# 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!

Let us fix the value of $\theta$ and vary $p$.

For $p=0$, $\cos^0(\theta)=\cos(0\theta)=1$.

For $p=1$, $\cos^1(\theta)=\cos(1\theta)=\cos(\theta)$.

Then, $$(\cos^p(\theta))''=(\log(\cos(\theta)))^2\cos^p(\theta)\ge0,$$ and $$(\cos(p\theta))''=-\theta^2\cos(p\theta)\le0.$$ The LHS function is concave down (negative exponential) and the RHS function is concave up (cosinusoid). They meet at endpoints without crossing.

With the straight line, this also establishes $$\color{blue}{\cos^p(\theta)}\le\color{magenta}{1-p(1-\cos(\theta))}\le\color{green}{\cos(p\theta)}.$$

• How did you get the log in the second derivative? Sep 3, 2014 at 8:39
• The variable is $p$, so you derive an exponential $a^p$.
– user65203
Sep 3, 2014 at 8:40
• I like it. I like it a lot. Sep 3, 2014 at 8:42
• You should like my other solution too.
– user65203
Sep 3, 2014 at 9:42

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.

• I should have thought of taking the derivative of the equation! It's a standard tool! Thank you. I will remember to try taking the derivative next time. Sep 3, 2014 at 8:22
• @anomaly:How can "$\cos^{p-1}\theta \ge 1$"? Jun 20, 2018 at 16:44
• @PKStyles: As indicated, $p - 1\leq 0$. Jun 20, 2018 at 19:19

Because cosine is concave on the interval $[0,\pi/2]$, we have $$\cos(p\theta)=\cos(p\theta+(1-p)0)\geq p\cos(\theta)+(1-p)\cos(0)=p\cos(\theta)+(1-p).$$ So our desired inequality follows if we can prove that $$\cos^p(\theta)\leq p\cos(\theta)+(1-p).\tag{*}$$ Clearly, (*) holds if $\theta=\frac{\pi}{2}$ (the LHS is $0$ while the RHS is nonnegative), so assume $\theta<\frac{\pi}{2}$. This assumption means $\cos(\theta)>0$ so that $\cos(\theta)-1>-1$, allowing us to apply Bernoulli's inequality: $$\cos^p(\theta)=[1+(\cos(\theta)-1)]^p\leq 1+p(\cos(\theta)-1)=p\cos(\theta)+(1-p).$$ This completes our proof.

• Didn't know Kim Jong Un is on math.SE! Please don't kill us sir, we are sorry! Sep 3, 2014 at 8:32

Set $f(\theta)=\cos^p(\theta)-\cos(p\theta).$ Then, $f(0)=0$. Differentiate $f$ w.r.t. $\theta$ you'll get \begin{align} f'(\theta)&=-p\cos^{p-1}(\theta)\sin(\theta)+p\sin(p\theta)\\ &=-p\cos^{p-1}(\theta)\sin(\theta)+p\sin(\theta)-p\sin(\theta)+p\sin(p\theta)\\ &=p\sin(\theta)[1-\cos^{p-1}(\theta)]+p[\sin(p\theta)-\sin(\theta)]. \end{align}

Note that for $\theta\in[0,\pi/2)$ we have $0<\cos(\theta)\le 1$. Since $0\le p\le 1$ then $\cos^{p-1}(\theta)\ge 1$ that implies $p\sin(\theta)[1-\cos^{p-1}(\theta)]\le0$.

On the other hands, we also have $\sin$ is increasing on $[0,\pi/2)$. Since $p\theta\le \theta$ then $\sin(p\theta)\le \sin(\theta)$ that implies $p[\sin(p\theta)-\sin(\theta)\le0$.

It follows that $f'(\theta)\le 0$ which means that $f$ is decresing. Since $f(0)=0$ then $f(\theta)\le 0$ for all $\theta\in[0,\pi/2]$. Of course, this leads to what you wanted.

$$\color{blue}{\cos^p(\theta)}\le\color{green}{\cos(p\theta)}.$$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$

For fixed $p$ ($0.5$ in the figure), take the logarithm $$p\log(\cos(\theta))\le\log(\cos(p\theta)),$$ and derive on $\theta$ $$-p\tan(\theta)\le-p\tan(p\theta).$$ The latter inequality is obviously true as the $\tan$ function is increasing. This shows that the $\color{blue}{LHS}$ of the original inequality decreases faster than the $\color{green}{RHS}$, and they are equal at $\theta=0$.