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

1 Answer
Dec 25, 2015

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

Explanation:

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!