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+...#