# How to prove that $(\cos\theta)^p\leqslant\cos(p\theta)$ for every $0≤p≤1$ and $0≤θ≤π/2$

How can I prove the following inequality:

$$\forall p\in[0,1], \theta\in [0,π/2];\cos^p(θ)≤\cos(pθ)$$

Please help me to complete this proof . I suppose $g(θ)=\cos^p(θ)-\cos(pθ)$ and I want show that $g(θ)≤0$. And I use derivative and integral theorems like mean value theorem ,…. but i couldn’t solve it.

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Your notation confused me. Are you talking about $\cos(\theta^p)-\cos(p\theta)$ or $(\cos\theta)^p-\cos(p\theta)$ or something else? – Patrick Li Oct 30 '12 at 12:57
This expression is meaningless in every mathematics I've ever studied: $$cos〖(θ)^p≤cos(pθ) 〗$$ What do you intend that to mean? – Thomas Andrews Oct 30 '12 at 13:02
I define this function and will show that this function is smaller than zero ( (cos(θ))^p≤cos(pθ) ))by use of integral or derivation or other ways – maisam hedyelloo Oct 30 '12 at 13:03
Please use LaTeX. – ᴊ ᴀ s ᴏ ɴ Oct 30 '12 at 13:03
@ThomasAndrews: I think the OP entered the function here exactly as it is written in a worksheet for example in Maple or Matlab. Any way he 'd better fix it. – S. Snape Oct 30 '12 at 13:06

If you take $f(\theta)=\log(\cos\theta)$, you simply have to show that: $$p\,f(\theta) \leq f(p\,\theta).$$ The relation clearly holds for $\theta=0$, so it is sufficient to show that: $$f'(\theta) \leq f'(p\,\theta),$$ that is obvious, since the function $-f'=\tan(\theta)$ is increasing over $[0,\pi/2]$.
Let $u(\theta)=\log\cos(p\theta)-p\log\cos\theta$, then $u(0)=0$ and $u'(\theta)=p\cdot(\tan\theta-\tan(p\theta))$. For every $0\leqslant\theta\lt\pi/2$, $0\leqslant p\theta\leqslant\theta\lt\pi/2$ hence $\tan\theta\geqslant\tan(p\theta)$ and $u'(\theta)\geqslant0$. Thus, $u(\theta)\geqslant0$ for every $0\leqslant\theta\leqslant\pi/2$, that is, $\cos(p\theta)\geqslant(\cos\theta)^p$.