Question

The general solution of the equation $\sum _{r=1}^n\cos r^2\theta \sin r\theta =\frac{1}{2}$ is

• A. $\frac{4k-1}{n(n+1)}.\frac{\pi }{2},k\in Z$
• B. $\frac{2k+1}{n(n+1)}.\frac{\pi }{2},k\in Z$
• C. $\frac{4k+1}{n(n+1)}.\frac{\pi }{2},k\in Z$
• D. None of these

Answer 1

The correct option is C $\frac{4k+1}{n(n+1)}.\frac{\pi }{2},k\in Z$

We have $\sum _{r=1}^n\cos r^2\theta \sin r\theta =\frac{1}{2}\Rightarrow \sum _{r=1}^{n}2\cos r^2\theta \sin r\theta =1$

$\Rightarrow \sum _{r=1}^n[\sin r(r+1)\theta -\sin r(r-1)\theta ]=1$

$[\sin 2\theta -0]+[\sin 6\theta -\sin 2\theta ]+..................+[\sin n(n+1)\theta -\sin n(n-1)\theta ]=1$

$\Rightarrow \sin n(n+1)\theta =1$

$\Rightarrow \theta =\frac{4k+1}{n(n+1)}.\frac{\pi }{2},k\in Z$