Question

If $x_i>0$, for $1\le i\le n$, and $x_1+x_2+.....+x_n=\pi$ then the greatest value of the sum $\sin x_i+\sin x_2+.....+\sin x_n=.........$

• A. $n$
• B. $\pi$
• C. $n\sin \left(\frac{\pi }{n}\right)$
• D. $0$

Answer 1

Answer: (C)

$n\sin \left(\frac{\pi }{n}\right)$

Since $x_i>0,1\le i\le n$ and $x_1+x_2+x_3+.....+x_n=\pi$

$\therefore x_i\le \pi$ for all $1\le i\le n$

$f\left(x\right)=\sin x$

$f^{\prime }\left(x\right)=\cos x$

$f^{\prime \prime }\left(x\right)=-\sin x$

in the interval $(0,\pi )$ and $f^{\prime \prime }\left(x\right)<0$

So graph of $f\left(x\right)=\sin x$ is concave downward graph

Applying Jensen's inequality for concave downward function we have

$f\left(\frac{x_1+x_2+x_3+.....+x_n}{n}\right)\ge \frac{f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+....+f\left(x_n\right)}{n}$

Since $f\left(x\right)=\sin x$

$\Rightarrow \sin \left(\frac{x_1+x_2+x_3+.....+x_n}{n}\right)\ge \frac{\sin x_1+\sin x_2+\sin x_3+....\sin x_n}{n}$

$\Rightarrow \sin \left(\frac{\pi }{n}\right)\ge \frac{\sin x_1+\sin x_2+\sin x_3+....\sin x_n}{n}$ since $x_1+x_2+x_3+.....+x_n=\pi$

$\therefore \sin x_1+\sin x_2+\sin x_3+....\sin x_n\le n\sin \left(\frac{\pi }{n}\right)$