What is cos^2thetasin^2theta 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 theta
Since this is of degree
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
cos 4 theta - 1 = -8(cos^2 theta - cos^4 theta)
Divide both sides by
1/8(1-cos 4 theta) = cos^2 theta - cos^4 theta
Voila!