Proof: $\cos^p (\theta) \le \cos(p\theta)$ I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!
Enough backstory. Time for math. The problem is as follows:
Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.
I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.
Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions!

Set $f(\theta) = \cos p\theta - \cos^p \theta$ for some fixed $0 \leq p\leq 1$. Then for $\theta\in [0, \pi/2]$,
$$f'(\theta) = p\left(\cos^{p-1}\theta \sin \theta - \sin p\theta\right) \geq p\left(\sin \theta - \sin p\theta\right) \geq 0,$$
since $\cos^{p-1}\theta \geq 1$ (as $p - 1\leq 0$) and $\sin \theta$ is increasing on $[0, \pi/2]$. Since $f(0) = 0$, it follows that $f$ is increasing and thus nonnegative on $[0, \pi/2]$, as required.
