Trigonometry / Unit 5: Inverse Trigonometric Functions
5.2 Evaluating Inverse Trig Functions
Notes
To evaluate $\arcsin x$, ask: which angle in the range has sine equal to $x$?
- $\displaystyle \arcsin\frac{1}{2} = \frac{\pi}{6}$ and $\displaystyle \arcsin\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$.
- $\displaystyle \arccos\frac{1}{2} = \frac{\pi}{3}$ and $\arccos(-1) = \pi$.
- $\displaystyle \arctan 1 = \frac{\pi}{4}$ and $\displaystyle \arctan\left(-\sqrt{3}\right) = -\frac{\pi}{3}$.
- Only the restricted range is allowed, so $\displaystyle \arcsin\frac{1}{2}$ is not $\displaystyle \frac{5\pi}{6}$.
Practice
What is $\displaystyle \arcsin\left(\frac{1}{2}\right)$?
$\displaystyle \sin\frac{\pi}{6} = \frac{1}{2}$ and $\displaystyle \frac{\pi}{6}$ is in $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
What is $\displaystyle \arccos\left(-\frac{\sqrt{2}}{2}\right)$?
Cosine is $\displaystyle -\frac{\sqrt{2}}{2}$ at $\displaystyle \frac{3\pi}{4}$, which is in $[0, \pi]$. The angle $\displaystyle \frac{5\pi}{4}$ is outside the range.
What is $\arctan(-1)$?
$\displaystyle \tan\left(-\frac{\pi}{4}\right) = -1$ and the angle is in $\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
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