I will assume that you mean a Taylor's series centered around pi, when you say x-pi. Also, I used enough derivatives to construct a Taylor series with at least three non-zero terms, but it continues indefinitely.
The Taylor approximation of a function involves using its derivatives to generate a series that is approximate to the function at a given point. Each term {u_n} follows a general form: (f^(n-1)(a)(x-a)^(n-1))/((n-1)!), where a is the centre of the series.
We start by finding the derivatives of different degrees of the function:
f(x)=sinx
f'(x)=cosx
f^2(x)=-sinx
f^3(x)=-cosx
f^4(x)=sinx
f^5(x)=cosx
and so on.
Now, we find the values of the derivatives at x=pi.
f(pi)=sinpi=0
f'(pi)=cospi=-1
f^2(pi)=-sinpi=0
f^3(pi)=-cospi=1
f^4(pi)=sinpi=0
f^5(pi)=cospi=-1
and so on.
We can now construct the Taylor approximation for f(x)=sinx centered around pi.
f(x)=f(pi)/(0!)+(f'(pi)(x-pi))/(1!)+(f^2(pi)(x-pi)^2)/(2!)+(f^3(pi)(x-pi)^3)/(3!)+(f^4(pi)(x-pi)^4)/(4!)+(f^5(pi)(x-pi)^5)/(5!)+...
=0+(-1)(x-pi)+0+((1)(x-pi)^3)/6+0+((-1)(x-pi)^5)/120+...
=pi-x+(x-pi)^3/6-(x-pi)^5/120+...
