`sinalpha*sin(60-alpha)sin(60+alpha) = 1/4*sin3alpha`

sin alpha * sin (60-alpha) sin (60 + alpha) = (1) / (4) * sin3 alpha

Prove that, 1/3 ( cos^(3) alpha sin 3 alpha + sin^(3) alpha cos 3 alpha) = 1/4 sin 4 alpha

If the expression 4sin5 alpha cos3 alpha cos2 alpha is expressed as the sum of threes sines then two of them are sin 4 alpha and sin10 alpha. The third one is (a)sin8 alpha(b)sin6 alpha(d)sin5 alpha(d)sin12 alpha

Find the sum of the series sinalpha+sin(alpha+beta)+sin(alpha+2beta)+sin(alpha+3beta)+.....+sin(alpha+(n-1)beta)

Sum of n terms of the series sinalpha-sin(alpha+beta)+sin(alpha+2beta)-sin(alpha+3beta)+….

Sum of n terms of the series sinalpha-sin(alpha+beta)+sin(alpha+2beta)-sin(alpha+3beta)+….

Sum of n terms of the series sinalpha-sin(alpha+beta)+sin(alpha+2beta)-sin(alpha+3beta)+….

sin 3 alpha = 4 sin alpha sin(x + alpha) sin(x-alpha)