① Concept Review
Topic 1
Angle Measures & the Unit Circle
To convert degrees → radians: multiply by $\dfrac{\pi}{180}$
To convert radians → degrees: multiply by $\dfrac{180}{\pi}$
On the unit circle, a point is $(\cos\theta,\, \sin\theta)$
Example
Convert $135°$ to radians:
$135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4}$
$135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4}$
★ Must-Memorize: Key Trig Values
| Angle | Radians | sin | cos | tan |
|---|---|---|---|---|
| 0° | $0$ | $0$ | $1$ | $0$ |
| 30° | $\dfrac{\pi}{6}$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{\sqrt{3}}{3}$ |
| 45° | $\dfrac{\pi}{4}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $1$ |
| 60° | $\dfrac{\pi}{3}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ |
| 90° | $\dfrac{\pi}{2}$ | $1$ | $0$ | undef. |
| 180° | $\pi$ | $0$ | $-1$ | $0$ |
Topic 2
Trigonometric Identities
Pythagorean: $\sin^2\!\theta + \cos^2\!\theta = 1$
$\tan^2\!\theta + 1 = \sec^2\!\theta$ | $1 + \cot^2\!\theta = \csc^2\!\theta$
$\tan^2\!\theta + 1 = \sec^2\!\theta$ | $1 + \cot^2\!\theta = \csc^2\!\theta$
Reciprocal: $\sec\theta = \dfrac{1}{\cos\theta}$, $\csc\theta = \dfrac{1}{\sin\theta}$, $\cot\theta = \dfrac{1}{\tan\theta}$
Double Angle: $\sin 2\theta = 2\sin\theta\cos\theta$
$\cos 2\theta = \cos^2\!\theta - \sin^2\!\theta$
$\cos 2\theta = \cos^2\!\theta - \sin^2\!\theta$
Sum/Difference:
$\cos(A\pm B) = \cos A\cos B \mp \sin A\sin B$
$\sin(A\pm B) = \sin A\cos B \pm \cos A\sin B$
$\cos(A\pm B) = \cos A\cos B \mp \sin A\sin B$
$\sin(A\pm B) = \sin A\cos B \pm \cos A\sin B$
Example
Given $\sin\theta = \tfrac{3}{5}$ in Q I, find $\cos\theta$:
$\cos\theta = \sqrt{1 - \sin^2\!\theta} = \sqrt{1 - \tfrac{9}{25}} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}$
$\cos\theta = \sqrt{1 - \sin^2\!\theta} = \sqrt{1 - \tfrac{9}{25}} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}$
Topic 3
Graphing Trig Functions
General form: $y = A\sin(Bx - C) + D$ or $y = A\cos(Bx - C) + D$
Amplitude $= |A|$
Period (sin/cos) $= \dfrac{2\pi}{|B|}$ ; Period (tan) $= \dfrac{\pi}{|B|}$
Phase shift $= \dfrac{C}{B}$ (right if $C>0$)
Vertical shift $= D$ (midline $y = D$)
Example
For $y = 2\sin(3x)$: Amplitude $= 2$, Period $= \dfrac{2\pi}{3}$
Topic 4
Triangles, Law of Cosines & Area
Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$
Area: $\text{Area} = \dfrac{1}{2}ab\sin C$
Inverse trig: $\arcsin\!\left(\tfrac{1}{2}\right) = \dfrac{\pi}{6}$, $\arctan(1) = \dfrac{\pi}{4}$
Reference angle: Q2 → $180°-\theta$, Q3 → $\theta-180°$, Q4 → $360°-\theta$
Example
Triangle with $a=5$, $b=7$, $C=60°$:
$c^2 = 25 + 49 - 2(5)(7)(0.5) = 74 - 35 = 39$, so $c = \sqrt{39}$
$c^2 = 25 + 49 - 2(5)(7)(0.5) = 74 - 35 = 39$, so $c = \sqrt{39}$
0
/ 20
—
0
Correct
0
Wrong
0
Skipped
—
Time Used