Notes
Fundamental identities
- Reciprocal: $\displaystyle \csc = \frac{1}{\sin}$, $\displaystyle \sec = \frac{1}{\cos}$, $\displaystyle \cot = \frac{1}{\tan}$. Quotient: $\displaystyle \tan = \frac{\sin}{\cos}$.
- Pythagorean: $\sin^2 x + \cos^2 x = 1$, $1 + \tan^2 x = \sec^2 x$, $1 + \cot^2 x = \csc^2 x$.
- To verify an identity, transform one side into the other. Convert to sine and cosine, factor, or use a conjugate. A test value never proves an identity.
Sum and difference
- $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$. $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$.
- $\displaystyle \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$.
Double and half angles
- $\sin 2\theta = 2\sin\theta\cos\theta$. $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$. $\displaystyle \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$.
- $\displaystyle \sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}$ and $\displaystyle \cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}$, with the sign set by the quadrant of $\displaystyle \frac{\theta}{2}$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$1 + \cot^2 x = \csc^2 x$, and $\sin x \csc^2 x = \csc x$.
$\displaystyle \sin(60^\circ + 45^\circ) = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2}$.
$\displaystyle \tan 2\theta = \frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{24}{7}$.
$\cos 2x = 2\cos^2 x - 1$. The expression $2\sin^2 x - 1$ is the negative of $\cos 2x$.
$1 - \cos^2 x = \sin^2 x$, so the expression is $\displaystyle \frac{\sin x}{\sin^2 x} = \frac{1}{\sin x}$.
This is $\cos(2 \cdot 15^\circ) = \cos 30^\circ$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Use identities to find exact values.
(a) Use a difference formula to find the exact value of $\cos 15^\circ$.
Show the answer and scoring guide
$\displaystyle \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$.
Check off the points you earned:
(b) Suppose $\sin\theta = \frac{5}{13}$ and $\theta$ is in Quadrant II. Find $\cos\theta$ and $\sin 2\theta$.
Show the answer and scoring guide
$\displaystyle \cos\theta = -\frac{12}{13}$ because cosine is negative in Quadrant II. Then $\displaystyle \sin 2\theta = 2 \cdot \frac{5}{13} \cdot \left(-\frac{12}{13}\right) = -\frac{120}{169}$.
Check off the points you earned:
(c) Find $\cos 2\theta$ for the same angle and state the quadrant of $2\theta$.
Show the answer and scoring guide
$\displaystyle \cos 2\theta = 1 - 2\sin^2\theta = 1 - 2 \cdot \frac{25}{169} = \frac{119}{169}$. Cosine is positive and sine is negative, so $2\theta$ is in Quadrant IV.
Check off the points you earned: