<s>[INST] How to explain this point transformation? Yesterday we were having a lecture on point coordinates after rotation. The prof. explained that the position of a point after a counterclockwise rotation is obtained from the following formula,
$x=x_0 \operatorname{Cos}(\theta)-y_0 \operatorname{Sin}(\theta), \qquad y=y_0 \operatorname{Cos}(\theta)+x_0\operatorname{Sin}(\theta)$,
where $\theta$ is the angle of rotation. 
And then, out of nowhere, for an exercise he used the following formula for the displacements of the point after rotation,
$u_x=(-x_0 \operatorname{Cos}(\frac{\pi-\theta}{2})-y_0\operatorname{Sin}(\frac{\pi-\theta}{2}))\times 2\operatorname{Sin}(\frac{\theta}{2}), \qquad \\ u_y=(x_0 \operatorname{Sin}(\frac{\pi-\theta}{2})-y_0\operatorname{Cos}(\frac{\pi-\theta}{2}))\times 2\operatorname{Sin}(\frac{\theta}{2}).$
Why the signs are different? why instead of just $\theta$ he has used $\frac{\pi-\theta}{2}$? and most importantly what is the role of $2\operatorname{Sin}(\frac{\theta}{2})$?
Thanks in advance.
 [/INST] You must use some trig equalities:
$$
\cos\frac{\pi-\theta}{2}=\sin\frac{\theta}{2},\quad
\sin\frac{\pi-\theta}{2}=\cos\frac{\theta}{2},\\
\cos\theta=1-2\sin^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}-1,\quad
\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}
$$
to obtain:
$$
\begin{align}
&\left(-x_0 \cos\frac{\pi-\theta}{2}-y_0\sin\frac{\pi-\theta}{2}\right)\cdot 2\sin\frac{\theta}{2}\\
=&\left(-x_0 \sin\frac{\theta}{2}-y_0\cos\frac{\theta}{2}\right)\cdot 2\sin\frac{\theta}{2}\\
=&-2x_0 \sin^2\frac{\theta}{2}-2y_0\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\
=&x_0 (1-\cos\theta)-y_0\sin\theta\\
=&x_0-x_0\cos\theta-y_0\sin\theta\\
=&x_0-x,
\end{align}
$$
and the analogous for the second equation.
</s>