Show that $\tan^{-1}(\cos(x)) > \frac{\pi}{4} \cos(x)$ for every $x \in (0,\pi/2)$

Question

This is something I made up like that only. Show that $$\tan^{-1}(\cos x) > \frac{\pi}{4} \cos x$$ for every $x \in (0,\pi/2)$

My method was that both are convex in given interval so we can just compare area under curve. $$A_2 = \frac{\pi}{4}\int_0^{\pi/2} \cos(x) dx = \frac{\pi}{4}$$ $$A_1 = \int_{0}^{\pi/2} \tan^{-1}(\cos x) dx \\ = \frac{1}{2}\int_{0}^{\pi/2} \tan^{-1}(\tfrac{\sin x+\cos x}{1-\sin x \cos x}) dx \\ = headbang$$

Evident this approach fails. Also I tried $f(x) = \tan^{-1}(\cos x) - \frac{\pi}{4} \cos x$ still now clue. I show this to miss but she said out of bounds of syllabus. Can you solve it easy way. You may use or not use calculus.

Answer 1

Since $0<\cos x<1$ for $0<x<\pi/2$ it suffices to show that (set $y=\cos x$) $$ \arctan y>\frac{\pi}{4}y $$ for $0<y<1$. In fact, the left-hand side and the right-hand side are equal for $y=0$ and $y=1$. It is also easy to see that $y\mapsto \arctan y$ is strictly concave for $0<y<1$ (just differentiate twice). It follows that $\pi y/4$ is a secant, and thus that it is strictly less than $\arctan y$.

Answer 2

Consider $$ f(t)=\log\frac{\arctan t}{t} $$ over $(0,1)$; its limit for $t\to0$ is $0$. The derivative is $$ f'(t)=\frac{1}{\arctan t}\frac{1}{1+t^2}-\frac{1}{t}= \frac{1}{t\arctan t}\left(\frac{t}{1+t^2}-\arctan t\right) $$ In order to study it, we can consider $$ g(t)=\frac{t}{1+t^2}-\arctan t $$ with $$ g'(t)=\frac{1-t^2}{(1+t^2)^2}-\frac{1}{1+t^2}=-\frac{2t^2}{(1+t^2)^2}<0 $$ which means $g$ is decreasing. Since $g(0)=0$, we conclude that $f'(t)<0$ for $t\in(0,1)$. Therefore $f$ is decreasing and the same can be said for $$ F(t)=\frac{\arctan t}{t} $$ Since $$ \lim_{t\to 1}F(t)=\frac{\pi}{4} $$ we conclude that $$ F(t)>\frac{\pi}{4} $$ for $t\in(0,1)$

Answer 3

It is straightforward to prove that $\frac{\arctan x}{x}$ is decreasing on $(0,1)$, and this is equivalent to $\frac{\tan x}{x}$ being increasing on $\left(0,\frac{\pi}{4}\right)$ - not surprising since all the coefficients of the Taylor series at the origin of $\tan x$ are non-negative. In particular $$ \forall x\in(0,1),\qquad \frac{\arctan x}{x}>\frac{\arctan 1}{1}=\frac{\pi}{4} $$ and your inequality follows by simply setting $x=\cos\theta$.