For example, I thought Taylor polynomial of cosine centered at $\frac{\pi}{2}$ meant $\cos (x-\frac{\pi}{2})$.
But when expanded with $(x-\frac{\pi}{2})^n$, of which $T_3$ becomes $-(x-\frac{\pi}{2}) + \frac{1}{3!}(x-\frac{\pi}{2})^3$, it is closer to $\cos (x)$ and not to $\cos (x-\frac{\pi}{2})$.
Am I doing something wrong?
The Taylor polynomial is a polynomial that is used to approximate differentiable functions. This approximation is made around one point $a$. This means that the approximation gets better and better the closer you are to $a$.
In this case you're approximating $\cos(x)$ around the point $a = \frac{\pi}{2}$. So in fact $T_3$ is an approximation of $\cos(x)$ near $\frac{\pi}{2}$ not an approximation of $\cos(x-\frac{\pi}{2}) = \sin(x)$.
You should really check out 3blue1brown's video on the subject https://www.youtube.com/watch?v=3d6DsjIBzJ4