How to compute taylor series for $f(x)=\frac{1}{1-x}$ about $a=3$ and $f(x)=\sin{x}$ about $a=\frac{\pi}{4}$? How to compute taylor series for $f(x)=\frac{1}{1-x}$ about $a=3$  and $f(x)=\sin{x}$ about $a=\frac{\pi}{4}$ ?
For the first one, using substitution, let $t=x-3$, then $x=t+3$. Then  $f(x)=\frac{1}{1-(t+3)}=\sum^{\infty}_{n=0}(t+3)^n=\sum^{\infty}_{n=0}(x)^n$. Apparently my solution wrong as it returns to the original series.
For the second one,  let $t=x-\frac{\pi}{4}$. Then $\sin({t+\frac{\pi}{4}})=\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)!}(t+\frac{\pi}{4})^{2n+1}$. Then I fail to countinue.
Could someone suggest a solution?

Hint. One may write
$$
f(x):=\frac{1}{1-x}=-\frac12\frac{1}{1+\frac{(x-3)}2}=-\frac12\sum_{n=0}^\infty(-1)^n\frac{(x-3)^n}{2^n}, \quad |x-3|<2,
$$ and one may write, for any real number $x$,
$$
\begin{align}
f(x)&:=\sin x
\\\\&=\frac{\sqrt{2}}2 \left(\sin\left(x-\frac{\pi}4\right)+ \cos\left(x-\frac{\pi}4\right)\right)
\\\\&=\frac{\sqrt{2}}2\left(\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)!}\left(x-\frac{\pi}{4}\right)^{2n+1}+\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n)!}\left(x-\frac{\pi}{4}\right)^{2n}\right).
\end{align}
$$
