# How to determine the period of composite functions?

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 $$2\pi$$?

• $f$ is the product of two functions, not the composition of two functions.
– lhf
Nov 13, 2018 at 17:22
• $2\pi$ is the LCM, the least common multiple, of the periods of the other functions. Nov 13, 2018 at 17:23
• @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... Nov 13, 2018 at 17:33

$$f(x)=\cos(2x)\sin(3x)=g(x)h(x)$$ 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 $$LCM(\pi,\frac{3\pi}{2})=2\pi$$
• 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 $f(x+\pi)=-f(x)$. Nov 13, 2018 at 19:51
• Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$. Nov 13, 2018 at 20:27