Question

If $\sum _{i-1}^{2n}{\sin }^{-1}{x}_i=n\pi$, then show that $\sum _{i=1}^{2n}{x}_i=2n$.

Answer 1

If $\sum _{i=1}^{2n}$${\sin }^{-1}{x}_i=n\pi$

then ${\sin }^{-1}{x}_i=\frac{\pi }{2}$ for all $i$

So, $x_i=1$ for all $i$ from $1$ to $2n$

Therefore, $\sum _{i=1}^{2n}{x}_i=\sum _{i=1}^{2n}1=2n$

Hence Proved.