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.

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

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 $$0 hence increasing $$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)

• 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? – Kay K. Apr 4 '19 at 21:59
• You are right, I think adding the symmetry argument around $\pi/4$ could do the trick. – Math-fun Apr 5 '19 at 18:22

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 $$a then $$f(a)+f(d) < f(c)+f(d)$$

Jensen's inequality

• 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 $x=\frac14$. – FredH Apr 4 '19 at 6:28
• @FredH found this counteexample, also. May be There should be a monotonous derivative for lemma to hold. – Lada Dudnikova Apr 4 '19 at 9:01

First find $$f(x +h) - f (x)$$. Then check if $$f(x) >0$$ or $$<0$$, the function will increase or decrease respectively.