Taylor series of f(x)=sin(x)f(x)=sin(x) at x=0x=0 is
sum_{n=0}^{infty}(-1)^n{x^{2n+1}}/{(2n+1)!}∞∑n=0(−1)nx2n+1(2n+1)!.
Taylor series for f(x)f(x) at x=ax=a can be found by
f(x)=sum_{n=0}^{infty}{f^{(n)}(a)}/{n!}x^nf(x)=∞∑n=0f(n)(a)n!xn
So, we need to find derivatives of f(x)=sin(x)f(x)=sin(x).
f(x)=sin(x) Rightarrow f(0)=0f(x)=sin(x)⇒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)!}
