How to find the period of $ (\tan x )^{0.45}$?

Question

How to find the period of $(\tan x)^{0.45}$?

I know the period of $\tan x$ is $\pi$. Can anyone please help me?

Answer 1

$\pi$. You do not change periodicity by raising to this particular power. Obviously $(\tan(x+\pi))^{0.45}= (\tan(x))^{0.45}$. So all you need to check if there is not a shorter period. But where $\tan$ is positive, so your function is defined, the function is increasing in each smaller than $\pi$ interval, so it cannot have smaller periodicity.

Answer 2

Yes, the period is indeed $\pi$:

The base function need not be monotonic. In general, if $f(x)$ has period $T$, then $g(f(x))$ will also have period $T$.

Answer 3

Given a periodic function $f$ with smallest period $p$ and an injective function $g$. The composition $g\circ f$ is a function with potentially smaller domain, but if that domain is nonempty, then $g\circ f$ is still periodic with smallest period $p$.

First of all, $p$ is clearly a period to $g\circ f$. Secondly, $g\circ f$ cannot have a smaller period. That would mean that there are two numbers in the new domain, $x,y$, whose distance is less than $p$ and $g(f(x))= g(f(y))$. But as the function $g$ is injective, this implies $f(x)=f(y)$, a contradiction.

As $x\mapsto x^{0.45}$ is injective, the period is $\pi$.