Q.Identify the function shown in the grap
(A) sin−1 𝑥
(B) sin−1(2𝑥)
(C) sin−1 ( 𝑥

Inverse Trigonometric Graphs

A trigonometric function such as $\sin x$ takes an angle and returns a ratio. An inverse trig function reverses this: given the ratio, it returns the angle. Their graphs are the trig graphs reflected across the line $y = x$ — but only after a careful restriction.

Why we must restrict first

On its full domain $\sin x$ repeats forever, so $\sin x = 0.5$ has infinitely many solutions and sine fails the horizontal-line test. To invert it we keep only a piece where it is one-to-one. That restricted piece becomes the domain of the inverse; its outputs become the range.

The three graphs

$\sin^{-1} x$ — restrict $\sin x$ to $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$ (strictly increasing).

• Domain $[-1,1]$, range $\left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$. An S-shaped curve from $(-1,-\tfrac{\pi}{2})$ up through $(0,0)$ to $(1,\tfrac{\pi}{2})$.

$\cos^{-1} x$ — restrict $\cos x$ to $[0,\pi]$ (strictly decreasing).

• Domain $[-1,1]$, range $[0,\pi]$. Falls from $(-1,\pi)$ through $(0,\tfrac{\pi}{2})$ to $(1,0)$.

$\tan^{-1} x$ — restrict $\tan x$ to $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$.

• Domain $(-\infty,\infty)$, range $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$. Passes through $(0,0)$ with horizontal asymptotes $y = \pm\tfrac{\pi}{2}$.