The graph of \(y = \sin^{-1} x\) can be obtained from the graph of \(y = \sin x\), \(-\displaystyle\frac{\pi}{2} \le x \le \displaystyle\frac{\pi}{2}\), by taking its mirror image (reflection) in
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → the line \(y = x\)
If \((a, b)\) is a point on the graph of \(y = \sin x\), then \(b = \sin a\), so \(a = \sin^{-1} b\). This means \((b, a)\) is a point on the graph of \(y = \sin^{-1} x\): the two coordinates are interchanged.
For example, \(\left(\frac{\pi}{6}, \frac{1}{2}\right)\) is on \(y = \sin x\), and
\( \displaystyle \left(\frac{1}{2}, \frac{\pi}{6}\right) \) is on \(y = \sin^{-1} x\)
Interchanging the coordinates of every point is the same as reflecting the graph in the line \(y = x\). So the graph of an inverse function is the mirror image of the original graph in the line \(y = x\).
Why the other options are incorrect
Option (2) the x-axis — Incorrect. Reflection in the x-axis sends \((a, b)\) to \((a, -b)\), which gives the graph of \(y = -\sin x\).
Option (3) the y-axis — Incorrect. Reflection in the y-axis sends \((a, b)\) to \((-a, b)\), which gives the graph of \(y = \sin(-x)\).
Option (4) the line \(x = 1\) — Incorrect. This sends \((a, b)\) to \((2 - a, b)\); for example the origin would go to \((2, 0)\), but \(\sin^{-1} 2\) is not defined.