Functions — Question 3

Question 3 - Solution

(a) Although $\sin x$ has period $2\pi$, the function $$f(x)=\cos(\sin x)$$ has a smaller period because cosine is even.

Using $$\sin(x+\pi)=-\sin x$$ and $$\cos(-u)=\cos(u),$$ we get $$f(x+\pi) = \cos(\sin(x+\pi)) = \cos(-\sin x) = \cos(\sin x) = f(x).$$ Therefore, $f(x)$ repeats after $\pi$. The reason the period changes from $2\pi$ to $\pi$ is that shifting by $\pi$ changes $\sin x$ to $-\sin x$, but cosine gives the same value for opposite inputs.

Thus, the period is $$\boxed{\pi}.$$

(b) To prove that $f(x)$ is even, compute $f(-x)$: $$f(-x) = \cos(\sin(-x)) = \cos(-\sin x) = \cos(\sin x) = f(x).$$ Therefore, $$\boxed{f(x)\text{ is even}.}$$

(c) Solve $$\cos(\sin x)=1.$$ Cosine equals $1$ when its input is $2k\pi$. Thus, $$\sin x = 2k\pi.$$ But since $-1\leq \sin x \leq 1$, the only possible value is $$\sin x=0.$$ On $[0,2\pi]$, this occurs at $$x=0,\ \pi,\ 2\pi.$$

Therefore, $$\boxed{x=0,\ \pi,\ 2\pi}.$$