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

1 Answer

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

Explanation:

Apply the trig identities:

  • #cos (a + b) = cos a * cos b - sin a * sin b#
  • #cos 2x = 2cos^2 x - 1#
  • #sin 2x = 2sin x * cos x#

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#