Algebra 2 / Unit 8: Trigonometric Functions
8.2 The Unit Circle
Notes
The unit circle has radius 1 and center $(0, 0)$. An angle $\theta$ meets it at the point $(\cos\theta, \sin\theta)$.
- Memorize the first-quadrant points: $0 \to (1, 0)$, $\displaystyle \frac{\pi}{6} \to \left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, $\displaystyle \frac{\pi}{4} \to \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, $\displaystyle \frac{\pi}{3} \to \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$, $\displaystyle \frac{\pi}{2} \to (0, 1)$.
- Other quadrants use the same numbers with signs from the quadrant.
- Example: $\displaystyle \frac{5\pi}{6}$ is in Quadrant II, so its point is $\displaystyle \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$.
Practice
Which point is on the unit circle at $\displaystyle \theta = \frac{\pi}{3}$?
$\displaystyle \cos\frac{\pi}{3} = \frac{1}{2}$ is the $x$-coordinate and $\displaystyle \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ is the $y$-coordinate.
Evaluate $\cos \pi$.
The angle $\pi$ lands at $(-1, 0)$.
Evaluate $\displaystyle \sin\frac{7\pi}{6}$.
$\displaystyle \frac{7\pi}{6}$ is in Quadrant III with reference angle $\displaystyle \frac{\pi}{6}$, and sine is negative there.
Ready for more? Take the Unit 8 test.