# Math Identities Proofs

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

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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 $$\frac{-1}{1}=\frac{1}{-1}=-\frac{1}{1}=-1$$ Distribution of two negatives $$\frac{-1}{-1}=-(-\frac{1}{1})=1$$

 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. $$x=a±b↔x=a+b \text{ and } x=a-b$$ 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. $$±x=a±b↔x=a+b \text{ and } -x=a-b$$ If ± and ∓ are used, then when one is positive, the other is negative. $$±x=a∓b↔x=a-b \text{ and } -x=a+b$$ If ± is used in an exponent, it is used to indicate multiplication or division and the same rules above apply. $$x·y^{±1}=a±b↔x·y=a+b \text{ and } x/y=a-b$$

 Real $$\R$$ All common numbers and multiples of one. Integer $$\Z$$ Real whole numbers that can be expressed without using a fraction. $$(-2, -1, 0, 1, 2)$$ Natural $$\N$$ Positive real integers. It is argued whether or not zero is included. Rational Any number that can be represented as a fraction inclusive of only natural numbers. Always either repeats or terminates as a decimal. $$1/2, 2/1, 1.2, 2.\overline{1}$$ Irrational Any number that cannot be represented as a fraction. Never repeats or terminates as a decimal. $$\sqrt{2}, e, \pi, \phi$$ 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 .9 equal a variable, $$x=.\overline{9}$$ Multiply by 10, $$10·x=9.\overline{9}$$ Separate 9.9 into 9 and .9, $$10·x=9+.\overline{9}$$ Substitute .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: nth 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. $$A·x^2+B·x·y+C·y^2+D·x+E·y+F=0$$

 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. mod 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.

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) $$f(x+P)=f(x)$$

$$a·f(b·x+c)+d$$
• If |a|>1, f stretches vertically.
• If |a|<1, f compresses vertically.
• If a<0, f flips vertically.
• If |b|>1, f compresses horizontally.
• If |b|<1, f stretches horizontally.
• If b<0, f flips vertically.
• If c>0, f shifts to the left.
• If c<0, f shifts to the right.
• If d>0, f shifts upwards.
• If d<0, f shifts downwards.

Square roots of negative numbers fail to yield real results. The imaginary unit satisfies the equation i²=–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. $$\frac{a+b·i}{c+d·i}=\frac{a+b·i}{c+d·i}·\frac{c-d·i}{c-d·i}=\frac{a·c+b·d+(b·c-a·d)·i}{c^2+d^2}$$

Exponentiation is repeated multiplication. An exponent is the number of times to "times" a number.
• Exponents do not commute: xaax
• Exponents do not associate: x^(ab) ≠ (xa)b
• Factoring is the condensing of terms: x3
• 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, xa is x multiplied a times, xb 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 (xa)c·(y±b)c. The terms are multiplied c times.

Example
Given 3·x–5·y=2, evaluate 8x/32y: $$\frac{8^x}{32^y}=\frac{{(2^3)}^x}{{(2^5)}^y}=\frac{2^{3·x}}{2^{5·y}}$$ 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. $$299792458=2.99792458·10^8$$ $$0.0820574=8.20574·10^{-2}$$

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·x²+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, subtract y, $$x^2 \pm b·x+ {\Big(\frac{b}{2} \Big)}^2=y-y_∘+{\Big(\frac{b}{2} \Big)}^2$$ Factor. $$\frac{{(2·x \pm b)}^2}{4}=y-y_∘+\frac{b^2}{4}$$

The (general) solutions of y=a·x²+b·x+y are $$x=\frac{-b \pm \sqrt{b^2+4·a·(y-y_∘)}}{2·a}, a≠0$$
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 b², 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<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

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
$${(a \pm b)}^2=a^2 \pm 2·a·b+b^2$$ $${(a \pm b)}^3=a^3 \pm 3·a^2·b+3·a·b^2+ \pm b^3$$

Given the rational function N(x)/D(x), the division results in the equality N(x)=D(xQ(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 (x³+2·x²+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 x², $$\begin{array}{r} x^2\phantom{+0·x+12}\\ x-2{\overline{\smash{\big)}\,x^3+2·x^2+0·x+12}}\\ \phantom{0}-x^3+2·x^2\phantom{+0·x+12}\\ {\overline{\phantom{-x^3+}4·x^2+0·x+12}}\\ \end{array}$$ 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(xQ(x)+R(x). $$x^3+2·x^2+12=(x-2)(x^2+4·x+8)+28$$

Partial Fraction Decomposition
Given a 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 original 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 xa=xb, 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, so 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=logb x. Rewrite in the exponential form x=ba, 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 $$x=\log_b M \land \pm y=\log_b N^{\pm 1}$$ So that $$M=b^x \land N^{\pm 1}=b^{\pm y}$$ Multiply the two equalities, $$M·N^{\pm 1}=b^x·b^{\pm y}$$ Apply the exponent power rule, $$M·N^{\pm 1}=b^{x \pm y}$$ Take the logarithm of base b, $$\log_b (M·N^{\pm 1})=\log_b b^{x \pm y}$$ Apply the base inverse property, $$\log_b (M·N^{\pm 1})=x \pm y$$ 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 logb(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=logbx so that bc=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 cn 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=log1/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, negate, $$-\log_b x=a$$ The equality –logbx=logbx–1 can be proven with the power and quotient rules.

$$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$$

A horizontal line is the line y equal to a constant for all x, having a defined slope of 0.

A vertical line is the line x equal to a constant for all yy, having an undefined slope of ±∞.

$$y(x)=m·x+y_∘$$

Given a line with point (x1, y1) and slope m. $$y-y_1=m·(x-x_1)$$

The point-slope equation with the slope formula substituted $$y-y_1=\frac{y_2-y_1}{x_2-x_1}·(x-x_1)$$

$$\frac{x}{x_∘}+\frac{y}{y_∘}=1, x_∘≠0 \land y_∘≠0$$
Proof
Use the coordinates for the intercepts in the slope formula, which are interchangeable, $$m=\frac{0-y_∘}{x_∘-0}=\frac{y_∘-0}{0-x_∘}=-\frac{y_∘}{x_∘}$$ Substitute it into the slope-intercept formula, $$y=y_∘-\frac{y_∘}{x_∘}·x$$ Solve.

$$A·x+B·y=C, x_∘=\frac{C}{A} \land y_∘=\frac{C}{B}$$

$$\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)$$

• Two lines are parallel if they have the same slope, i.e. m1=m2
• Two lines intercept at (x, y), which is found by first isolating x in each, and setting them equal to each other to solve for y.
• Two lines are perpendicular if the slope of one is the negative reciprocal to the slope of the other, i.e. m1=–1/m2

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) $$\frac{1}{2}·b·h$$ 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

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)

$$h^2=x^2+y^2$$
Proof
The square c² 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 (x1, y1) and (x2, y2)

In an equilateral triangle, the sum of distances from any inner point to the sides is equal to the height of the triangle.
Proof
Given side s and height h, the area of the triangle equals the sum of the areas of the inner triangles with altitudes a, b, and c, $$\frac{s·h}{2}=\frac{s·a}{2}+\frac{s·b}{2}+\frac{s·c}{2}$$ Simplify.

 $$\sin(\theta)=\frac{\text{opp}}{\text{hyp}}$$ $$\cos(\theta)=\frac{\text{adj}}{\text{hyp}}$$ $$\csc(\theta)=\frac{\text{hyp}}{\text{opp}}$$ $$\sec(\theta)=\frac{\text{hyp}}{\text{adj}}$$ $$\tan(\theta)=\frac{\text{opp}}{\text{adj}}$$ $$\cot(\theta)=\frac{\text{adj}}{\text{opp}}$$

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 $$(\text{sides}-2)·\pi$$

Equilateral

Equiangular

Regular

Irregular

Convex & Concave

• A = area
• n = number of sides
• s = length of sides
• h = apothem
 Perimeter $$n·s$$ Area $$\frac{n·s·h}{2}$$ Sum of angles $$(n-2)·\pi$$ Angle (h,r) $$\frac{\pi}{n}$$ 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 $$A=n·r^2·\sin \Big(\frac{\pi}{n}\Big)·\cos \Big(\frac{\pi}{n}\Big)$$

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.
 $$\sec \Big(\frac{\pi}{n}\Big)=\frac{r}{h}$$ $$\csc \Big(\frac{\pi}{n}\Big)=\frac{r}{s/2}$$

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. The result is equal to the basic area function. $$n·h^2·\tan \Big(\frac{\pi}{n}\Big)=n·\frac{s·h}{2}$$

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
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<0, the function is an ellipse
• If Δ<0, B=0, and A<C, the function is an ellipse with a horizontal major axis
• If Δ<0, B=0, and A>C, the function is an ellipse with a vertical major axis
• If Δ<0, B=0, and A=C, the function is a circle
• If Δ<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>0, the function is a hyperbola
• If Δ>0 and A+C=0, the function is a rectangular hyperbola, meaning the asymptotes are perpendicular

Eccentricity
Eccentricity is a function's deviation from being circular
• 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 $$\frac{\theta·r^2}{2}$$ Chord length (k) $$2·r·\sin\Big(\frac{\theta}{2}\Big)$$ Segment area $$\frac{r^2}{2}·\Big(\theta-\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)\Big)$$ 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 $$\text{None}$$

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 proportion 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 proportion to it and its angle $$\frac{A_S}{\pi·r^2}=\frac{\theta}{2·\pi}$$

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

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 trig definitions with respect to the radius to find the base and height
 $$b=r·\sin\Big(\frac{\theta}{2}\Big)$$ $$h=r·\cos\Big(\frac{\theta}{2}\Big)$$
Substitute the triangular area using these values (b·h)/2 $$A_S=\frac{\theta·r^2}{2}-\frac{r^2}{2}·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)$$ Factor $$A_S=\frac{r^2}{2}·\Big(\theta-\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)\Big)$$

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 $${\Big(x+\frac{D}{2·A}\Big)}^2+{\Big(y+\frac{E}{2·A}\Big)}^2=\frac{D^2+E^2-4·A·F}{4·A^2}$$ Apply the coefficient method of solving

Annulus

Universal Properties

Standard-Conic Conversion

Conic-Standard Conversion

Universal Properties

Conic-Standard Conversion

Standard-Vertex Conversion

Standard-Intercept Conversion

Universal Properties

Conversions

Cofunctions

Complementary Angles

Inverse Cofunctions

Inverse Complementary Angles

Examples

Proof of Sine & Cosine

Proof of Tangent & Secant

Proof of Cotangent & Cosecant

Proof

Proof

Proof of Sine & Cosine

Proof of Tangent

Proof of Cotangent

Proofs

Proof

Proof

Proof of Sine

Proof of Cosine

Proof of Tangent

Proof of Cotangent

Proof

Proof of Sine Limit

Proof of Cosine Limit

e Exponential Limit

Natural Logarithm

Limits of Natural Exponents & Logarithms

Definition

Proof of Ratios

Fibonacci Series & φ Limit

Golden Ratio Powers

Algebraic Functions for the Fibonacci Series

Proof

Proof

Proof

Proof

Proof

Proof

Proof

The indeterminate form 0·∞

The indeterminate form ∞–∞

Exponential Indeterminate Forms

Proof

Proof

Proof

Proof

Proof

Higher Order Sine & Cosine Derivatives

Proof of Sine

Proof of Cosine

Proof of Tangent

Proof of Cotangent

Proof of Secant

Proof of 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

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

xn
1/x
ba•x & ea•x
logbx & ln(x)

sin(a·x)
cos(a·x)
csc(a·x)
sec(a·x)
tan(a·x)
cot(a·x)

sin–1(a·x)
cos–1(a·x)
csc–1(a·x)
sec–1(a·x)
tan–1(a·x)
cot–1(a·x)

sin2(a·x)
cos2(a·x)
csc2(a·x)
sec2(a·x)
tan2(a·x)
cot2(a·x)

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)

Exponential Reduction for xn·sin(a·x)
Exponential Reduction for xn·cos(a·x)

x²+a²
x²–a²
a²–x²

x·√x²+a²
x·√x²–a²
x·√a²–x²

x²·√x²+a²
x²·√x²–a²
x²·√a²–x²

x²+a²/x
x²–a²/x
a²–x²/x

x²+a²/x²
x²–a²/x²
a²–x²/x²

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 February 2024
Bill Liam East