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Question: The sum \(S = \sin\theta + \sin 2\theta + .... + \sin n\theta,\) equals...

The sum S=sin⁡θ+sin⁡2θ+....+sin⁡nθ,S = \sin\theta + \sin 2\theta + .... + \sin n\theta, equals

A

sin⁡12(n+1) θsin⁡12n θ/sin⁡θ2\sin\frac{1}{2}(n + 1)\ \theta\sin\frac{1}{2}n\ \theta/\sin\frac{\theta}{2}

B

cos⁡12(n+1) θsin⁡12nθ/sin⁡θ2\cos\frac{1}{2}(n + 1)\ \theta\sin\frac{1}{2}n\theta/\sin\frac{\theta}{2}

C

sin⁡12(n+1)θcos⁡12nθ/sin⁡θ2\sin\frac{1}{2}(n + 1)\theta\cos\frac{1}{2}n\theta/\sin\frac{\theta}{2}

D

cos⁡12(n+1)θcos⁡12nθ/sin⁡θ2\cos\frac{1}{2}(n + 1)\theta\cos\frac{1}{2}n\theta/\sin\frac{\theta}{2}

Answer

sin⁡12(n+1) θsin⁡12n θ/sin⁡θ2\sin\frac{1}{2}(n + 1)\ \theta\sin\frac{1}{2}n\ \theta/\sin\frac{\theta}{2}

Explanation

Solution

S=sin⁡θ+sin⁡2θ+sin⁡3θ+.....+sin⁡nθS = \sin\theta + \sin 2\theta + \sin 3\theta + ..... + \sin n\theta

We know, sin⁡θ+sin⁡(θ+β)+sin⁡(θ+2β)+.......nterm\sin\theta + \sin(\theta + \beta) + \sin(\theta + 2\beta) + .......n\text{term}

= sin⁡nβ2sin⁡β2sin⁡[θ+θ+(n−1)β2]\frac{\sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}}\sin\left\lbrack \frac{\theta + \theta + (n - 1)\beta}{2} \right\rbrack

Put β=θ\beta = \theta, then S=sin⁡nθ2.sin⁡θ(n+1)2sin⁡θ2S = \frac{\sin\frac{n\theta}{2}.\sin\frac{\theta(n + 1)}{2}}{\sin\frac{\theta}{2}}.