Calculate periodicity of $f(x) = \cos(\sin(x))$

Question

Given the function:

$$f(x) = \cos(\sin(x))$$

Calculate the periodicity of $f(x)$.

I understand it equals $\pi$ but how would a solid answer look like?

Thank you.

Answer 1

If $r(x)$ has period $k$, then $r(s(x))$ has a period less than or equal to $k$

$$r(x)=\cos x, \ s(x)=\sin x$$

For $s(x)$ period as $2 \pi$, $r(s(x))$ has period less than $2 \pi$. So, checking for half-period i.e., $\pi$,

$$\cos(\sin (x+\pi))=\cos(\sin (-x))=\cos( \sin x)$$

For the fundamental period check,

$$f(x+\frac{ \pi}{t})=\cos (\sin(x+\frac{\pi}{t}))$$ for integer $t$

Clearly for $t>1$, no integer satisfy the periodicity property. So, $\pi$ is the fundamental period.

It is worth noting that $t$ is generally to be considered as a rational no. and not an integer(i.e., period of $\cos \frac{px}{q}$ is $\frac{2 \pi}{k}, k=\frac{p}{q}$), but here since coefficient of $x=1$, we consider only integers.

Answer 2

Basically, a function $f(x) $ has a period of $L$ if: $$f(x+L) = f(x) $$

Now, note that the period of $\sin x$ is $2\pi$, where it is negative for half of the time. And since $\cos(-x) =\cos x$, we obtain: $$\text{ Period of} \cos(\sin x) = \frac12\left[\text{Period of} \sin x \right] = \pi$$

Answer 3

$$\cos\sin x=\cos \sin y\iff \frac{\sin x-\sin y}{2\pi}\in\Bbb Z\vee\frac{\sin x+\sin y}{2\pi}\in\Bbb Z$$

Now, $\lvert\sin x\pm\sin y\rvert\le2<2\pi$ and therefore the two cases are equivalent to $$\sin x=\sin y\vee \sin x=-\sin y$$ or, $\sin^2x=\sin^2y$. In other words, if and only if $\cos(2x)=\cos (2y)$. Which means that the periods of $\cos\sin x$ are exactly the periods of $\cos(2x)$. In other words, the multiples of $\frac{2\pi}2=\pi$ by some positive integer.