The problem is such:
Show that $$\cos{px}\ge \cos^p{x}$$ given that $0\le x\le \pi/2$ and $0<p<1$
I'm pretty sure this is straight forward given the sum formula for cosine...
$$\cos{px}=\cos{[(p-1)x+x]}=\cos{[(p-1)x]}\cos{x}-\sin{[(p-1)x]}\sin{x}$$
Since $\sin{[(p-1)x]}<0$, then
$$\cos{px}=\cos{[(p-1)x]}\cos{x}+\sin{[(1-p)x]}\sin{x}\ge\cos{[(p-1)x]}\cos{x}$$
So if i do this $p-1$ more times I will end up with
$$\cos{px}\ge \cos{[(p-p)x]}\cos^p{x}=\cos^p{x}$$
My question, is there a better way to do this? Not that it is not good, but I was wondering if there were more indirect approaches or other more subtle ways. I was thinking DeMoivre, but was trying to stay within the Real realm and not delve into the Complex system.