Question

What is `cos^5theta-6sin^3theta` in terms of non-exponential trigonometric functions?

Answer 1

Use these trig identities to transform `cos^5 x` and `sin^3 x`
`cos 2x = 2cos^2 x - 1 = 1 - 2sin^2 x`
`sin^2 x + cos ^2 x = 1`

(1) --> `cos^5 x = cos x(cos^2 x)(cos^2 x) =`
`= cos x(1 - sin x)(1 + sin x)((1 + cos 2x)/2)`
(2) --> `6sin^3 x = 6sin x(sin^2 x) = 6sin x(1 - cos x)(1 + cos x).`
f(x) = (1) + (2)
`f(x) = (1/2)cos x(1 - sin x)(1 + sin x)(1 + cos 2x) - 6sin x(1 - cos x)(1 + cos x)`