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.
Best Answer
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.