Question

The value of $\sum _{x=0}^4{\sin }^{-1}\left(\sin x\right)$ is equal to

• A. $3\pi -8$
• B. $3\pi -7$
• C. $3\pi -9$
• D. $3\pi -6$

Answer 1

The correct option is A $3\pi -8$

Given: $\sum _{x=0}^4{\sin }^{-1}\left(\sin x\right)={\sin }^{-1}\left(\sin 0\right)+{\sin }^{-1}\left(\sin 1\right)+{\sin }^{-1}\left(\sin 2\right)+{\sin }^{-1}\left(\sin 3\right)+{\sin }^{-1}\left(\sin 4\right)$
We know, ${\sin }^{-1}\left(\sin x\right)=\{\begin{array}{cc}x,& -\frac{\pi }{2}\le x\le \frac{\pi }{2}\\ \pi -x,& \frac{\pi }{2}<x\le \frac{3\pi }{2}\end{array}$
$\therefore {\sin }^{-1}\left(\sin 0\right)+{\sin }^{-1}\left(\sin 1\right)+{\sin }^{-1}\left(\sin 2\right)+{\sin }^{-1}\left(\sin 3\right)+{\sin }^{-1}\left(\sin 4\right)$
$=\left(0\right)+\left(1\right)+(\pi -2)+(\pi -3)+(\pi -4)$
$=3\pi -8$