Question

If $\sum _{i=1}^{2n}{\sin }^{-1}{x}_i=n\pi$, then $\sum _{i=1}^{2n}{x}_i$ is equal to

• A. $n$
• B. $2n$
• C. $\frac{n(n+1)}{2}$
• D. None of these

Answer 1

Answer: (B)

$2n$

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

$\Rightarrow {\sin }^{-1}{x}_1+{\sin }^{-1}{x}_2+...+{\sin }^{-1}{x}_{2n}=n\pi$

Let ${\sin }^{-1}{x}_1={\alpha }_1,{\sin }^{-1}{x}_2={\alpha }_2,...,{\sin }^{-1}{x}_{2n}={\alpha }_n$

$\therefore {\alpha }_1+{\alpha }_2+...+{\alpha }_{2n}=n\pi$ ....(1)

$\Rightarrow x_1=\sin {\alpha }_1,x_2=\sin {\alpha }_2,...,x_{2n}=\sin {\alpha }_{2n}$

$\therefore \sum _{i=1}^{2n}{x}_i=x_1+x_2+...+x_{2n}=\sin {\alpha }_1+\sin {\alpha }_2+...+\sin {\alpha }_{2n}$

Clearly, from (1) ${\alpha }_i=\frac{\pi }{2},\forall i=1,2,...2n$

$\therefore {\sum }_{i=1}^{2n}{x}_i=1+1+...+1=2n$