Answer
\begin{aligned}
\sin x &= x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\cdots+(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !}
\end{aligned}
Work Step by Step
Since
\begin{array}{rlrl}
{f(x)} & {=\sin x} & {f(0)} & {=0} \\
{f^{\prime}(x)} & {=\cos x} & {f^{\prime}(0)} & {=1} \\
{f^{\prime \prime}(x)} & {=-\sin x} & {f^{\prime \prime}(0)} & {=0} \\
{f^{\prime \prime \prime}(x)} & {=-\cos x} & {f^{\prime \prime \prime}(0)} & {=-1} \\
{f^{(4)}(x)} & {=\sin x} & {f^{(4)}(0)} & {=0}
\end{array}
Then
\begin{aligned}
\sin x &=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\ldots \ldots+\frac{f^{(n)}(0)}{n !} x^{n} \\
&=\frac{1}{1 !} x-\frac{1}{3 !} x^{3}+\ldots \ldots+\frac{1}{n !} x^{n} \\
&=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots \\
&=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\cdots+(-1)^{n} \frac{x^{2 n+1}}{(2 n+1) !}
\end{aligned}
