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

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