# Showing that $\cos^p{\Theta} \le \cos{p\Theta}$, for $0<\Theta<\frac{\pi}{2}$ and $0<p<1$, by analyzing $f(\theta)=\frac{\cos^p\Theta}{\cos p\Theta}$

Given $$0 < \Theta < \frac{\pi}{2}$$ and $$0 < p < 1$$, show that $$\cos^p{\Theta} \le \cos{p\Theta}$$

Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.

Proof. Let $$f(\Theta) = \frac{\cos^p{\Theta}}{\cos{p\Theta}}$$considering $$p$$ fixed. The derivative is

$$f'(\Theta) = \frac{p*\cos^{p-1}{\Theta}*{(-\sin{\Theta})*\cos{p\Theta} -\cos^p{\Theta}*(-\sin{p\Theta}*p) } } {\cos^2{p\Theta}} = \frac{p*\cos^{p-1}{\Theta}*\sin{(p-1)\Theta}}{\cos^2{p\Theta}}$$

We have $$f(0)=1$$, and because $$0, the term $$\sin(p-1)\Theta$$ makes the derivative negative: $$f'(\Theta) < 0$$ on $$(0,\frac{\pi}{2})$$.

At $$0$$ we have $$f'(0)=0$$, so we cannot immediately deduce that $$f$$ is decreasing on all of $$[0,\frac{\pi}{2})$$.

However, considering that $$f(0)=1, f(\frac{\pi}{2})=0$$, and both $$f$$ and $$f'$$ are continuous on $$[0,\frac{\pi}{2})$$, if it were the case that $$f$$ rises above $$1$$ at some $$\Theta \in (0,\frac{\pi}{2})$$, $$f$$ would have to have a local maximum somewhere in $$(0,\frac{\pi}{2})$$, with the derivative vanishing at that point, which cannot happen. Therefore $$f(\Theta) \le 1$$ throughout $$[0,\frac{\pi}{2}]$$, Q.E.D.

• if it were the case that $f$ rises above $1$ at some $\Theta \in (0,\frac{\pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,\frac{\pi}{2})$. Why??? Jan 22, 2019 at 9:39
• As a continuous function on a compact interval $[0,\frac{\pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(\Theta)>1$ somewhere in $(0,\frac{\pi}{2})$, then that maximum must lie within $(0, \frac{\pi}{2})$ because the values on the borders are 0 and 1. Jan 22, 2019 at 9:52
• Yes, it is right. Jan 22, 2019 at 9:56

suppose $$f(x)$$ is continuous on $$[a,b]$$ and has derivative in $$(a,b)$$. If $$f'(x)<0$$ for $$x\in(a,b)$$, then $$f(x)$$ is strictly decreasing on $$[a,b]$$.
So you need only $$f'(\Theta) < 0$$ for $$\Theta\in(0,\pi/2)$$ in your situation!
Also we can deal this one as follows: For fix $$p\in(0,1)$$, let $$f(x)=\cos px-\cos^px,x\in[0,\pi/2],$$ then $$f$$ is continuous on $$[0,\pi/2]$$ and is differentiable in $$(0,\pi/2)$$. Its derivative is $$f'(x)=p\sin x\left(\cos^{p-1}x-\frac{\sin px}{\sin x}\right),x\in(0,\pi/2).$$ Because $$p\in(0,1)$$, $$0, we have $$0<\frac{\sin px}{\sin x}<1<\frac{1}{\cos^{1-p}x}=\cos^{p-1}x.$$ So $$f'(x)>0,\ \forall x\in(0,\pi/2).$$ This implies $$f(x)>f(0)=0,\forall x\in(0,\pi/2).$$