Show that operator is normal and determine its Singular Value Decomposition could anybody please help me with the following task?

Consider the operator
  $$
Af(x):=\int\limits_{-\pi}^{\pi}\sin(x-y)f(y)\, dy, x\in [-\pi,\pi], f\in L_2(-\pi,\pi).
$$
  Show that the operator $A\in\mathcal{L}(L_2(-\pi,\pi))$ is normal. Determine the Singular Value Decomposition (SVD) of A.

In order to check if A is normal, I determined the adjoint operator with the result that
$$
A^* f(x)=\int\limits_{-\pi}^{\pi}\overline{\sin(x-y)} f(y)\, dy.
$$
Then I calculated $AA^*$ and $A^*A$. Here are my results:
$$
AA^* f(x)=\int\limits_{-\pi}^{\pi}\sin(x-y) \int\limits_{-\pi}^{\pi}\overline{\sin(y-z)} f(z)\, dz\, dy
$$
$$
A^*Af(x)=\int\limits_{-\pi}^{\pi}\overline{\sin(x-y)}\int\limits_{-\pi}^{\pi}\sin(y-z)f(z)\, dz\, dy
$$
And this is identical because $\sin(x)=\overline{\sin(x)}$.
Could you please write me in a comment if it is okay until now?
Thanks a lot.
Edit:
Concerning the SVD:
Is it right, that I have to determine the eigenvalues (resp. eigenfunctions) of $AA^*$? 
Does one need the convolution theorem of the Fouriertransformation?
Explicitly:
To my opinion it is
$$
AA^*f(x)=(\sin\star A^*f)(x)
$$
and therefore
$$
AA^* f(x)=\lambda f(x)\Leftrightarrow \mathcal{F}(\sin\star A^*f)=(2\pi)^{1/2}\mathcal{F}(\sin)\cdot\mathcal{F}(A^*f)=\lambda\mathcal{F}(f),
$$
i.e.
$$
(2\pi)^{1/2}\mathcal{F}(\sin)\cdot\mathcal{F}(A^*f)=\lambda\mathcal{F}(f).
$$
Can one use that equation to determine now $\lambda$ resp. $f$?
Greetings
 A: Yes, Fourier transform is the way to go.
Recall the spectral theorem: a bounded operator on a Hilbert space is normal if and only it is unitarily equivalent to a multiplication operator. What you have here is a special case, with the implementing unitary being the Fourier transform $\mathcal{F}$.
As you noted, the convolution theorem says $\mathcal{F}$ maps convolution to multiplication. The Pontryagin dual in this case, $\mathbb{Z}$, is discrete. This means $A$ is diagonalizable, in particular normal. The $\mathcal{F}(h)$ for any $h \in L^1(\mathbb{T})$ is a sequence vanishing at infinity. So $A$ is compact and SVD makes sense.
In Fourier domain, your answer is very simple: $A$ is projection onto the basis element $\sin x$ (up to a scaling factor, possibly). So the eigenvalues of $A^*A$ is same as $A$: $1$ for the basis element projected on and $0$'s for every other element (again, you might need to adjust for a scaling factor that makes $\mathcal{F}$ unitary).
A: Note that, if 
$$ K \equiv \int_{a}^{b}k(x,t) dt $$ and 
$$ H \equiv \int_{a}^{b}h(x,t) dt, $$ then

$$ KH \equiv \int_{a}^{b}k(x,z)h(z,t) dz.$$

Now, apply this result to your operators $A$ and $A^{*}.$
