Notes
Identities
- Pythagorean: $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, $1 + \cot^2\theta = \csc^2\theta$.
- To verify an identity, work on one side, often by rewriting in terms of $\sin$ and $\cos$.
- Sum: $\sin(A + B) = \sin A\cos B + \cos A\sin B$ and $\cos(A + B) = \cos A\cos B - \sin A\sin B$.
- Double angle: $\sin 2\theta = 2\sin\theta\cos\theta$, $\cos 2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$.
Solving equations
- Isolate the trig function, use reference angles and quadrant signs, and list all solutions in the interval.
- Factor instead of dividing by a trig expression. For multiple angles like $2x$, solve over the wider interval and then divide.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$1 + \cot^2 x = \csc^2 x$, and $\displaystyle \sin^2 x \cdot \frac{1}{\sin^2 x} = 1$.
$\displaystyle \sin(45^\circ - 30^\circ) = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2}$.
$\displaystyle \sin\theta = \frac{12}{13}$, so $\displaystyle \sin 2\theta = 2 \cdot \frac{12}{13}\cdot\left(-\frac{5}{13}\right) = -\frac{120}{169}$.
The correct form uses $\cos^2\theta$, not $\cos\theta$. At $\displaystyle \theta = \frac{\pi}{3}$ the left side is $\displaystyle -\frac{1}{2}$ but the right side is $0$.
Cosine is negative in quadrants II and III with reference angle $\displaystyle \frac{\pi}{3}$.
$\sin x(2\sin x - 1) = 0$, so $\sin x = 0$ or $\displaystyle \sin x = \frac{1}{2}$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Suppose $\cos\theta = \frac{4}{5}$ and $\theta$ is in quadrant IV.
(a) Find $\sin\theta$ and $\tan\theta$.
Show the answer and scoring guide
$\displaystyle \sin\theta = -\sqrt{1 - \frac{16}{25}} = -\frac{3}{5}$, so $\displaystyle \tan\theta = \frac{-3/5}{4/5} = -\frac{3}{4}$.
Check off the points you earned:
(b) Find $\sin 2\theta$ and $\cos 2\theta$.
Show the answer and scoring guide
$\displaystyle \sin 2\theta = 2\left(-\frac{3}{5}\right)\left(\frac{4}{5}\right) = -\frac{24}{25}$. $\displaystyle \cos 2\theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.
Check off the points you earned:
(c) Find $\tan 2\theta$ and verify with the double-angle tangent formula.
Show the answer and scoring guide
$\displaystyle \tan 2\theta = \frac{-24/25}{7/25} = -\frac{24}{7}$. Formula: $\displaystyle \frac{2(-3/4)}{1 - 9/16} = \frac{-3/2}{7/16} = -\frac{24}{7}$.
Check off the points you earned: