Prove that, `sintheta+sin2theta+sin3theta+ . . .sinntheta=(sin""(ntheta)/(2)"sin"(n+1)/(2)theta)/("sin"(theta)/(2))` for all `ninN`.

Explanation

To prove the statement

\[ \sin \theta + \sin 2\theta + \sin 3\theta + \ldots + \sin n\theta = \frac{\sin \left( \frac{n\theta}{2} \right) \cdot \sin \left( \frac{(n+1)\theta}{2} \right)}{\sin \left( \frac{\theta}{2} \right)} \]

for all \( n \in \mathbb{N} \), we will use the principle of mathematical induction.

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