The sum S = sin theta + sin 2theta + dots + s | vedclass

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(A) The sum of the series is given by $S = \sum_{k=1}^{n} \sin(k	heta)$.Using the formula for the sum of sines in arithmetic progression: $\sum_{k=0}

Vedclass · Accepted Answer

$\frac{\sin \frac{1}{2}(n + 1)	heta \sin \frac{1}{2}n	heta}{\sin \frac{	heta}{2}}$

Vedclass · Answer

(A) The sum of the series is given by $S = \sum_{k=1}^{n} \sin(k	heta)$.Using the formula for the sum of sines in arithmetic progression: $\sum_{k=0}^{n-1} \sin(\alpha + k\beta) = \frac{\sin(n\beta/2)}{\sin(\beta/2)} \sin(\alpha + \frac{(n-1)\beta}{2})$.Here,$\alpha = 	heta$ and $\beta = 	heta$.Substituting these values,we get:$S = \frac{\sin(n	heta/2)}{\sin(	heta/2)} \sin(	heta + \frac{(n-1)	heta}{2})$$S = \frac{\sin(n	heta/2)}{\sin(	heta/2)} \sin(\frac{2	heta + n	heta - 	heta}{2})$$S = \frac{\sin(n	heta/2) \sin(\frac{(n+1)	heta}{2})}{\sin(	heta/2)}$.

The sum $S = \sin \theta + \sin 2\theta + \dots + \sin n\theta$ equals

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