Question

Sketch the graph for $y={\sin }^{-1}\left(\sin x\right)$

Answer 1

As, $y={\sin }^{-1}\left(\sin x\right)$ is periodic with period $2\pi$
$\therefore$ to draw this graph we should draw the graph for one interval of length $2\pi$ and repeat for entire values of $x$
As we know that $y={\sin }^{-1}\left(\sin x\right)=\{\begin{array}{c}x;-\frac{\pi }{2}\le x\le \frac{\pi }{2}\\ (\pi -x);-\frac{\pi }{2}\le \pi -x<\frac{\pi }{2}\end{array}$
$\Rightarrow y={\sin }^{-1}\left(\sin x\right)=\{\begin{array}{c}x;-\frac{\pi }{2}\le x\le \frac{\pi }{2}\\ (\pi -x);\frac{\pi }{2}\le x<\frac{3\pi }{2}\end{array}$
which is defined for the interval of length $2\pi$, plotted as;
Thus the graph for $y={\sin }^{-1}\left(\sin x\right)$ is a straight line up and a straight line down with slopes $1$ and $-1$ respectively lying between $[-\frac{\pi }{2},\frac{\pi }{2}]$
Note: $y={\sin }^{-1}\left(\sin x\right)$ can also be defined as
$=\{\begin{array}{c}x+2\pi ;-\frac{5\pi }{2}\le x\le -\frac{3\pi }{2}\\ -\pi -x;-\frac{3\pi }{2}\le x\le -\frac{\pi }{2}\\ x;-\frac{\pi }{2}\le x\le \frac{\pi }{2}\\ \pi -x;\frac{\pi }{2}\le x\le \frac{3\pi }{2}\\ x-2\pi ;\frac{3\pi }{2}\le x\le \frac{5\pi }{2}\end{array}$...and so on.