Notes
Domains and ranges
- $\arcsin x$: domain $[-1, 1]$, range $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. $\arccos x$: domain $[-1, 1]$, range $[0, \pi]$. $\arctan x$: domain all reals, range $\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
- The output is always an angle in the restricted range. $\sin^{-1}x$ is not $\displaystyle \frac{1}{\sin x}$.
Evaluating and composing
- Find the angle in the range with the given value: $\displaystyle \arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.
- $\sin(\arcsin x) = x$ for $x$ in $[-1, 1]$, but $\arcsin(\sin x) = x$ only for $x$ in $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
- For expressions like $\displaystyle \cos(\arcsin\frac{3}{5})$, draw a right triangle and use the range to choose signs.
- $\arccos(-x) = \pi - \arccos x$ and $\arcsin(-x) = -\arcsin x$.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \sin\left(-\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$ and $\displaystyle -\frac{\pi}{3}$ is in the range.
$\displaystyle \cos\frac{\pi}{2} = 0$.
$\displaystyle \sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$ and $\displaystyle \arcsin\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$.
Opposite 5, adjacent 12, hypotenuse 13, so cosine is $\displaystyle \frac{12}{13}$.
The angle is in Quadrant II (range $[0, \pi]$), where sine is positive. The missing side is 4, so $\displaystyle \sin = \frac{4}{5}$.
$\displaystyle \tan\theta = \frac{1.5}{6} = 0.25$, so $\theta = \tan^{-1}(0.25) \approx 14.0^\circ$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
Evaluate several inverse trigonometric expressions exactly.
(a) Find $\arcsin\left(-\frac{1}{2}\right)$ and $\arccos\left(-\frac{1}{2}\right)$.
Show the answer and scoring guide
$\displaystyle \arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ and $\displaystyle \arccos\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$.
Check off the points you earned:
(b) Find the exact value of $\cos\left(\arcsin\left(-\frac{4}{5}\right)\right)$.
Show the answer and scoring guide
Let $\displaystyle \theta = \arcsin\left(-\frac{4}{5}\right)$, an angle in Quadrant IV. Then $y = -4$, $r = 5$, $x = 3$, so $\displaystyle \cos\theta = \frac{3}{5}$.
Check off the points you earned:
(c) Find $\arcsin\left(\sin\frac{7\pi}{6}\right)$ and explain why the answer is not $\frac{7\pi}{6}$.
Show the answer and scoring guide
$\displaystyle \sin\frac{7\pi}{6} = -\frac{1}{2}$, so the result is $\displaystyle \arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$. The answer is not $\displaystyle \frac{7\pi}{6}$ because arcsin only outputs angles in $\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Check off the points you earned: