Finding fundamental period of $f(x)=\sec($sin(x)$)$ 
Find the fundamental period of $f(x)=\sec($sin(x)$)$. Choose the correct option.
(A)$\frac{\pi}{2}$  (B)$2\pi$   (C)$\pi$ (D)Aperiodic

We know the fundamental period of $\sin(x)$ is $2\pi$. Since $\sec(x)$ is an even function, both negative and positive values of x having same magnitude give same value of $\sec(x)$. Since |$\sin(x)$| has a period $\pi$, I concluded the fundamental period of $f(x)$ is $\pi$. The answer is correct. Is there any other disciplined way to approach this question, since I used some logic to solve this problem.
 A: First observe that $\pi$ is a period of this function because $\sec (\sin (\pi +x))=\sec (-\sin \, x)=\sec (\sin \, x))$. Next you have to prove that no number in $(0,\pi)$ is a period. Let $0<p<\pi$. If possible suppose $\sec (\sin (p +x))=\sec (\sin \, x)$ for all $x$. Put $x=0$ to see that $\sec (\sin \, p )=1$. This means $\cos (\sin \, p )=1$. But $t=\sin \, p$ lies strictly between $0$ and $1$ so $\cos \, t$ cannot be $1$. This proves that $\pi$ is the fundamental period.  
A: This is just a ruling out procedure assuming one of the answers is correct.
Consider
$f(x)=\cos (\sin x);$
$f(x+T)=\cos (\sin (x+T))=$
$\cos ( \sin x \cos T +\cos x \sin T);$
1) $T=π/2$:
$f(x+π/2)= \cos (\cos x);$
$f(x)= \cos (\sin x) \not =f(x+π/2)= \cos( \cos x).$
2) Check $T=π√$
3) $T= 2π$ is not the fundamental period (answer $2$)
4) Rule out .
Show that $T=π$ is the fundamental period.
Since $π$ is a period (possibly not fundamental) we conclude that any possible fundamental period needs to be of the form $π/k$, where $k >2$ , e.g. $π/3, π/5, π/7, π/9, ......$ (Rule out denominators with multiples of $2$(why?)).
Then 
$\sin x=\sin x \cos π/k + \cos x \sin π/k$
$0 < \sin π/k =:b <1$, $0< \cos π/k =:a <1$.
$\sin x =a \sin x +b \cos x$;
This is impossible for $x \in \mathbb{R}$.
We conclude $T=π$ is the fundamental period.
