$$f(x)=a \cos(bx-c)+d$$
where
-$a$ is the amplitude
-$b$ is the $x$-dilation
-$c$ is the $y$-shift
+$a$ is the $y$-dilation (amplitude)
+$b$ is the $x$-dilation (period)
+$c$ is the $x$-shift (phase)
+$d$ is the $y$-shift (equilibrium position)
-Period is ${2 \pi} \over b$
Domain is $\mathbb{R}$
Range is $[-b+c, b+c]$;
Graph of $\cos(x)$ starts at $(0,1)$. Graph of $\sin(x)$ starts at $(0,0)$.
+<<<<<<< HEAD
**Mean / equilibrium:** line that the graph oscillates around ($y=d$)
## Solving trig equations
$2\theta=\sin^{-1}{\sqrt{3} \over 2}$
$2\theta={\pi\over 3}, {2\pi \over 3}, {7\pi \over 3}, {8\pi \over 3}$
$\therefore \theta = {\pi \over 6}, {\pi \over 3}, {7 \pi \over 6}, {4\pi \over 3}$
+=======
+### 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
+
+$$\sin(\theta+{\pi\over 2})=\sin\theta$$
+$$\sin(\theta+\pi)=-\sin\theta$$
+
+$$\cos(\theta+{\pi \over 2})=-\cos\theta$$
+$$\cos(\theta+\pi)=-cos(\theta+{3\pi \over 2})=\cos(-\theta)$$
+
+## Pythagorean identity
+
+$$\cos^2\theta+\sin^2\theta=1$$
+
+## Complementary relationships
+
+$$\sin({\pi \over 2} - \theta)=\cos\theta$$
+$$\cos({\pi \over 2} - \theta)=\sin\theta$$
+
+$$\sin\theta=-\cos(\theta+{\pi \over 2})$$
+$$\cos\theta=\sin(\theta+{\pi \over 2})$$
+
+## $tan$ graph
+
+$$y=a\tan(nx)$$
+
+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 $x={{(2k+1)\pi}\over 2},\quad k \in \mathbb{Z}$
+>>>>>>> 924c0548b3e7564d4015e879c56a46a5606807fe