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(N/A) $P(n): \sin 	heta + \sin 2	heta + \ldots + \sin n	heta = \frac{\sin \frac{n	heta}{2} \sin \frac{(n+1)	heta}{2}}{\sin \frac{	heta}{2}}, n \

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(N/A) $P(n): \sin 	heta + \sin 2	heta + \ldots + \sin n	heta = \frac{\sin \frac{n	heta}{2} \sin \frac{(n+1)	heta}{2}}{\sin \frac{	heta}{2}}, n \in N$For $n=1$,$L.H.S. = \sin 	heta$.$R.H.S. = \frac{\sin \frac{	heta}{2} \sin \frac{2	heta}{2}}{\sin \frac{	heta}{2}} = \sin 	heta$.Since $L.H.S. = R.H.S.$,$P(1)$ is true.Assume $P(k)$ is true for some $k \in N$:$\sin 	heta + \sin 2	heta + \ldots + \sin k	heta = \frac{\sin \frac{k	heta}{2} \sin \frac{(k+1)	heta}{2}}{\sin \frac{	heta}{2}}$.For $n=k+1$,$L.H.S. = (\sin 	heta + \ldots + \sin k	heta) + \sin(k+1)	heta$$= \frac{\sin \frac{k	heta}{2} \sin \frac{(k+1)	heta}{2}}{\sin \frac{	heta}{2}} + \sin(k+1)	heta$$= \frac{\sin \frac{k	heta}{2} \sin \frac{(k+1)	heta}{2} + \sin(k+1)	heta \sin \frac{	heta}{2}}{\sin \frac{	heta}{2}}$Using $2\sin A \sin B = \cos(A-B) - \cos(A+B)$:$= \frac{\frac{1}{2} [\cos(\frac{k	heta}{2} - \frac{(k+1)	heta}{2}) - \cos(\frac{k	heta}{2} + \frac{(k+1)	heta}{2})] + \frac{1}{2} [\cos((k+1)	heta - \frac{	heta}{2}) - \cos((k+1)	heta + \frac{	heta}{2})]}{\sin \frac{	heta}{2}}$$= \frac{\cos \frac{	heta}{2} - \cos \frac{(2k+1)	heta}{2} + \cos \frac{(2k+1)	heta}{2} - \cos \frac{(2k+3)	heta}{2}}{2 \sin \frac{	heta}{2}}$$= \frac{\cos \frac{	heta}{2} - \cos \frac{(2k+3)	heta}{2}}{2 \sin \frac{	heta}{2}}$Using $\cos C - \cos D = 2 \sin \frac{C+D}{2} \sin \frac{D-C}{2}$:$= \frac{2 \sin \frac{(k+2)	heta}{2} \sin \frac{(k+1)	heta}{2}}{2 \sin \frac{	heta}{2}} = \frac{\sin \frac{(k+1)	heta}{2} \sin \frac{(k+2)	heta}{2}}{\sin \frac{	heta}{2}} = R.H.S.$Thus,$P(k+1)$ is true. By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in N$.

Use the Principle of Mathematical Induction to prove that for all $n \in N$:
$\sin \theta + \sin 2\theta + \ldots + \sin n\theta = \frac{\sin \frac{n\theta}{2} \sin \frac{(n+1)\theta}{2}}{\sin \frac{\theta}{2}}$

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Use the Principle of Mathematical Induction to prove that for all $n \in N$:$\sin \theta + \sin 2\theta + \ldots + \sin n\theta = \frac{\sin \frac{n\theta}{2} \sin \frac{(n+1)\theta}{2}}{\sin \frac{\theta}{2}}$