Q. The sum$\ S = sin\theta + \sin{2\theta + ... + \sin{n\theta\ }}$equals
A) $\sin{\frac{1}{2}(n + 1)\theta\sin{\frac{n\theta}{2}/\sin\frac{\theta}{2}}}$
B) $\cos{\frac{1}{2}(n + 1)\theta}\sin{\frac{n\theta}{2}/\sin\frac{\theta}{2}}$
C) $\sin{\frac{1}{2}(n + 1)\theta\cos{\frac{n\theta}{2}/\sin\frac{\theta}{2}}}$
D) $\cos{\frac{1}{2}(n + 1)\theta\cos\frac{n\theta}{2}}/\sin\frac{\theta}{2}$
Explanation
(a)
We have$,\ S = \sin{\theta + \sin{2\theta + \sin{3\theta + ... + \sin{n\theta}}}}$
We know that,
$\sin{\theta + \sin{(\theta + \beta) + \sin{(\theta + 2\beta) + ...n\ terms}}}$
$= \frac{\sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}}\sin\left\lbrack \frac{\theta + \theta + (n - 1)\beta}{2} \right\rbrack$
Put,$\ \beta = \theta$
$\therefore S = \frac{\sin{\frac{n\theta}{2} \bullet \sin\frac{(n + 1)\theta}{2}}}{\sin\frac{\theta}{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.