More elegant way to prove $\cos(\phi)\leq 1-\frac{2\phi^2}{\pi^2} $?

Question

Today I came across the inequality $$ \cos(\phi)\leq 1-\frac{2\phi^2}{\pi^2}\tag{1} $$ which holds for (actually, iff) $\phi\in[-\pi,\pi]$; this can be used, e.g., to show that for all $z\in\mathbb C\setminus\{0\}$ $$ \operatorname{Arg}(z)\leq \frac\pi2\left|\frac{z}{|z|}-1\right|\,. $$ I believe most calculus students would prove (1) by showing that the function \begin{align*} f:[0,\pi]&\to\mathbb R\\ \phi&\mapsto 1-\frac{2\phi^2}{\pi^2}-\cos(\phi) \end{align*} is non-negative. As $f$ is differentiable it attains its minimum either at the boundary or at a critical point. As $f(0)=0=f(\pi)$ it suffices to show that $f$ restricted to $(0,\pi)$ has no local minima -- this can be done via some basic, yet somewhat tedious calculations (e.g., "$\sin(\phi)=\frac{4}{\pi^2}\phi$ has exactly one solution in $(0,\infty)$, and this solution is in $(\frac{\pi}2,\pi)")$.

This seems overly complicated and it feels like there should exist a more "elegant" approach to this problem, but so far I could not come up with anything of sorts. For example similar inequalities can be cast into inequalities of certain integrals (e.g., here), but I was not yet able to cast the above problem into such a framework. Any comments and ideas are welcome!

Answer 1

It is easy to show that $\sin(\phi)\ge 2\phi/\pi$ for $0\le \phi\le \pi/2$. Then, we have

$$\cos(\phi)\le \sqrt{1-(2\phi/\pi)^2}\le 1- \frac{2\phi^2}{\pi^2}\tag1$$

for $\phi\in [0,\pi/2]$ since $\sqrt{1-x}\le 1-\frac12x$ for $0\le x\le1$.

Now, use analogous analysis to show that the inequality applies for $-\pi/2\le \phi\le 0$.

Finally, for $\pi/2\le |x|\le\pi$, $\cos(\phi)\le 0$ while $1-2\phi^2/\pi^2\ge 0$.

---

EDIT: This edit is in response to a comment regarding the proof of the inequality that $\sin(x)\ge 2x/\pi$ on $[0,\pi/2]$ without relying on calculus.

To show that $\sin(\phi)\ge 2\phi/\pi$ without relying on calculus, we make use of the identity

$$\sin(2a)+\sin(2b)-\sin(a+b)=-4\sin^2(a-b)\sin(a+b)$$

Then for any $\phi\in [0,\pi/2]$ and $\theta\in[0,\pi/2]$, it is easy to see that

$$\frac{\sin(\phi)+\sin(\theta)}2\le \sin\left(\frac{\phi+\theta}{2}\right)$$

Therefore, the sine function is concave on $[0,\pi/2]$. Inasmuch as $\sin(0)=0$ and $\sin(\pi/2)=1$, it is evident that

$$\sin(\phi)\ge 2\phi/\pi$$

as was to be shown.

Answer 2

I cannot resist the pleasure of using my favored $\large 1,400$ years old approximation

$$\cos(\phi) \simeq\frac{\pi ^2-4\phi^2}{\pi ^2+\phi^2}\qquad \text{for}\qquad -\frac \pi 2 \leq \phi\leq\frac \pi 2$$

$$ 1-\frac{2\phi^2}{\pi^2}-\cos(\phi) \simeq 1-\frac{2\phi^2}{\pi^2}-\frac{\pi ^2-4\phi^2}{\pi ^2+\phi^2}=\frac{\phi ^2 \left(3 \pi ^2-2 \phi ^2\right)}{\pi ^2 \left(\phi ^2+\pi ^2\right)} ~> ~0.$$

Answer 3

By means of the half angle formula $\cos\phi=1-2\sin^2\frac\phi 2$ the claim is that $|\sin\frac\phi 2|\geq\frac{|\phi|}{\pi}$ on the interval $[-\pi,\pi]$. Now, it is enough to show that $\sin x\geq\frac{2x}{\pi}$ on the interval $[0,\frac\pi 2]$. But this follows from the comparison of the graphs of $y=\sin x$ and $y=\frac{2x}\pi$ on the interval $[0,\frac\pi 2]$. Or it may follow from Bhaskara's method with some help from WA.