Trigonometry – Proving cos(x) * cos(x-60°) * cos(x+60°) = 1/4 cos(3x)

Question

I should prove this trigonometric identity.
I think I should get to this point :
$\cos(3x) = 4\cos^3 x – 3\cos x $
But I don't have any idea how to do it (I tried solving $\cos(x+60^{\circ})\cos(x-60^{\circ})$ but I got nothing)

Answer 1

Using sum of angles identity, \begin{align*} \cos x\cos (x-60^{\circ})&\cos(x+60^{\circ})\\ & = \cos x(\cos x\cos 60^{\circ}+\sin x\sin 60^{\circ})(\cos x\cos 60^{\circ}-\sin x\sin 60^{\circ})\\ \\ & = \cos x(\cos^2x\cos^260^{\circ}-\sin^2x\sin^260^{\circ})\\\\ & = \cos x\left(\frac{1}{4}\cos^2x-\frac{3}{4}\sin^2x\right)\\\\ & = \frac{1}{4}(\cos^3x-3\cos x(1-\cos^2x))\\\\ & = \frac{1}{4}(4\cos^3x-3\cos x) \end{align*}