# How do you find the Taylor series of f(x)=sin(x) ?

Aug 31, 2014

Taylor series of $f \left(x\right) = \sin \left(x\right)$ at $x = 0$ is
sum_{n=0}^{infty}(-1)^n{x^{2n+1}}/{(2n+1)!}.

Taylor series for $f \left(x\right)$ 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 $f \left(x\right) = \sin \left(x\right)$.
$f \left(x\right) = \sin \left(x\right) R i g h t a r r o w f \left(0\right) = 0$
$f ' \left(x\right) = \cos \left(x\right) R i g h t a r r o w f ' \left(0\right) = 1$
$f ' ' \left(x\right) = - \sin \left(x\right) R i g h t a r r o w f ' ' \left(0\right) = 0$
$f ' ' ' \left(x\right) = - \cos \left(x\right) R i g h t a r r o w f ' ' ' \left(0\right) = - 1$
${f}^{\left(4\right)} \left(x\right) = \sin \left(x\right) R i g h t a r r o w {f}^{\left(4\right)} \left(0\right) = 0$
$\cdots$

Since ${f}^{\left(4\right)} \left(x\right) = f \left(x\right)$, 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)!}