Precalculus / Unit 4: Trigonometric Functions
4.6 Inverse Trigonometric Functions
Notes
To make trig functions invertible we restrict their domains.
- $\arcsin x$ has range $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. $\arccos x$ has range $[0, \pi]$. $\arctan x$ has range $\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
- $\displaystyle \arcsin\frac{1}{2} = \frac{\pi}{6}$, $\displaystyle \arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$, $\displaystyle \arctan(-1) = -\frac{\pi}{4}$.
- $\sin(\arcsin x) = x$ for $-1 \le x \le 1$, but $\arcsin(\sin x) = x$ only when $x$ is in the restricted range.
Practice
Evaluate $\displaystyle \arccos\left(-\frac{\sqrt{2}}{2}\right)$.
The angle in $[0, \pi]$ with cosine $\displaystyle -\frac{\sqrt{2}}{2}$ is $\displaystyle \frac{3\pi}{4}$.
Evaluate $\displaystyle \arcsin\left(\sin\frac{5\pi}{6}\right)$.
$\displaystyle \sin\frac{5\pi}{6} = \frac{1}{2}$, and $\displaystyle \arcsin\frac{1}{2} = \frac{\pi}{6}$ lies in the allowed range.
Evaluate $\displaystyle \cos\left(\arcsin\frac{3}{5}\right)$.
Let $\displaystyle \theta = \arcsin\frac{3}{5}$ (quadrant I); the adjacent side is 4, so $\displaystyle \cos\theta = \frac{4}{5}$.
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