What is #cos^2thetasin^2theta# in terms of non-exponential trigonometric functions?

First express purely in terms of #cos theta#:

#cos^2 theta sin^2 theta = cos^2 theta (1 - cos^2 theta) = cos^2 theta - cos^4 theta#

Since this is of degree #4#, a good place to start might be #cos 4 theta# and #sin 4 theta#.

Using De Moivre's formula:

#cos 4 theta + i sin 4 theta#

#=(cos theta + i sin theta)^4#

#=cos^4 theta + 4i cos^3 theta sin theta -6 cos^2 theta sin^2 theta - 4i cos theta sin^3 theta + sin^4 theta#

#=(cos^4 theta -6 cos^2 theta sin^2 theta + sin^4 theta) + (4 cos^3 theta sin theta - 4 cos theta sin^3 theta)i#

So looking at the Real part, we find:

#cos 4 theta#

#=cos^4 theta -6 cos^2 theta sin^2 theta + sin^4 theta#

#=cos^4 theta - 6 cos^2 theta (1-cos^2 theta) + (1-cos^2 theta)(1-cos^2 theta)#

#=cos^4 theta - 6 cos^2 theta+6 cos^4 theta+1 - 2 cos^2 theta +cos^4 theta#

#=8 cos^4 theta-8cos^2 theta + 1#

#=-8(cos^2 theta - cos^4 theta) + 1#

Subtract #1# from both ends to get:

#cos 4 theta - 1 = -8(cos^2 theta - cos^4 theta)#

Divide both sides by #-8# to get:

#1/8(1-cos 4 theta) = cos^2 theta - cos^4 theta#

Voila!