# 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 $e^x\in L^1_{loc}(\Bbb R)\setminus \mathcal S'(\Bbb R)$.

In your case the straightforward argument that continuous periodic functions define tempered distributions is more than sufficient.

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.