$$\lim\limits_{x→0^\pm} \frac{c}{x}=\pm ∞,\medspace c\ne 0$$ $$\lim\limits_{x→\pm ∞} \frac{c}{x}=0,\medspace c\ne 0$$
In regular polygons, as the number of sides → ∞, the perimeter approaches a circumference, wherein $π$ can be calculated.
Continually compounded growth with 100% (1) return at a continuous rate yields a limit at $e$ $$e=\lim_{x→∞} {\bigg( 1+\frac{1}{x} \bigg)}^x=\lim_{x→0} {(1+x)}^{1/x}≈2.7 \medspace 1828 \medspace 1828 \medspace 459$$
$e$ Limit
$$\lim_{x→0}\frac{e^x-1}{x}=1$$ From $$e=\lim_{x→0}{(1+x)}^{1/x}$$

Natural Logarithm
The inverse function of $e^x$ is $\log_e x$, represented as $\ln(x)$

Limits of Natural Exponents & Logarithms
$$\lim_{x→\pm ∞}e^{\pm x}=∞$$ $$\lim_{x→\pm ∞}e^{\mp x}=0$$
Definition
Two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger two quantities $$\frac{x+a}{x}=\frac{x}{a}$$
Proof of Ratios
Given the ratio equality, substitute 1 for $a$ $$\frac{x+1}{x}=x$$ Multiply by $x$ $$x+1=x^2$$ Rearrange to appear as a quadratic equation $$x^2-x-1=0$$
Fibonacci Series & $φ$ Limit
Golden Ratio Powers
Algebraic Functions for the Fibonacci Series
A function’s instantaneous rate of change along any given point. The derivative function is essentially the slope formula with one point.

PEEKSURE $$\frac{d}{dx}f(x)=\lim\limits_{x→a}\frac{f(x)-f(a)}{x-a}=\lim\limits_{\Delta x→0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
Notation First Second
Leibniz $$\frac{dy}{dx} \medspace\text{or}\medspace \frac{d}{dx}f{x}$$ $$\frac{d^2y}{dx^2} \medspace\text{or}\medspace \frac{d^2}{dx^2}f{x}$$
Langrange $$f'(x)$$ $$f''(x)$$
Newton $$\dot{x}$$ $$\ddot{x}$$
Euler $$Df$$ $$D^2f$$
The point-slope equation for a line parallel to and touching a function at any given point $$y-f(a)=f'(a)(x-a)$$
  • If $f'(a)=0$, then $f(a)$ is a critical point (local maxima or minima)
  • If $f'(a) \gt 0$, then $f(a)$ is increasing
  • If $f'(a) \lt 0$, then $f(a)$ is decreasing
  • If $f'(x)$ doesn’t change signs, then $f(x)$ is monotonic (only increasing or decreasing
  • If $f''(a)=0$, then $f(a)$ is an inflection point (change in concavity)
  • If $f''(a)\lt 0$, then $f(a)$ is concave (downwards)
  • If $f''(a)\gt 0$, then $f(a)$ is convex (concave upwards)
$$\frac{d}{dx}(C)=0$$
Proof
A constant has a slope of zero at every point, therefore its derivative is zero at every point
Proof
$$\frac{d}{dx}\big(f(x)\pm g(x)\big)=f'(x)\pm g'(x)$$
Proof
$$\frac{d}{dx}\big(f(x)·g(x)\big)=f'(x)·g(x)+f(x)·g'(x)$$
Proof
Proof
Proof
Proof
Proof
The indeterminate form 0·∞
The indeterminate form ∞–∞
Exponential Indeterminate Forms
Proof
$$\frac{d}{dx}e^x=e^x$$
Proof
Proof
$$\frac{d}{dx}b^x=b^x·\ln(b)$$
Proof
$$\frac{d}{dx}x^x=x^x·\big(1+\ln(x)\big)$$
Proof
Higher Order Sine & Cosine Derivatives
Proof of Sine
Proof of Cosine
Proof of Tangent
Proof of Cotangent
Proof of Secant
Proof of Cosecant
$$\frac{d}{dx}\sin\bigg(\pm\frac{x}{a}\bigg)=\pm\frac{1}{\sqrt{a^2-x^2}},\medspace |x| \lt 1$$ $$\frac{d}{dx}\cos\bigg(\pm\frac{x}{a}\bigg)=\mp\frac{1}{\sqrt{a^2-x^2}},\medspace |x| \lt 1$$ $$\frac{d}{dx}\tan\bigg(\pm\frac{x}{a}\bigg)=\pm\frac{a}{x^2+a^2}$$ $$\frac{d}{dx}\cot\bigg(\pm\frac{x}{a}\bigg)=\mp\frac{a}{x^2+a^2}$$ $$\frac{d}{dx}\sec\bigg(\pm\frac{x}{a}\bigg)=\pm\frac{a}{\sqrt{x^2·(x^2-a^2)}},\medspace |x| \gt 1$$ $$\frac{d}{dx}\csc\bigg(\pm\frac{x}{a}\bigg)=\mp\frac{a}{\sqrt{x^2·(x^2-a^2)}},\medspace |x| \gt 1$$
Proof of Inverse Sine
Proof of Inverse Cosine
Proof of Inverse Tangent
Proof of Inverse Cotangent
Proof of Inverse Secant
Proof of Inverse Cosecant
Proof of Euler's Formula
Proof of Negative Natural Logarithms
Proof of Imaginary Natural Logarithms
Proof of Sine & Cosine
Proof of Tangent, Cotangent, Secant & Cosecant
Proof of Inverse Sine
Proof of Inverse Cosine
Proof of Inverse Tangent
Proof of Inverse Cotangent
Proof of Inverse Secant
Proof of Inverse Cosecant
$$\int_a^b u·dv=u·v \bigg|_a^b-\int_a^b v·du$$
Proof
$\int x^n dx$
$\int (1/x)dx$
ba·x & ea·x
logbx & ln(x)

$$\int \sin(a·x) dx$$
$$\int \cos(a·x) dx$$
$$\int \csc(a·x) dx$$
$$\int \sec(a·x) dx$$
$$\int \tan(a·x) dx$$
$$\int \cot(a·x) dx$$
$$\int \sin^{–1}(a·x) dx$$
$$\int \cos^{–1}(a·x) dx$$
$$\int \csc^{–1}(a·x) dx$$
$$\int \sec^{–1}(a·x) dx$$
$$\int \tan^{–1}(a·x) dx$$
$$\int \cot^{–1}(a·x) dx$$
$$\int \sin^2(a·x) dx$$
$$\int \cos^2(a·x) dx$$
$$\int \csc^2(a·x) dx$$
$$\int \sec^2(a·x) dx$$
$$\int \tan^2(a·x) dx$$
$$\int \cot^2(a·x) dx$$
sin3(a·x)
cos3(a·x)
csc3(a·x)
sec3(a·x)
tan3(a·x)
cot3(a·x)

secn(a·x)·tan(a·x)
cscn(a·x)•cot(a·x)

1/(1±sin(a·x))
1/(1±cos(a·x))

sin(a·x)·sin(b·x)
sin(a·x)·cos(b·x)
cos(a·x)·cos(b·x)

Reduction for sinn(a·x)
Reduction for cosn(a·x)
Reduction for cscn(a·x)
Reduction for secn(a·x)
Reduction for tann(a·x)
Reduction for cotn(a·x)

Sine Reduction for sinm(a·x)·cosn(b·x)
Cosine Reduction for sinm(a·x)·cosn(b·x)

$$\int x^n·\sin(a·x) dx$$
$$\int x^n·\cos(a·x) dx$$
$$\int \sqrt{x^2+a^2} dx$$
$$\int \sqrt{x^2-a^2} dx$$
$$\int \sqrt{a^2-x^2} dx$$
$$\int x·\sqrt{x^2+a^2} dx$$
$$\int x·\sqrt{x^2-a^2} dx$$
$$\int x·\sqrt{a^2-x^2} dx$$
$$\int x^2·\sqrt{x^2+a^2} dx$$
$$\int x^2·\sqrt{x^2-a^2} dx$$
$$\int x^2·\sqrt{a^2-x^2} dx$$
$$\int \frac{\sqrt{x^2+a^2}}{x} dx$$
$$\int \frac{\sqrt{x^2-a^2}}{x} dx$$
$$\int \frac{\sqrt{a^2-x^2}}{x} dx$$
$$\int \frac{\sqrt{x^2+a^2}}{x^2} dx$$
$$\int \frac{\sqrt{x^2-a^2}}{x^2} dx$$
$$\int \frac{\sqrt{a^2-x^2}}{x^2} dx$$
1/√x²+a²
1/√x²–a²
1/√a²–x²

x/√x²+a²
x/√x²–a²
x/√a²–x²

x²/√x²+a²
x²/√x²–a²
x²/√a²–x²

1/(x·√x²+a²)
1/(x·√x²–a²)
1/(x·√a²–x²)

1/(x²·√x²+a²)
1/(x²·√x²–a²)
1/(x²·√a²–x²)
Divergence
Integral
Ratio
Root
Comparison

Updated October 2024
Bill Liam East
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