Question

Prove that : $sin\Theta +sin2\Theta +sin3\Theta +....+sinn\Theta =\frac{\sin \frac{(n+1)\theta }{2}\sin \frac{n\theta }{2}}{\sin \frac{\theta }{2}}$

Answer 1

L.H.S.

$f\left(x\right)=\sin \theta +\sin 2\theta +\sin 3\theta +-----+\sin n\theta =\frac{\sin \frac{(n+1)\theta }{2}\sin \frac{n\theta }{2}}{\sin \frac{\theta }{2}}$

Step (1):-

For $n=1$ L.H.S.

$\sin n\theta =\sin \theta$

For $n=1$ R.H.S.

$\frac{\sin \left(\frac{n+1}{2}\right)\theta \sin \frac{n\theta }{2}}{\sin \frac{\theta }{2}}=\frac{\sin \frac{2\theta }{2}\sin \frac{\theta }{2}}{\sin \frac{\theta }{2}}$

$=\sin \theta$

Thus then the result is true for $n=1$

Step (2):-

Let it is true for $n=k.$

Therefore,

$\sin \theta +\sin 2\theta +\sin 3\theta +---+\sin k\theta =\frac{\sin \frac{(k+1)}{2}\theta \sin \frac{k\theta }{2}}{\sin \frac{\theta }{2}}$

Now in the step (3):-

To prove that, It is true for $n=(k+1)$

Then,

$\sin \theta +\sin 2\theta +\sin 3\theta +---+\sin k\theta +\sin \frac{(k+1)}{2}\theta =\frac{\sin \left(\frac{k+1}{2}\right)\theta \sin \frac{k\theta }{2}}{\sin \frac{\theta }{2}}+\sin (k+1)\theta$

$=\frac{\sin \left(\frac{k+1}{2}\right)\theta \sin \frac{k\theta }{2}}{\sin \frac{\theta }{2}}+2\sin \frac{(k+1)}{2}\theta \cos \frac{(k+1)}{2}\theta$

$=\frac{\sin \left(\frac{k+1}{2}\right)\theta }{\sin \frac{\theta }{2}}[\sin \frac{k\theta }{2}+2\cos \frac{(k+1)}{2}\theta \sin \frac{\theta }{2}]$

$=\frac{\sin \left(\frac{k+1}{2}\right)\theta }{\sin \frac{\theta }{2}}[\sin \frac{k\theta }{2}+2\cos [\left(\frac{k+1}{2}\right)\theta +\frac{\theta }{2}]-\sin [\left(\frac{k+1}{2}\right)\theta -\frac{\theta }{2}]]$

$=\frac{\sin \left(\frac{k+1}{2}\right)\theta }{\sin \frac{\theta }{2}}[\sin \frac{k\theta }{2}+\sin [\left(\frac{k+2}{2}\right)\theta -\sin \frac{k}{2}]]$

$=\frac{\sin \left(\frac{k+1}{2}\right)\theta }{\sin \frac{\theta }{2}}.\sin \frac{(k+1)+1}{2}x$

Then it is true for $n=k+1$ therefore $n\in N$

Hence proved.