Find extrema of $\cos\left(\frac\pi2\cos x\right)+\cos\left(\frac\pi2\sin x\right)$ without differentiation 
Question is: find minimum and maximum of $f(x)$:
  $$f(x)=\cos\left(\frac\pi2\cos x\right)+\cos\left(\frac\pi2\sin x\right)$$
  without differentiation.

This problem is supposed to be solved only with pre-calculus knowledge, but I have no idea how to do it.
$f(x)$ decreases monotonically from $\frac{n\pi}2$ to $\frac{n\pi}2+\frac\pi4$, and monotonically increases from $\frac{n\pi}2+\frac\pi4$ to $\frac{(n+1)\pi}2$, but how can it be proved the monotonicity without calculus?
I also tried to transform the expression into
\begin{align}f(x)=&2\cos\left(\frac\pi4(\cos x+\sin x)\right)\cos\left(\frac\pi4(\cos x-\sin x)\right)\\=&2\cos\left(\frac{\sqrt2\pi}4\sin \left(x+\frac\pi4\right)\right)\cos\left(\frac{\sqrt2\pi}4\sin\left(-x+\frac\pi4\right)\right)\end{align}
but found it has the same issue of proving the monotonicity.
 A: I really like this question!
Consider $f(x)=\cos(\frac{\pi}2\cos x)$ and $g(x)=\cos(\frac{\pi}2\sin x)$. Since $\cos$ is positive over $(-\frac{\pi}2,\frac{\pi}2)$ both of these functions are positive. Further as $\cos(x-\pi/2)=\sin(x)$, $g(x)$ decreases when $f(x)$ increases and vice versa, leading to the fact the minimum happens when $f(x)=g(x)$. For the maximum you may use the phase shift of $\pi/2$ and the unit circle to see that $\cos x$ is larger than $\sin x$ for say $0<x<\pi/4$ hence increasing $x$ from zero towards $\pi/4$, the increase in $f(x)$ is much slower than the drop in $g(x)$ hence the maximum must be $1$ (you loose more than what you get :-) hahaha, so the sum must drop) 
A: First find $f(x +h) - f (x)$. Then check if $f(x) >0$ or $<0$, the function will increase or decrease respectively. 
A: I think you can use a lemma that if $f(x)$ is continuous and monotonic, then $f(x)+f(1-x)$ is monotonic on the interval  $\left(0,\frac{1}{2}\right)$
UPD: $\cos(x)$ is a convex function on $(0, \pi)$, so you can use some kind of Jensen's inequality. If you have $a<b<c<d$ then $f(a)+f(d) < f(c)+f(d)$
Jensen's inequality
