For exemple, take :$f\left(x\right)=$cos$\left(2x\right)\cdot \:$sin$\left(3x\right)$. Period of cos$\left(2x\right)$ is $\pi$ and that of sin$\left(3x\right)$ is $\frac{2\pi }{3}$. But why is the period of $f\left(x\right)$ $2\pi$?

For good measure, here's another example: How do I prove that the period of $f\left(x\right)=\frac{tan\left(x\right)}{1+sin\left(x\right)}$ is $2\pi$?

  • 5
    $\begingroup$ $f$ is the product of two functions, not the composition of two functions. $\endgroup$
    – lhf
    Nov 13, 2018 at 17:22
  • 3
    $\begingroup$ $2\pi$ is the LCM, the least common multiple, of the periods of the other functions. $\endgroup$
    – Mason
    Nov 13, 2018 at 17:23
  • $\begingroup$ @Mason. Ok, but why is it like that? I mean, why do we use the LCM? I was kind of searching to build a strong intuitive notion... $\endgroup$ Nov 13, 2018 at 17:33

1 Answer 1


$$ f(x)=\cos(2x)\sin(3x)=g(x)h(x)$$ Since $g$ is $\pi-$periodic and $h$ is $\frac{3\pi}{2}-$periodic and $\pi\neq\frac{3\pi}{2}$, then it is clear that the period $T$ of $f$ is a multiple of $\pi$ (to guaranty that the periodicity of $g$) and a multiple of $\frac{3\pi}{2}$ (to guaranty that the periodicity of $h$) at the same time. Hence, $T$ is the $LCM(\pi,\frac{3\pi}{2})=2\pi$

You can use the same logic to answer the second question.

  • 3
    $\begingroup$ However this proves only that $2\pi$ is a period, not that it is the smallest period. For example, while $\sin(x)$ and $\cos(x)$ have smallest period $2\pi$, their product has smallest period $\pi$.In this case, looking at the zeros of $f(x)$ yields that a smaller period would have to be $\pi$, but we have $f(x+\pi)=-f(x)$. $\endgroup$
    – Ingix
    Nov 13, 2018 at 19:51
  • $\begingroup$ Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$. $\endgroup$ Nov 13, 2018 at 20:27

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