The smallest positive period of the function $\sin{(k\cos{x})}$ let $$f(x)=\sin{(k\cos{x})},k&gt;0$$
  Find the minimum of $T&gt;0$,such
  $$f(x+T)=f(x),\forall x\in R$$


I have know $$f(x+2\pi)=\sin{(k\cos(x+2\pi))}=f(x)$$
maybe $T_{min}=2\pi$?,I can't proof why$2\pi$  is smallest? Thanks
Let $T$ be the minimum period of $f$. Since the period of $\sin(x)$ is $2\pi$, it must be true that

$$k\cos(x+T)=k\cos(x)+2\pi n\\
\implies \cos(x+T)=\cos(x)+\frac{2\pi n}{k}$$

for some $n \in \Bbb{Z}$.  The range of $k\cos(x+T)$ is $[-1,1]$. The range of $\cos(x)+\frac{2\pi n}{k}$ is $\big[-1+\frac{2\pi n}{k}, 1+\frac{2\pi n}{k}\big]$.  These can only be equal if $n=0$.  Hence,

$$\cos(x+T)=\cos(x)$$

But we know the minimum period of $\cos(x)$ is $2\pi$, so $T=2\pi$.
If
$f(x)=\sin(k\cos(x))
$
then
$f(x+2\pi)
=\sin(k\cos(x+2\pi))
=\sin(k\cos(x))
=f(x)
$
since
$\cos(x+2\pi)
=\cos(x)
$.