prove that : $\cos x \cdot \cos(x-60^{\circ}) \cdot \cos(x+60^{\circ})= \frac14 \cos3x$
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)
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*}
$ \cos (x-60^\circ ) \times \cos ( x + 60^\circ ) = \frac{1}{2} ( \cos (2x) + \cos 120^\circ) = \frac{1}{2} \cos (2x) - \frac{1}{4}$
$ (\frac{1}{2} \cos (2x) - \frac{1}{4}) \times \cos x = \frac{1}{4} ( \cos (3x) + \cos (x) ) - \frac{1}{4} \cos (x) = \frac{1}{4} \cos (3x) $