does the function $|\sin(x) |$ define a tempered distribution? if so compute the fourier transform

I need to check if the function $|\sin(x)|$ defines a tempered distribution and find the fourier transform of the distribution.

I think it defines it because it is summable on every compact subset and I have also found its first distributional derivative, which is: $$f^\prime(x) = \frac{\sin(x)\cos(x)}{|\sin(x)| }$$ but I am clueless on how to find the $F$ transform.

I know the for a tempered distribution, if $\phi$ is a test function then: $$\left\langle {\hat T},\phi\right\rangle= \left\langle T,{\hat \phi} \right\rangle$$ and $$\widehat{ D^{\alpha}\phi} = (i 2 \pi k )^\alpha {\hat \phi}$$ but I am not able to use these rules to complete the problem.

First, and quite important thing: if the function is $L^1_{loc}$, it does not imply that this function is a tempered distribution. Classic counterexample is $e^x\in L^1_{loc}(\Bbb R)\setminus \mathcal S'(\Bbb R)$.
Second, in order to find the Fourier transform, one of possible approaches would be to introduce the function $$f(x)=\begin{cases} \sin x,&x\in[0,\pi ),\\0,&\text{otherwise},\end{cases}$$ and then take $$|\sin x| = \sum _{k\in\Bbb Z}(-1)^kf(x-\pi k).$$
You should be able to find the Fourier transform of $f(x)$ and then use the formula for Fourier transform of a translated function.