Prove that, $\sin \theta + \sin 2\theta + \sin 3\theta+...+ \sin n\theta = \large\frac{\Large\frac{\sin n \theta}{2} \sin \Large\frac{(n+1)}{2} \theta}{\sin \Large\frac{\theta}{2}}$, for all $n \in N$ by using principle of mathematical induction.

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