Graph of $\cos(x)$ starts at $(0,1)$. Graph of $\sin(x)$ starts at $(0,0)$.
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+**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 $n$
+
+$\sin2\theta={\sqrt{3}\over2}, \quad \theta \in[0, 2\pi] \quad(\therefore 2\theta \in [0,4\pi])$
+$2\theta=\sin^{-1}{\sqrt{3} \over 2}$
+$2\theta={\pi\over 3}, {2\pi \over 3}, {7\pi \over 3}, {8\pi \over 3}$
+$\therefore \theta = {\pi \over 6}, {\pi \over 3}, {7 \pi \over 6}, {4\pi \over 3}$
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+ ### 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 $x={{(2k+1)\pi}\over 2},\quad k \in \mathbb{Z}$
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