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 $$\frac 13+\frac 23\left(\frac 32\cos^2a-\frac 12\right)\left(\frac 32\cos^2b-\frac 12\right)$$$$=\frac 13+\frac 23\left(\frac94\cos^2\cos^2b-\frac{3}{4}\cos^2a-\frac 34\cos^2b+\frac 14\right)$$ $$=\frac 13+\frac 32\cos^2a\cos^2b-\frac 12\cos^2a-\frac12\cos^2b+\frac 16$$ $$=\frac 12-\frac 12\cos^2a-\frac12\cos ^2b+\frac 32\cos^2a\cos^2b.$$



Observe that the Right Hand Side does not involve sine, so use $\cos^2x+\sin^2x=1$ to eliminate $\sin a,\sin b$ from the Left Hand Side


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