Question

prove the following I am not understanding please prove

Answer 1

[tex]sin(x)-sin(3x) = 2cos(2x)sin(-x)=-2cos(2x)sin(x) \\ \\ sin^{2}x-cos^{2}x = -(cos^{2}x-sin^{2}x) = -cos(2x) \\ \\ Hence \\ \\ \frac{sin(x)-sin(3x)}{sin^{2}x-cos^{2}x} = \frac{-2cos(2x)sin(x)}{-cos(2x)} = 2sinx[/tex]