What is -sin^4thetasin4θ in terms of non-exponential trigonometric functions?

1 Answer
Mar 18, 2016

= -3/8 - (cos 4t)/8 + (cos 2t)/2=38cos4t8+cos2t2

Explanation:

2sin^2 t = (1 - cos 2t)2sin2t=(1cos2t).
Square the 2 sides:
4sin^4 t = (1 - cos 2t)^2 = 1 + cos^2 2t - 2cos 2t4sin4t=(1cos2t)2=1+cos22t2cos2t
Replace cos^2 2tcos22t by (1 + cos 4t)/21+cos4t2
4sin^4t = 1 + 1/2 + cos (4t)/2 - 2cos 2t = 4sin4t=1+12+cos(4t)22cos2t=
3/2 + (cos (4t)/2) - 2cos 2t32+(cos(4t)2)2cos2t
Divide both sides by -4 -->

- sin^4 t = - 3/8 - (cos 4t)/8 + (cos 2t)/2sin4t=38cos4t8+cos2t2