Notes
Angles and the unit circle
- $180^\circ = \pi$ radians. Arc length is $s = r\theta$ with $\theta$ in radians.
- On the unit circle the point at angle $\theta$ is $(\cos\theta, \sin\theta)$. Key values: $\displaystyle \sin\frac{\pi}{6} = \frac{1}{2}$, $\displaystyle \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\displaystyle \tan\frac{\pi}{3} = \sqrt{3}$.
- Use reference angles and quadrant signs for other angles. In a right triangle, SOH-CAH-TOA.
Graphs
- For $y = A\sin(B(x - C)) + D$: amplitude $|A|$, period $\displaystyle \frac{2\pi}{|B|}$, phase shift $C$, midline $y = D$.
- $\tan x$ has period $\pi$ and vertical asymptotes at $\displaystyle \frac{\pi}{2} + k\pi$.
Inverse trig
- $\arcsin$ has range $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\arccos$ has range $[0, \pi]$, $\arctan$ has range $\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle 300 \cdot \frac{\pi}{180} = \frac{5\pi}{3}$.
The reference angle is $\displaystyle \frac{\pi}{6}$ and sine is negative in quadrant IV.
$\displaystyle \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.
$\displaystyle \frac{2\pi}{4} = \frac{\pi}{2}$.
$\displaystyle \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\displaystyle \frac{\pi}{6}$ is in $[0, \pi]$.
$\displaystyle s = r\theta = 8 \cdot \frac{\pi}{4} = 2\pi$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Consider $f(x) = 2\sin(3x - \pi) + 1$.
(a) Rewrite $f$ in the form $2\sin(3(x - C)) + 1$ and give the amplitude, period, phase shift and midline.
Show the answer and scoring guide
$\displaystyle 3x - \pi = 3\left(x - \frac{\pi}{3}\right)$. Amplitude 2, period $\displaystyle \frac{2\pi}{3}$, phase shift $\displaystyle \frac{\pi}{3}$ to the right, midline $y = 1$.
Check off the points you earned:
(b) State the maximum and minimum values of $f$.
Show the answer and scoring guide
The maximum is $1 + 2 = 3$ and the minimum is $1 - 2 = -1$.
Check off the points you earned:
(c) Evaluate $f\left(\frac{\pi}{2}\right)$ and $f\left(\frac{\pi}{3}\right)$.
Show the answer and scoring guide
$\displaystyle f\left(\frac{\pi}{2}\right) = 2\sin\left(\frac{3\pi}{2} - \pi\right) + 1 = 2\sin\frac{\pi}{2} + 1 = 3$. $\displaystyle f\left(\frac{\pi}{3}\right) = 2\sin(\pi - \pi) + 1 = 1$.
Check off the points you earned: