Question

\[\sin x + \sin 3x + . . . + \sin (2n - 1)x = \frac{\sin^2 nx}{\sin x}\]

 

Answer 1

Let P(n) be the given statement.

\[P(n): \sin x + \sin 3x + . . . + \sin\left( 2n - 1 \right)x = \frac{\sin^2 nx}{\sin x}\]
\[\text{ Step } 1: \]
\[P(1): \sin x = \frac{\sin^2 x}{\sin x}\]
\[\text{ Thus, P(1) is true } . \]
\[\text{ Step 2: } \]
\[\text{ Let P(m) be true .}  \]
\[ \therefore \sin x + \sin 3x + . . . + \sin\left( 2m - 1 \right)x = \frac{\sin^2 mx}{\sin x}\]
\[\text{ We shall show that P(m + 1) is true .}  \]
\[\text{ We know that P(m) is true } . \]
\[ \therefore \sin x + \sin 3x + . . . + \sin (2m - 1) = \frac{\sin^2 mx}{\sin x}\]
\[ \Rightarrow \sin x + \sin 3x + . . . \sin (2m - 1)x + \sin (2m + 1)x = \frac{\sin^2 mx}{\sin x} + \sin (2m + 1)x \left( \text{ Adding }  \sin (2m + 1)x \text{ to both the sides } \right)\]
\[ \Rightarrow P(m + 1)x = \frac{\sin^2 mx + \sin x\left[ \sin mx\cos\left( m + 1 \right)x + \sin\left( m + 1 \right)x \cos x \right]}{\sin x}\]
\[ = \frac{\sin^2 mx + \sin x\left( \sin mx\cos mxcos x - \sin^2 mx\sin x + \sin mx\cos x\cos mx + \cos^2 mx\sin x \right)}{\sin x}\]
\[ = \frac{\sin^2 mx + 2\sin x\cos x\cos mx - \sin^2 x \sin^2 mx + \cos^2 mx \sin^2 x}{\sin x}\]
\[ = \frac{\sin^2 mx\left( 1 - \sin^2 x \right) + 2\sin x\cos x\cos mx + \cos^2 mx \sin^2 x}{\sin x}\]
\[ = \frac{\sin^2 mx \cos^2 x + 2\sin x\cos x\cos mx + \cos^2 mx \sin^2 x}{\sin x}\]
\[ = \frac{\left( \sin mx \cos x + \cos mx \sin x \right)^2}{\sin x}\]
\[ = \frac{\left[ \sin\left( m + 1 \right) \right]^2}{\sin x}\]
\[\text{ [Hence, P(m + 1) is true } . \]
\[ \text{ By the principle of mathematical induction, the given statement P(n) is true for all } n \in N . \]