Math Identities Proofs

Repeating values can always be represented as a fraction with the repeated value over an equal number of digits of 9’s. $$.\overline{2}=\frac{2}{9}\qquad.\overline{137}=\frac{137}{999}$$ $$.\overline{142857}=\frac{142857}{999999}=\frac{1}{7}$$

Where a number starts repeating is a matter of multiples of 10, which can be isolated. $$433.\overline{3}=400+\frac{3}{9}·100=\frac{1300}{3}$$ $$.35\overline{7}=\frac{35}{100}+\frac{7}{9}·\frac{1}{100}=\frac{161}{450}$$ $$3.8\overline{3}=.5+3.\overline{33}=\frac{1}{2}+\frac{30}{9}=\frac{23}{6}$$

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 $9$

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) $$x=\frac{-b \pm \sqrt{b^2+4·a·(y-y_∘)}}{2·a}$$

The conic sections general formula describes ellipses, circles, parabolas, hyperbolas, and lines. $$A·x^2+B·x·y+C·y^2+D·x+E·y+F=0$$

Multiplication is reversible with division, therefore exponential reduction yields inverses once reaching negatives.

$$x^3=1·x·x·x$$	$$3^3=27$$
$$x^2=1·x·x$$	$$3^2=9$$
$$x=1·x$$	$$3^1=3$$
$$x^0=1$$	$$3^0=1$$
$$x^{-1}=1/x$$	$$3^{-1}=1/3$$
$$x^{-2}=1/(x·x)$$	$$3^{-2}=1/9$$
$$x^{-3}=1/(x·x·x)$$	$$3^{-3}=1/27$$

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 $$2^3·2^{-5}=2^{3-5}=2^{-2}=1/4$$

Distribute the exponential $c$ by expanding into ${(x^a)}^c·{(y^{±b})}^c$. The terms are multiplied $c$ times

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 $3·x–5·y=2$ $$\frac{2^{3·x}}{2^{5·y}}=2^{3·x-5·y}=2^2$$

Even though $1^2=1$ and $(-1)^2=1$, radicands cannot be factored into two negatives, because it creates the logical fallacy in the example below

$\qquad 2=2$
$\qquad 2=1+1$
$\qquad 2=1+\sqrt{1}$
$\qquad 2=1+\sqrt{-1·-1}$
$\qquad 2=1+\sqrt{-1}·\sqrt{-1}$
$\qquad 2=1+(\sqrt{-1})^2$
$\qquad 2=1-1$
$\qquad 2=0$
$\qquad $No solution

$$5-3·i+(-7+9·i)=-2+6·i$$

$$\frac{(x-4)·(x+2)}{(x-4)·(x-2)}=\frac{(x+2)}{(x-2)}$$ $$\frac{(x-4)·(x+2)}{(4-x)·(x-2)}=-\frac{(x+2)}{(x-2)}$$

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 $2·a$

Solve for $x$ and $y$ $$x+y=1\medspace\land\medspace x·y=1$$ Isolate $y$ in the first equation $$y=1-x$$ Substitute for $y$ in the second equation $$x·(1-x)=1$$ Expand $$x-x^2=1$$ Rearrange into a standard quadratic equation $$-x^2+x-1=0$$ Negate $$x^2-x+1=0$$ Use the quadratic formula to find the zeros of $x$ $$x=\frac{-(-1) \pm \sqrt{(-1)^2-4·1·1}}{2·1}$$ Simplify in expanded form $$x=\frac{1}{2}\pm\frac{\sqrt{3}}{2}·i$$ Substitute for $x$ in the first equation with $y$ isolated $$y=1-\bigg(\frac{1}{2}\pm\frac{\sqrt{3}}{2}·i\bigg)$$ Simplify in expanded form $$y=\frac{1}{2}\mp\frac{\sqrt{3}}{2}·i$$ The solutions for $(x,y)$ are $$\bigg(\frac{1}{2}\pm\frac{\sqrt{3}}{2}·i,\frac{1}{2}\mp\frac{\sqrt{3}}{2}·i\bigg)$$

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

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 $N(x)=D(x)·Q(x)+R(x)$ $$x^3+2·x^2+12=(x-2)(x^2+4·x+8)+28$$

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 $$\frac{5·x}{(x-1)(x^2+2){(x+7)}^2}=\frac{A}{x-1}+\frac{B·x+C}{x^2+2}+\frac{D}{x+7}+\frac{E}{{(x+7)}^2}$$

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

Logarithms are the inverse functions of exponents, therefore operating a logarithm in an exponent with the same base cancels both

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

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

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

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

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

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

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 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	$m_1=-1/m_2$

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.

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 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 a matrix to display the values for $(1·x,1·y)$ in the column with constants. If this is unattainable, then the lines are parallel.

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 $(x,y)=(3,8)$

Given side length $L$ and height $h$, the triangle area equals the sum of the inner triangle areas with altitudes $a$, $b$, and $c$ $$\frac{L·h}{2}=\frac{L·a}{2}+\frac{L·b}{2}+\frac{L·c}{2}$$ Multiply by $2/L$

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

Given a triangle with sides $\{a,b,c\}$, divide it with an orthogonal line $h$ to form two right triangles, dividing line $b$ into lines $x$ and $b-x$ Using the right angle theorem to determine the relationships and length of $x$ $$x^2=c^2-h^2\therefore x=\sqrt{c^2-h^2}$$ Apply the right angle theorem to determine the relationships of $b-x$ $$(b-x)^2=a^2-h^2$$ Expand $$b^2-2·b·x+x^2-h^2$$ Substitute $x$ and $x^2$ $$b^2-2·b·\sqrt{c^2-h^2}+c^2-h^2=a^2-h^2$$ Simplify $$b^2-2·b·\sqrt{c^2-h^2}+c^2=a^2$$ Isolate $a^2$, $b^2$ and $c^2$ to one side $$b^2+c^2-a^2=2·b·\sqrt{c^2-h^2}$$ Square $$\big(b^2+c^2-a^2\big)^2=4·b^2·(c^2-h^2)$$ Isolate $h^2$ $$h^2=c^2-\frac{\big(b^2+c^2-a^2\big)^2}{4·b^2}$$ Factor the right into a single fraction $$h^2=\frac{4·b^2·c^2-\big(b^2+c^2-a^2\big)}{4·b^2}$$ Factor the numerator as a squared binomial $$h^2=\frac{(2·b·c+b^2+c^2-a^2)(2·b·c-b^2-c^2+a^2)}{4·b^2}$$ Factor the terms with $b$ and $c$ as squared binomials $$h^2=\frac{\big((b+c)^2-a^2\big)\big(a^2-(b-c)^2\big)}{4·b^2}$$ Factor each term in the numerator as squared binomials $$h^2=\frac{(b+c+a)(b+c-a)(a+b-c)(a-b+c)}{4·b^2}$$ For each negative variable within a term, both add and subtract the variable in the term $$h^2=\frac{(b+c+a)(b+c+a-2·a)(a+b+c-2·c)(a+b-2·b+c)}{4·b^2}$$ Take the square root $$h=\frac{\sqrt{(b+c+a)(b+c+a-2·a)(a+b+c-2·c)(a+b-2·b+c)}}{2·b}$$ Multiply by b/2 to obtain the triangle area $$A=\frac{\sqrt{(b+c+a)(b+c+a-2·a)(a+b+c-2·c)(a+b-2·b+c)}}{4}$$ Factor the $4$ into the radicand and expand it into the terms $$A=\sqrt{\bigg(\frac{b+c+a}{2}\bigg)\bigg(\frac{b+c+a}{2}-a\bigg)\bigg(\frac{a+b+c}{2}-c\bigg)\bigg(\frac{a+b+c}{2}-b\bigg) }$$ Substitute $(a+b+c)/2$ for $s$

Apply the right angle theorem to $(x_1,y_1)$ and $(x_2,y_2)$

Two simple triangles help identify certain values

Function	$$0$$	$$\pi/6$$	$$\pi/4$$	$$\pi/3$$	$$\pi/2$$
$\sin(\theta)$	$$0$$	$$1/2$$	$$1/\sqrt{2}$$	$$\sqrt{3}/2$$	$$1$$
$\cos(\theta)$	$$1$$	$$\sqrt{3}/2$$	$$1/\sqrt{2}$$	$$1/2$$	$$0$$
$\csc(\theta)$	$$∞$$	$$2$$	$$\sqrt{2}$$	$$2/\sqrt{3}$$	$$1$$
$\sec(\theta)$	$$1$$	$$2/\sqrt{3}$$	$$\sqrt{2}$$	$$2$$	$$∞$$
$\tan(\theta)$	$$0$$	$$1/\sqrt{3}$$	$$1$$	$$\sqrt{3}$$	$$∞$$
$\cot(\theta)$	$$∞$$	$$\sqrt{3}$$	$$1$$	$$1/\sqrt{3}$$	$$0$$

With respect to the center angle between the apothem and circumradius, use the corresponding right angle definitions for $h$ = adjacent, $b/2$ = opposite, and $r$ = hypotenuse, and solve for $r$ in each case

$$\sec \Big(\frac{\pi}{n}\Big)=\frac{r}{h}$$

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

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 right angle 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 $$n·h^2·\tan \Big(\frac{\pi}{n}\Big)=n·\frac{s·h}{2}$$

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 right angle 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 $$A=n·r^2·\sin\Big(\frac{\pi}{n}\Big)·\cos\Big(\frac{\pi}{n}\Big)$$

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 right angle 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 $$A=\frac{n·s^2}{4}·\cot\Big(\frac{\pi}{n}\Big)$$

$$A⋅x^2+B⋅x⋅y+C⋅y^2+D⋅x+E⋅y+F=0$$ All conic sections will be represented without rotation until the section on rotations. In other words, $B=0$.

$$\Delta=B^2-4⋅A⋅C$$ 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

A focus/foci is a point or set of points around which a curve is guided A Directrix is a fixed line perpendicular to the (major) axis of a function, which is determined by its focus and curvature Eccentricity is a function's deviation from being circular, and a constant ratio 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

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	$A⋅x^2+A⋅y^2+1=0$
Point	$A⋅x^2+A⋅y^2=0$
Line	$D⋅x+E⋅y+F=0$
Intersecting lines	$x^2-y^2=0$
Parallel lines	$x^2-1=0$

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	$x_∘=-D/(2·A)$ $y_∘=-E/(2·A)$ $r^2=(D^2+E^2-4·A·F)/(4·A^2)$
Focus Coordinates	$(x_∘,y_∘)$
Eccentricity	$0$
Directrix	None

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

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

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

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 $2·r$

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 right angle 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 $$A_S=\frac{r^2}{2}·\Big(\theta-2·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)\Big)$$

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

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	$\theta·(R^2-r^2)/2$

Closed ovular curves whose points are the result of a constant sum between two internal points

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	$A=b^2$ $C=a^2$ $D=-2·b^2·x_∘$ $E=-2·a^2·y_∘$ $F=b^2·{x_∘}^2+a^2·{y_∘}^2-a^2·b^2$ $a^2=C$ $b^2=A$ $x_∘=-D/(2·A)$ $y_∘=-E/(2·C)$
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	$x=\pm a/e,a>b$ $y=\pm b/e,b>a$

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

Open mirrored curves whose points are the same distance between a common internal point and an exterenal line.

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	$y=-f$

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

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

Given the quadratic formula with $y=0$, find the zeros of $x$

A mirrored set of open mirrored curvers whose points are the difference between two common internal points.

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	$A=b^2$ $C=a^2$ $D=-2·b^2·x_∘$ $E=-2·a^2·y_∘$ $F=b^2·{x_∘}^2+a^2·{y_∘}^2-a^2·b^2$ $a^2=C$ $b^2=A$ $x_∘=-D/(2·A)$ $y_∘=-E/(2·C)$
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	$y= \pm b·(x-x_∘)/a+y_∘$

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.

$$\sin^{-1}(-\theta)=-\sin^{-1}(\theta)\qquad\cos^{-1}(-\theta)=\pi-\cos^{-1}(\theta)$$ $$\csc^{-1}(-\theta)=-\csc^{-1}(\theta)\qquad\sec^{-1}(-\theta)=\pi-\sec^{-1}(\theta)$$ $$\tan^{-1}(-\theta)=-\tan^{-1}(\theta)\qquad\cot^{-1}(-\theta)=\pi-\cot^{-1}(\theta)$$

$$\sin(\theta)=\frac{1}{\csc(\theta)} \qquad \cos(\theta)=\frac{1}{\sec{\theta}}$$ $$\tan(\theta)=\frac{\sin{\theta}}{\cos{\theta}}=\frac{1}{\cot{\theta}}$$

$$\sin\bigg(\frac{\pi}{2}\pm\theta\bigg)=\pm\cos(\theta)\qquad\cos\bigg(\frac{\pi}{2}\pm\theta\bigg)=\mp\sin(\theta)$$ $$\csc\bigg(\frac{\pi}{2}\pm\theta\bigg)=\pm\sec(\theta)\qquad\sec\bigg(\frac{\pi}{2}\pm\theta\bigg)=\mp\csc(\theta)$$ $$\tan\bigg(\frac{\pi}{2}\pm\theta\bigg)=\mp\cot(\theta)\qquad\cot\bigg(\frac{\pi}{2}\pm\theta\bigg)=\mp\tan(\theta)$$

$$\sin^{-1}(1/\theta)=\csc^{-1}(\theta),\medspace |\theta| \ge 1$$ $$\cos^{-1}(1/\theta)=\sec^{-1}(\theta),\medspace |\theta| \ge 1$$ $$\tan^{-1}(1/\theta)=\cot^{-1}(\theta),\medspace \phantom{|}\theta\phantom{|} \gt 0$$

$$\sin^{-1}(\theta)+\cos^{-1}(\theta)=\pi/2,\medspace |\theta| \le 1$$ $$\csc^{-1}(\theta)+\sec^{-1}(\theta)=\pi/2 \phantom{,\medspace |\theta| \le 1}$$ $$\tan^{-1}(\theta)+\cot^{-1}(\theta)=\pi/2,\medspace |\theta| \ge 1$$

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 definition

Given the triangle area, substitute the height orthogonal to 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 $2/(a·b·c)$

Insert an orthogonal 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 $c^2$

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 unit circle 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 identity 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 $a$ $$\cos\bigg(\frac{\pi}{2}-a-b\bigg)=\cos\bigg(\frac{\pi}{2}-a\bigg)·\cos(b)-\sin\bigg(\frac{\pi}{2}-a\bigg)·\sin(b)$$ 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 $$\sin(a+b)=\sin(a)·\cos(b)+\cos(a)·\sin(b)$$

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 for sine and cosine $$\tan(a\pm b)=\frac{\sin(a)·\cos(b)\pm\cos(a)·\sin(b)}{\cos(a)·\cos(b)\mp\sin(a)·\sin(b)}$$ 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 $$\tan(a\pm b)=\frac{\sin(a)/\cos(a)\pm\sin(b)/\cos(b)}{1\mp\big(\sin(a)·\sin(b)\big)/\big(\cos(a)·\cos(b)\big)}$$ Substitute the tangent cofunction for the sines over cosines

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 for cosine and sine $$\cot(a\pm b)=\frac{\cos(a)·\cos(b)\mp\sin(a)·\sin(b)}{\sin(a)·\cos(b)\pm\cos(a)·\sin(b)}$$ Divide the numerator and denominator by $\sin(a)·\sin(b)$ $$\cot(a\pm b)= \frac{\big(\cos(a)·\cos(b)\mp\sin(a)·\sin(b)\big)/\big(\sin(a)·\sin(b)\big)} {\big(\sin(a)·\cos(b)\pm\cos(a)·\sin(b)\big)/\big(\sin(a)·\sin(b)\big)}$$ Cancel like terms $$\cot(a\pm b)=\frac{\big(\cos(a)·\cos(b)\big)/\big(\sin(a)·\sin(b)\big)\mp 1}{\cos(b)/\sin(b)\pm\cos(a)/\sin(a)}$$ Substitute the cotangent cofunction for the cosines over sines

In each case, substitute the sum and difference identities for the terms on the right, and simplify

Using the angles from the product-to-sum identities, let $a=(u+v)/2$ and $b=(u-v)/2$ so that $$a+b=\frac{u+v}{2}+\frac{u-v}{2}=u$$ $$a-b=\frac{u+v}{2}-\frac{u-v}{2}=v$$ Substitute the terms with $a$ and $b$ for the terms with $u$ and $v$ into the product-to-sum identities

In each case, start with the sum identity for the corresponding function, and use the same angle for both angles. For secant and cosecant, use $\cos(a\pm b)^{-1}$ and $\sin(a\pm b)^{-1}$ respectively.

The circle segment area expression can be simplified by using the sine double angle identity $$\frac{r^2}{2}·\Big(\theta-2·\sin\Big(\frac{\theta}{2}\Big)·\cos\Big(\frac{\theta}{2}\Big)\Big)$$ $$\therefore\frac{r^2}{2}·\big(\theta-\sin(\theta)\big)$$

Given the cosine double angle identity, use $\theta$ for $2·\theta$ and $\theta/2$ for $\theta$ $$\cos(\theta)=\cos^2\bigg(\frac{\theta}{2}\bigg)-\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Substitute the right angle identity for $\cos^2(\theta)$ $$\cos(\theta)=1-2·\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Subtract 1 $$\cos(\theta)-1=-2·\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Divide by –2 $$\frac{1-\cos(\theta)}{2}=\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Take the square root

Given the cosine double angle identity, use $\theta$ for $2·\theta$ and $\theta/2$ for $\theta$ $$\cos(\theta)=\cos^2\bigg(\frac{\theta}{2}\bigg)-\sin^2\bigg(\frac{\theta}{2}\bigg)$$ Substitute the right angle identity for $\sin^2(\theta)$ $$\cos(\theta)=2·\cos^2\bigg(\frac{\theta}{2}\bigg)-1$$ Add 1 $$1+\cos(\theta)=2·\cos^2\bigg(\frac{\theta}{2}\bigg)$$ Divide by 2 $$\frac{1+\cos(\theta)}{2}=\cos^2\bigg(\frac{\theta}{2}\bigg)$$ Take the square root

Divide the sine half angle identity by the cosine half angle identity $$\bigg|\sin\bigg(\frac{\theta}{2}\bigg)\bigg|\bigg/\bigg|\cos\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1-\cos(\theta)}{2}}\bigg/\sqrt{\frac{1+\cos(\theta)}{2}} $$ Substitute sine over cosine for cofunction $$\bigg|\tan\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1-\cos(\theta)}{2}}\bigg/\sqrt{\frac{1+\cos(\theta)}{2}}$$ Factor the root $$\bigg|\tan\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1-\cos(\theta)}{2}\bigg/\frac{1+\cos(\theta)}{2}}$$ Simplify

Divide the cosine half angle identity by the sine half angle identity $$\bigg|\cos\bigg(\frac{\theta}{2}\bigg)\bigg|\bigg/\bigg|\sin\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1+\cos(\theta)}{2}}\bigg/\sqrt{\frac{1-\cos(\theta)}{2}} $$ Substitute cosine over sine for its cofunction $$\bigg|\cot\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1+\cos(\theta)}{2}}\bigg/\sqrt{\frac{1-\cos(\theta)}{2}}$$ Factor the root $$\bigg|\cot\bigg(\frac{\theta}{2}\bigg)\bigg|=\sqrt{\frac{1+\cos(\theta)}{2}\bigg/\frac{1-\cos(\theta)}{2}}$$ Simplify

In each case, start with the half angle identity for the corresponding function, use $\theta$ for $2·\theta$ and $\theta/2$ for $\theta$, and square the equations

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 $$\lim_{\theta→0^-}\frac{\sin(\theta)}{\theta}=\lim_{\theta→0^+}\frac{\sin(\theta)}{\theta}=\lim_{\theta→0}\frac{\sin(\theta)}{\theta}=1$$

$$\lim_{x→0}\frac{e^x-1}{x}=1$$ From $$e=\lim_{x→0}{(1+x)}^{1/x}$$

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

$$\lim_{x→\pm ∞}e^{\pm x}=∞$$ $$\lim_{x→\pm ∞}e^{\mp x}=0$$

Two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger two quantities $$\frac{x+a}{x}=\frac{x}{a}$$

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 $$x^2-x-1=0$$

A constant has a slope of zero at every point, therefore its derivative is zero at every point