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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!

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5 Answers 5

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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.

enter image description here

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

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

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  • $\begingroup$ 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. $\endgroup$ Sep 3, 2014 at 8:22
  • $\begingroup$ @anomaly:How can "$\cos^{p-1}\theta \ge 1$"? $\endgroup$
    – P.Styles
    Jun 20, 2018 at 16:44
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    $\begingroup$ @PKStyles: As indicated, $p - 1\leq 0$. $\endgroup$
    – anomaly
    Jun 20, 2018 at 19:19
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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.

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

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

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $enter image description here

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 $\theta=0$.

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