Trigonometric Identity involving $\sin^2$ and $\cos^2$ I was trying to prove this trigonometric identity, it looks like using the elementary relations should be enough, but I still can't find how:
$$\frac{1}{2}\sin^2 a \ \sin^2 b + \cos^2a \ \cos^2 b  =  \frac{1}{3} + \frac{2}{3} \biggl( \frac{3}{2}\cos^2 a - \frac{1}{2} \biggr)\biggl( \frac{3}{2}\cos^2 b - \frac{1}{2}\biggr)$$
Thank you!
(taken from Celestial Mechanics)

The left hand side is
$$\begin{align}\frac 12\sin^2 a\sin^2b+\cos^2 a\cos^2 b&=\frac 12(1-\cos^2 a)(1-\cos^2b)+\cos^2a\cos ^2b\\&=\frac 12-\frac 12\cos^2a-\frac12\cos ^2b+\frac 32\cos^2a\cos^2b.\end{align}$$
The right hand side is
$$\frac 13+\frac 23\left(\frac 32\cos^2a-\frac 12\right)\left(\frac 32\cos^2b-\frac 12\right)$$$$=\frac 13+\frac 23\left(\frac94\cos^2\cos^2b-\frac{3}{4}\cos^2a-\frac 34\cos^2b+\frac 14\right)$$
$$=\frac 13+\frac 32\cos^2a\cos^2b-\frac 12\cos^2a-\frac12\cos^2b+\frac 16$$
$$=\frac 12-\frac 12\cos^2a-\frac12\cos ^2b+\frac 32\cos^2a\cos^2b.$$
