Domain and Range of Each Inverse Trigonometric Function

3. Domain and Range of Each Inverse Trigonometric Function

Because an inverse function's domain equals the range of the original function (restricted to its chosen branch), and its range equals that branch itself (Section 1), the domain and range of every inverse trigonometric function follow immediately once the branches of Section 2 are fixed.

Domain-range table.

Why $\sin^{-1}x$ and $\cos^{-1}x$ need $x\in[-1,1]$. For every real $\theta$, the fundamental identity gives $\sin^2\theta\le\sin^2\theta+\cos^2\theta=1$, so $-1\le\sin\theta\le1$ always -- sine never leaves $[-1,1]$, and the same bound holds for cosine. Consequently there is no real angle whose sine or cosine is, say, $\dfrac{3}{2}$, so $\sin^{-1}\!\left(\dfrac32\right)$ and $\cos^{-1}\!\left(\dfrac32\right)$ are simply undefined -- $\dfrac32$ is outside the domain, not merely "hard to compute".

Why $\sec^{-1}x$ and $\text{cosec}^{-1}x$ exclude $(-1,1)$. From $1+\tan^2\theta=\sec^2\theta$ (Class XI), $\sec^2\theta\ge1$ for every $\theta$ at which it is defined, so $|\sec\theta|\ge1$ always; secant never takes a value strictly between $-1$ and $1$. The same argument, using $1+\cot^2\theta=\text{cosec}^2\theta$, shows $|\text{cosec}\,\theta|\ge1$. So the domain of $\sec^{-1}x$ and $\text{cosec}^{-1}x$ is exactly the set of real numbers outside the open interval $(-1,1)$. …