Confused with $x$ and $a$ of Taylor Series Given general formula of Taylor Series:

$$T_{n}(x)=\sum_{n=0}^{\infty}{\frac{f^{(n)}(a)}{n!}(x-a)^{n}}$$

Why Taylor Series for $f(x)=\sin{x}$ is always evaluated at $x=0$ (the evalution means $a=0$) so we can get desired $\sin{x}$ value  like $\sin{\pi}?$

The derivatives of increasing order at $a$ are
$$\sin(a),\cos(a),-\sin(a),-\cos(a),\sin(a),\cos(a),-\sin(a),-\cos(a),\cdots$$periodically, and the development is
$$T_a(x)=\sin(a)+\cos(a)(x-a)-\sin(a)\frac{(x-a)^2}2-\cos(a)\frac{(x-a)^3}{3!}\cdots.$$
If you evaluate it at $x=a$, all terms but the first vanish:
$$T_a(a)=\sin(a)+\cos(a)(a-a)-\sin(a)\frac{(a-a)^2}2-\cos(a)\frac{(a-a)^3}{3!}\cdots=\sin(a).$$
