Question

Prove that induction that
sin x + sin 3x + .. + sin(2n - 1) x = $\frac{si{n}^{2}nx}{sinx}$

Answer 1

sin x = $\frac{si{n}^{2}x}{sinx}=sinx\therefore ,$ p(1) is true
Assume p(m) holds good
p ( m + 1) = p(m) + sin (2m + 1) x
= $\frac{si{n}^{2}mx}{sinx}+sin(2m+1)x$
= $\frac{1}{2sinx}[2si{n}^2$ mx + 2 sin x sin (2m +1)x]
= $\frac{1}{2sinx}$[(1 - cos 2mx) + cos2mx - cos (2m + 2)x]
= $\frac{1}{2sinx}$ [1 - cos (2m + 2)x]
= $\frac{1}{2sinx}\left[2si{n}^2\right(m+1\left)x\right]=\frac{si{n}^2(m+1)x}{sinx}$
Thus p(m +1) also holds good