-# Circular functions
-
-## Radians and degrees
+---
+geometry: a4paper, margin=2cm
+columns: 2
+author: Andrew Lorimer
+header-includes:
+- \usepackage{setspace}
+- \usepackage{fancyhdr}
+- \usepackage{graphicx}
+- \pagestyle{fancy}
+- \fancyhead[LO,LE]{Year 12 Methods}
+- \fancyhead[CO,CE]{Andrew Lorimer}
+---
+
+\setstretch{1.2}
+\pagenumbering{gobble}
-$$1 \thinspace \operatorname{rad}={{180 \operatorname{deg}}\over \pi}$$
+# Circular functions
## Exact values
+\includegraphics[width=0.2\textwidth]{./graphics/exact-values-1.png}
+\includegraphics[width=0.2\textwidth]{./graphics/exact-values-2.png}
+$$1 \thinspace \operatorname{rad}={{180 \operatorname{deg}}\over \pi}$$
## $\sin$ and $\cos$ graphs
$$f(x)=a \cos(bx-c)+d$$
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)
+
+- $a$ 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
+Graph oscillates between $+a$ and $-a$ in $y$-axis
$a=0$ produces straight line
-$a\lt0$ inverts the phase ($\sin$ becomes $\cos$, vice vera)
+
+$a < 0$ 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
+
+$b<0$ 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
$$\sin\theta=-\cos(\theta+{\pi \over 2})$$
$$\cos\theta=\sin(\theta+{\pi \over 2})$$
-## $tan$ graph
+## $\tan$ 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 $x={{(2k+1)\pi}\over 2n},\quad k \in \mathbb{Z}$
+
+- $a$ 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**
+
+## Solving trig equations
+
+1. Solve domain for $n\theta$
+2. Find solutions for $n\theta$
+3. Divide solutions by $n$
+
+$\sin2\theta={\sqrt{3}\over2}, \quad \theta \in[0, 2\pi] \quad(\therefore 2\theta \in [0,4\pi])$
+
+$2\theta=\sin^{-1}{\sqrt{3} \over 2}$
+
+$2\theta={\pi\over 3}, {2\pi \over 3}, {7\pi \over 3}, {8\pi \over 3}$
+
+$\therefore \theta = {\pi \over 6}, {\pi \over 3}, {7 \pi \over 6}, {4\pi \over 3}$