sin x+ sin 3x+...+sin

Solution:

Let P(n) be the given statement. 

$P(n): \sin x+\sin 3 x+\ldots+\sin (2 n-1) x=\frac{\sin ^{2} n x}{\sin x}$

Step 1:

$P(1): \sin x=\frac{\sin ^{2} x}{\sin x}$

Thus, $P(1)$ is true.

Step $2:$

Let $P(m)$ be true.

$\therefore \sin x+\sin 3 x+\ldots+\sin (2 m-1) x=\frac{\sin ^{2} m x}{\sin x}$

We shall show that $P(m+1)$ is true.

We know that $P(m)$ is true.

$\therefore \sin x+\sin 3 x+\ldots+\sin (2 m-1)=\frac{\sin ^{2} m x}{\sin x}$

$\Rightarrow \sin x+\sin 3 x+\ldots \sin (2 m-1) x+\sin (2 m+1) x=\frac{\sin ^{2} m x}{\sin x}+\sin (2 m+1) x$

(Adding $\sin (2 m+1) x$ to both the sides)

$\Rightarrow P(m+1) x=\frac{\sin ^{2} m x+\sin x[\sin m x \cos (m+1) x+\sin (m+1) x \cos m x]}{\sin x}$

$=\frac{\sin ^{2} m x+\sin x\left(\sin m x \cos m x \cos x-\sin ^{2} m x \sin x+\sin m x \cos x \cos m x+\cos ^{2} m x \sin x\right)}{\sin x}$

$=\frac{\sin ^{2} m x+2 \sin x \cos x \cos m x-\sin ^{2} x \sin ^{2} m x+\cos ^{2} m x \sin ^{2} x}{\sin x}$

$=\frac{\sin ^{2} m x\left(1-\sin ^{2} x\right)+2 \sin x \cos x \cos m x+\cos ^{2} m x \sin ^{2} x}{\sin x}$

$=\frac{\sin ^{2} m x \cos ^{2} x+2 \sin x \cos x \cos m x+\cos ^{2} m x \sin ^{2} x}{\sin x}$

$=\frac{(\sin m x \cos x+\cos m x \sin x)^{2}}{\sin x}$

$=\frac{[\sin (m+1)]^{2}}{\sin x}$

Hence, $P(m+1)$ is true.

By the principle of mathematical induction, the given statement $P(n)$ is true for all $n$ $\in N$.