Show that $\tan^{-1}(\cos(x)) > \frac{\pi}{4} \cos(x)$ for every $x \in (0,\pi/2)$ 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.
 A: 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 $\arctan y$.
A: 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 $t\in(0,1)$
A: 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 $x=\cos\theta$.
