# Upper and lower bound of $f(x)=(\tan x)^{\sin 2x}$ for $x \in (0, \frac{\pi}{2})$

Let define $f(x)=(\tan x)^{\sin 2x}$ for $x \in (0, \frac{\pi}{2})$

Please help me prove, that $f$ reaches its lower bound in only one point $x_1$ and reaches its upper bound $x_2$ also in only one point of domain.

Calculate $x_1 + x_2$.

What I have done:

I've checked the derivative, but it's looking horrible:

http://www.wolframalpha.com/input/?i=%28tg+x%29%5E%28sin+2x%29

The key to find the extremum is to calculate $x$, that satisfy $\cos 2x \cdot \ln (\tan x) = -1$

I don't have any idea how to do this. All help and hints are appreciated. Thanks in advance!

Under the transformation $x \mapsto \frac{\pi}{2} - x$,
• $\tan(x) \mapsto \frac{1}{\tan(x)}$
• $\sin(2x) \mapsto \sin(2x)$
This means if $(\tan x)^{\sin(2x)}$ reaches minimum at $x_1$, then it reaches maximum at $\frac{\pi}{2} - x_1$.
As a result, $x_1 + x_2 = \frac{\pi}{2}$.