What is the period of $f(x) = \cos (x) \cos(2x) \cos(3x)$?

Question

What is the period of $f(x) = \cos(x) \cos(2x) \cos(3x)$? Please tell me the method plus the logic behind solving these kind of problems .. Plus is there any property for even functions like even functions are always onto ?

Answer 1

\begin{align} \cos(3x - x) &= \cos 3x \cos x + \sin 3x \sin x \\ \cos(3x + x) &= \cos 3x \cos x - \sin 3x \sin x \\ \hline \cos x \cos 3x &= \dfrac 12 \cos 2x + \dfrac 12 \cos 4x \\ \cos x \cos 3x &= \dfrac 12 \cos 2x + \dfrac 12 (2 \cos^2 2x - 1) \\ \cos x \cos 3x &= \dfrac 12 \cos 2x + \cos^2 2x - \dfrac 12 \\ \hline \cos x \cos 2x \cos 3x &= \dfrac 12 \cos^2 2x + \cos^3 2x - \dfrac 12 \cos 2x \\ \end{align}

The period is $\pi$.

Answer 2

That the product has period $\pi$ is easily seen, once we prove $$\cos x\cos2x\cos3x =\frac{1}{4}\left(\cos6x+\cos4x+\cos2x+1\right).$$There are at least two quick ways to do this. One uses $\cos nx =\tfrac{1}{2}\left(z^n+z^{-n}\right)$ with $z:=e^{ix}$; the other uses $\cos A\cos B=\tfrac{1}{2}\left(\cos\left(A+B\right)+\cos\left(A-B\right)\right)$.

Edited for a more detailed explanation: the product is $$\tfrac{1}{8}\left(z+z^{-1}\right)\left(z^2+z^{-2}\right)\left(z^3+z^{-3}\right)=\tfrac{1}{8}\left(z^6+z^{-6}+z^4+z^{-4}+z^2+z^{-2}+2\right).$$The right-hand side is $\tfrac{1}{8}\left(2\cos 6x+2\cos 4x+2\cos 2x+2\right).$