What is cos^2thetasin^2thetacos2θsin2θ in terms of non-exponential trigonometric functions?
1 Answer
Explanation:
First express purely in terms of
cos^2 theta sin^2 theta = cos^2 theta (1 - cos^2 theta) = cos^2 theta - cos^4 thetacos2θsin2θ=cos2θ(1−cos2θ)=cos2θ−cos4θ
Since this is of degree
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+4icos3θsinθ−6cos2θsin2θ−4icosθsin3θ+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=(cos4θ−6cos2θsin2θ+sin4θ)+(4cos3θsinθ−4cosθsin3θ)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=cos4θ−6cos2θsin2θ+sin4θ
=cos^4 theta - 6 cos^2 theta (1-cos^2 theta) + (1-cos^2 theta)(1-cos^2 theta)=cos4θ−6cos2θ(1−cos2θ)+(1−cos2θ)(1−cos2θ)
=cos^4 theta - 6 cos^2 theta+6 cos^4 theta+1 - 2 cos^2 theta +cos^4 theta=cos4θ−6cos2θ+6cos4θ+1−2cos2θ+cos4θ
=8 cos^4 theta-8cos^2 theta + 1=8cos4θ−8cos2θ+1
=-8(cos^2 theta - cos^4 theta) + 1=−8(cos2θ−cos4θ)+1
Subtract
cos 4 theta - 1 = -8(cos^2 theta - cos^4 theta)cos4θ−1=−8(cos2θ−cos4θ)
Divide both sides by
1/8(1-cos 4 theta) = cos^2 theta - cos^4 theta18(1−cos4θ)=cos2θ−cos4θ
Voila!
