Question

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

Answer 1

`= -3/8 - (cos 4t)/8 + (cos 2t)/2`

Explanation:

`2sin^2 t = (1 - cos 2t)`.
Square the 2 sides:
`4sin^4 t = (1 - cos 2t)^2 = 1 + cos^2 2t - 2cos 2t`
Replace `cos^2 2t` by `(1 + cos 4t)/2`
`4sin^4t = 1 + 1/2 + cos (4t)/2 - 2cos 2t =`
`3/2 + (cos (4t)/2) - 2cos 2t`
Divide both sides by -4 -->

`- sin^4 t = - 3/8 - (cos 4t)/8 + (cos 2t)/2`