Prove that #cos 3x = 4cos^3 x - 3cos x# ?

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Answer 1 · 2015-11-08T00:17:37

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

Apply the trig identities:

We get:

#cos 3x = cos (x + 2x) = cos x * cos 2x - sin x * sin 2x#

#= cos x(2cos^2 x - 1) - 2sin^2 x * cos x #

#= 2cos^3 x - cos x - 2cos x(1 - cos^2 x) #

#= 2cos^3 x - cos x - 2cos x + 2cos^3 x #

So

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