Circular functions

Radians and degrees

Exact values

$\sin$ and $\cos$ graphs

where
$a$ is the $y$-dilation (amplitude)
$b$ is the $x$-dilation (period)
$c$ is the $x$-shift (phase)
$d$ is the $y$-shift (equilibrium position)

Domain is $\mathbb{R}$
Range is $[-b+c, b+c]$;

Graph of $\cos(x)$ starts at $(0,1)$. Graph of $\sin(x)$ starts at $(0,0)$.

Mean / equilibrium: line that the graph oscillates around ($y=d$)

Solving trig equations

1. Solve domain for $n\theta$
2. Find solutions for $n\theta$
3. Divide solutions by $n$

$\sin2\theta={\sqrt{3}\over2}, \quad \theta \in[0, 2\pi] \quad(\therefore 2\theta \in [0,4\pi])$
$2\theta=\sin^{-1}{\sqrt{3} \over 2}$
$2\theta={\pi\over 3}, {2\pi \over 3}, {7\pi \over 3}, {8\pi \over 3}$
$\therefore \theta = {\pi \over 6}, {\pi \over 3}, {7 \pi \over 6}, {4\pi \over 3}$

Amplitude

Amplitude of $a$ means graph oscillates between $+a$ and $-a$ in $y$-axis

$a=0$ produces straight line
$a\lt0$ inverts the phase ($\sin$ becomes $\cos$, vice vera)

Period

Period $T$ is ${2 \pi}\over b$
$b=0$ produces straight line
$b\lt0$ inverts the phase

Phase

$c$ moves the graph left-right in the $x$ axis.
If $c=T={{2\pi}\over b}$, the graph has no actual phase shift.

Symmetry

Pythagorean identity

Complementary relationships

$\tan$ graph

where
$a$ is $x$-dilation (period)
$n$ is $y$-dilation ($\equiv$ amplitude)
period $T$ is $\pi \over n$
range is $R$
roots at $x={k\pi \over n}$
asymptotes at $x={{(2k+1)\pi}\over 2n},\quad k \in \mathbb{Z}$
Asymptotes should always have equations and arrow pointing up