Question

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

Answer 1

`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!