Taylor series of #f(x)=sin(x)# at #x=0# is
#sum_{n=0}^{infty}(-1)^n{x^{2n+1}}/{(2n+1)!}#.
Taylor series for #f(x)# at #x=a# can be found by
#f(x)=sum_{n=0}^{infty}{f^{(n)}(a)}/{n!}x^n#
So, we need to find derivatives of #f(x)=sin(x)#.
#f(x)=sin(x) Rightarrow f(0)=0#
#f'(x)=cos(x) Rightarrow f'(0)=1#
#f''(x)=-sin(x) Rightarrow f''(0)=0#
#f'''(x)=-cos(x) Rightarrow f'''(0)=-1#
#f^{(4)}(x)=sin(x) Rightarrow f^{(4)}(0)=0#
#cdots#
Since #f^{(4)}(x)=f(x)#, the cycle of {0, 1, 0, -1} repeats itself, which means that every derivative of even degree gives #0# and that every derivative of odd degree alternates between 1 and -1. So, we have
#f(x)={1}/{1!}x^1+{-1}/{3!}x^3+{1}/{5!}x^5+cdots#
#=sum_{n=0}^{infty}(-1)^n{x^{2n+1}}/{(2n+1)!}#
