Where does $\vec e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ go to? Rotate it by an angle $\theta$. Its new coordinates are $\Bigl(\cos\bigl(\theta + \frac {\pi}{2}\bigr), \sin\bigl(\theta + \frac {\pi}{2}\bigr)\Bigr)$.
Any reason we add $\frac {\pi}{2}$ to $\theta$ instead of say, $\pi$? I am just wondering why we get specifically $\Bigl(\cos\bigl(\theta + \frac {\pi}{2}\bigr), \sin\bigl(\theta + \frac {\pi}{2}\bigr)\Bigr)$ after rotating $\vec e_2$ around the unit circle.
On
General formula for rotation on a plane about angle $\theta$ is given by the formula: $$R_\theta(\begin{bmatrix} x \\ y \end{bmatrix})=\begin{bmatrix} \cos\theta &-\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix}x\cos\theta - y\sin\theta \\ x\sin\theta+y\cos\theta \end{bmatrix}$$ Which coincides with the intuiton how rotation should behave. Now when you consider point $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ then you get that $R_\theta(\begin{bmatrix} 0 \\ 1 \end{bmatrix})=\begin{bmatrix} -\sin\theta \\ \cos\theta \end{bmatrix}.$ Now you have to remember from school (or derive simply using complex analasis or from wiki) that $-\sin\theta=\cos(\theta+\pi/2)$ and $\cos\theta=\sin(\theta+\pi/2).$ And voila
Note that $$\left[\begin{array}{c}\cos\frac{\pi}{2}\\ \sin\frac{\pi}{2}\end{array}\right] = \left[\begin{array}{c} 0 \\ 1\end{array}\right].$$ That is, the point on the unit circle which makes an angle of $\frac{\pi}{2}$ with the positive real axis is $(0, 1)$. So if we rotate by an angle of $\theta$, we obtain the point $$\left[\begin{array}{c}\cos\left(\theta + \frac{\pi}{2}\right)\\ \sin\left(\theta + \frac{\pi}{2}\right)\end{array}\right].$$