Everything is in the question!
I've seen loads examples like "centered at $\pi$, $\pi/2$,... But $1$ would make everything much different...
I've tried to work this way:
$\sin(z) = \sin((z-1)+1) = \sin(1)\cos(z-1) + \cos(1)\sin(z-1)$ and then expand... But I get a series in two parts and can't put them back together or exploit it (to find a radius of convergence or anything)...
Can you help me? Thank you very much!
On
Just as 5xum answered, the development of $\sin(x)$ at $x=a$ write $$\sin(x)=\sin (a)+(x-a) \cos (a)-\frac{1}{2} (x-a)^2 \sin (a)-\frac{1}{6} (x-a)^3 \cos (a)+\frac{1}{24} (x-a)^4 \sin (a)+\frac{1}{120} (x-a)^5 \cos (a)-\frac{1}{720} (x-a)^6 \sin (a)+O\left((x-a)^7\right)$$
This can be useful for a good estimation of values at bizarre angles. Say, for example, that you want $\sin(\frac{23\pi}{66})$. Select $a=\frac{\pi}{3}$ and using the successive terms, you will get $0.8898253481$, $0.8888442499$, $0.8888352624$, $0.8888354477$, $0.8888354487$, $0.8888354487$.
You can always use the standard construction of the power series, i.e.
$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n$$ to find the series. You already know what $f^{(n)}$ is in the case of $f(x) = \sin (x)$.