What is cos^2thetasin^2thetacos2θsin2θ in terms of non-exponential trigonometric functions?

1 Answer
Dec 25, 2015

cos^2 theta sin^2 theta = 1/8(1-cos 4 theta)cos2θsin2θ=18(1cos4θ)

Explanation:

First express purely in terms of cos thetacosθ:

cos^2 theta sin^2 theta = cos^2 theta (1 - cos^2 theta) = cos^2 theta - cos^4 thetacos2θsin2θ=cos2θ(1cos2θ)=cos2θcos4θ

Since this is of degree 44, a good place to start might be cos 4 thetacos4θ and sin 4 thetasin4θ.

Using De Moivre's formula:

cos 4 theta + i sin 4 thetacos4θ+isin4θ

=(cos theta + i sin theta)^4=(cosθ+isinθ)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=cos4θ+4icos3θsinθ6cos2θsin2θ4icosθsin3θ+sin4θ

=(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=(cos4θ6cos2θsin2θ+sin4θ)+(4cos3θsinθ4cosθsin3θ)i

So looking at the Real part, we find:

cos 4 thetacos4θ

=cos^4 theta -6 cos^2 theta sin^2 theta + sin^4 theta=cos4θ6cos2θsin2θ+sin4θ

=cos^4 theta - 6 cos^2 theta (1-cos^2 theta) + (1-cos^2 theta)(1-cos^2 theta)=cos4θ6cos2θ(1cos2θ)+(1cos2θ)(1cos2θ)

=cos^4 theta - 6 cos^2 theta+6 cos^4 theta+1 - 2 cos^2 theta +cos^4 theta=cos4θ6cos2θ+6cos4θ+12cos2θ+cos4θ

=8 cos^4 theta-8cos^2 theta + 1=8cos4θ8cos2θ+1

=-8(cos^2 theta - cos^4 theta) + 1=8(cos2θcos4θ)+1

Subtract 11 from both ends to get:

cos 4 theta - 1 = -8(cos^2 theta - cos^4 theta)cos4θ1=8(cos2θcos4θ)

Divide both sides by -88 to get:

1/8(1-cos 4 theta) = cos^2 theta - cos^4 theta18(1cos4θ)=cos2θcos4θ

Voila!