`(sin A + 2 sin 3 A + sin 5 A) / (sin 3 A + 2 sin 5 A + sin 7 A) = (sin3A) /(sin 5 A)`

(sin A+2sin3A+sin5A)/(sin3A+2sin5A+sin7A)=(sin3A)/(sin5A)

(sin A sin2A + sin3A sin6A) / (sin A cos2A + sin3A cos6A) = tan5A

int sin3x sin5xdx

(sin theta+2sin3 theta+sin5 theta)/(sin3 theta+2sin5 theta+sin7 theta)=(sin3 theta)/(sin5 theta)

(sin theta+2sin3 theta+sin5 theta)/(sin3 theta+2sin5 theta+sin7 theta)=(sin3 theta)/(sin5 theta)

Prove that :(sin theta+2sin3 theta+sin5 theta)/(sin3 theta+2sin5 theta+sin7 theta)=(sin3 theta)/(sin5 theta)

(sin3A+sin5A+sin7A+sin9A)/(cos3A+cos5A+cos7A+cos9A)

sin A sin2A + sin3A sin6A-sin4A sin5A = 0

sin A+sin3A+sin5A+sin7A=

sin3A+sin2A-sin A is