Question

prove: cos (3x) = cos^3(x) - 3cos(x) sin^2(x)

Answer 1

go backward from the right side: \[\cos^3 x-3\cos x \sin^2 x = \cos ^2 x \cos x-3cosxsin^2x\]
\[= \cos x(\cos^2 x - \sin^2 x - 2\sin^2x) = \cos x(\cos 2x -2 \sin^2 x) \]
\[= cosxcos2x - 2sinxcosx sinx = cosx \cos 2x - \sin2x sinx = \] left side

Answer 2

because cos (3x) = cos (2x +x) and we have formula exactly what I did from right side

Answer 3

hope this helps