How do you prove #sin4x = 4sinxcos^3x - 4sin^3xcosx#?

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Answer 1 · 2015-05-28T22:57:02

#cos 2theta = cos^2 theta - sin^2 theta#

#sin 2theta = 2sin theta cos theta#

Then:

#sin 4x = 2sin 2x cos 2x#

#=2(2 sin x cos x)(cos ^2x - sin^2 x)#

#= 2(2sin x cos x cos^2 x - 2sinx cos x sin^2 x)#

#=4sin x cos^3 x - 4 sin^3 x cos x#