Question: If $R\_\theta=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ is the rotation in $\mathbb{R}^2$ through angle $\theta$ then for $f\in\mathcal{S}(\mathbb{R}^2)$ I get

$$(\mathcal{F}(f\circ R\_\theta))(\xi)=\mathcal{F}(f(R\_\theta(\xi))) = \int\_{\mathbb{R}^2} f(x)e^{ix\cdot R\_{\theta}(\xi)}\,dx$$

I don't immediately see why this would be equal to $(\mathcal{F}f)\circ R\_\theta$ though?

Answer: 
[![](https://i.stack.imgur.com/qgPvF.png)](https://i.stack.imgur.com/qgPvF.png)
Here's a fun little proof without words. Sorry for the horrible art skills.
The first angle is $\pi /4$ because his tangent is equal to $1$ so we must show that $a+b=3\pi /4$. Looking to the picture we have $\tan a= 2$ and $\tan b= 3$ so:

$$\tan(a+b)=\frac{\tan a + \tan b}{1-\tan a \tan b}=\frac{2+3}{1-2\cdot3}=-1$$

and once $a+b \le \pi$ then $a+b=3\pi /4$.