Trig identity Q

[T-7]

Could someone tell me which trig identity this chap (in a Physics textbook) is using to turn

sin(Kx)sin(2Kx)

into

sin^2(3Kx/2) - sin^2(Kx/2)

Cheers.

[nicktacik]

\sin{kx}\sin{2kx}

=\frac{\cos{(-kx)}-\cos{(3kx)}}{2}

=\frac{\cos{(kx)}}{2} - \frac{\cos{(3kx)}}{2}

=\frac{-1 + \cos{(kx)}}{2} - \frac{-1 + \cos{(3kx)}}{2}

=\sin^2{(3kx/2) - \sin^2{(kx/2)}

[T-7]

\sin{kx}\sin{2kx}

=\frac{\cos{(-kx)}-\cos{(3kx)}}{2}

=\frac{\cos{(kx)}}{2} - \frac{\cos{(3kx)}}{2}

=\frac{-1 + \cos{(kx)}}{2} - \frac{-1 + \cos{(3kx)}}{2}

=\sin^2{(3kx/2) - \sin^2{(kx/2)}

Ah... Thank you very much.