What is the fundamental period of $\cos(\sin x) + \cos(\cos x)$? If we let $f(x) = \cos(\sin x) + \cos(\cos x)$, then it is easy to show that $f(x+ \pi/2)=f(x)$, this shows that $\pi/2$ is a period of $f$, but the problem is that how we can show that $\pi/2$ is the fundamental period (means for no real number $T$ less than $\pi/2$ we have $f(x+T)=f(x)$ for all $x\in\mathbb{R}$ holds).
There are many authors who claim that $\pi/2$ is the fundamental period of $f(x)$, I tried to prove it but couldn't do it.
Please help.
Edit: Non calculus solution is preferable because this problem is usually tackled in Algebra/Trigonometry which is usually taught before Calculus. I am not sure whether it could be solved without Calculus at last.
 A: Whatever the value of $f(0)=f(\pi/2)$, we will prove that this value is reached nowhere else in $(0,\pi/2)$, and this will prove that $\pi/2$ must be the fundamental period.
Represent $f(x)=g(t)+g(1-t)$, where:
$$g(t)=\cos\sqrt{t}$$
$$t=\sin^2x$$
for $x\in[0, \pi/2]$, i.e. $t\in[0,1]$. It is fairly easy to prove that $g$ is strictly convex on $[0,1]$: its second derivative on $(0,1)$ is:
$$\frac{d^2g}{dt^2}=\frac{\sin\sqrt{t}-\sqrt{t}\cos\sqrt{t}}{4t^{3/2}}$$
and (the numerator) $\sin\sqrt{t}-\sqrt{t}\cos\sqrt{t}\gt 0\iff \tan\sqrt{t}\gt\sqrt{t}$ which is true for $t\in[0,1]$.
Now, from strict convexity of $g$ follows strict convexity of $f$ (as a function of $t$) - so $f$ can only achieve a maximum on $t\in [0,1]$ at the end-points (which correspond to $x=0, \pi/2$).
A: For $x\in[0, \pi/2]$:
$$\begin{array}{rcl}\frac{\sin x + \cos x}{2}&=&\frac{\sin x + \sin(\pi/2-x)}{2}\\&=&\sin(\pi/4)\cos(x-\pi/4)\\&\le&\sin(\pi/4)\end{array}$$
with equality satisfied only for $x=\pi/4$.
In fact, from the same expression, it is also easy to see that, for $x\in [0, \pi/2]$, we actually have
$$\frac{\sin x+\cos x}{2}\in\left[\sin(\pi/4)\cos(\pi/4), \sin(\pi/4)\right]=\left[\frac{1}{2}, \frac{1}{\sqrt{2}}\right]$$
Notice that $x\mapsto \cos x$ is (strictly) decreasing on this interval.
Now (again for $x\in[0, \pi/2]$):
$$\begin{array}{rcl}\cos(\sin x)+\cos(\cos x)&=&2\cos\frac{\sin x + \cos x}{2}\cos\frac{\sin x-\cos x}{2}\\&\ge&2\cos\frac{\sin x + \cos x}{2}\\&\ge&2\cos\left(\sin(\pi/4)\right)\\&=&\cos(\sin(\pi/4))+\cos(\cos(\pi/4))\end{array}$$
with the first inequality turning into equality also only when $\cos\frac{\sin x-\cos x}{2}=1$, i.e. when $x=\pi/4$. (In the second inequality we used the inequality established above and the fact that $\cos x$ is strictly decreasing.)
Thus, on $x=[0,\pi/2$] the function $f(x)=\cos(\sin x)+\cos(\cos x)$  reaches the value $f(\pi/4)$ only once (for $x=\pi/4$) while in all the other cases we have $f(x)>f(\pi/4)$.
That proves that the fundamental period is $\pi/2$, or otherwise there would be multiple points (on $[0,\pi/2]$) on which $f$ takes that value. Namely, if there was a smaller period $T$ of $f$, $0<T<\pi/2$, WLOG we can assume that $T\le\pi/4$ (or otherwise we can take $\pi/2-T$ instead of $T$ as a period), and this would give us additional points $\pi/4\pm T$ with the same value $f(\pi/4\pm T)=f(\pi/4)$.
