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

1 Answer
Nghi N.
Dec 18, 2015

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)

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