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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.

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  • $\begingroup$ It might help seeing that the function has period $\pi/2$ and is even. $\endgroup$ – egreg Apr 4 '19 at 13:03
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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)

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    $\begingroup$ Thanks. I agree with the maximum, but for the minimum, although $f(x)+g(x)$ is symmetric relative to $x=\frac{\pi}4$ and it decreases when it goes off from either $x=0$ or $x=\frac\pi2$, the minimum could still happen at somewhere else in $(0,\frac\pi2)$ not exactly at $x=\frac{\pi}4$ (for example $x=\frac18\pi$ or $x=\frac38\pi$) and $x=\frac{\pi}4$ could be a local maximum? $\endgroup$ – Kay K. Apr 4 '19 at 21:59
  • $\begingroup$ You are right, I think adding the symmetry argument around $\pi/4$ could do the trick. $\endgroup$ – Math-fun Apr 5 '19 at 18:22
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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

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  • $\begingroup$ This lemma is false: Consider $f(x)$ given by $2x^2$ on $[0,\frac12]$ and by $x$ on $[\frac12,1]$; then $f(x) + f(1-x) = 2x^2 - x + 1$ on $[0,\frac12]$, which has a minimum at $x=\frac14$. $\endgroup$ – FredH Apr 4 '19 at 6:28
  • $\begingroup$ @FredH found this counteexample, also. May be There should be a monotonous derivative for lemma to hold. $\endgroup$ – Lada Dudnikova Apr 4 '19 at 9:01
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First find $f(x +h) - f (x)$. Then check if $f(x) >0$ or $<0$, the function will increase or decrease respectively.

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