The domain of the function $f(x) = \sin^{-1}(x)$ is
A$[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$
B$\mathbb{R}$
C$[-1, 1]$
D$[0, \pi]$
Answer & Solution
Correct answer: C. $[-1, 1]$
$\sin^{-1}$ is the inverse of sine restricted to $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, where sine takes every value in $[-1,1]$ exactly once.
So the inverse is defined precisely on $x \in [-1, 1]$.
**Option D is the *range* of $\sin^{-1}$, not the domain.**
Keep practising Unit 1A2: Trigonometric Functions & Inverse Trig
Browse all Unit 1A2: Trigonometric Functions & Inverse Trig questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
In a right triangle, for an acute angle θ, sin θ is defined as:A relation $f$ from $A$ to $B$ is a function if every element of $A$ has:The domain of $f(x)= qrt{4-x^2}$ is:If $(x+1,\,y-2)=(3,\,1)$, then $(x,y)$ equals:The domain of $f(x)= qrt{x-1}$ is:The range of the function $f(x)=x^2,\; x\in\mathbb{R}$ is:The domain of the function $f(x)=\dfrac{1}{x-2}$ is:The reciprocal of sin θ is: