<s>[INST] Taylor series expansion of sin(x) I understand that Taylor series expansion for $\sin(x)$ is derived as follow: 
$$
\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-...
$$
Now, what exactly is the first, second, and third term?
Is the first term just $\sin(x) = x$?
Is the second term $\sin(x) = x-\frac{x^3}{3!}$?
 [/INST] In the Taylor expansion at $0$ of the function $\sin(x)$, the even powers of $x$, i.e. the "missing" terms,  are  zero because  $\sin(x)$ is an odd function:
$$
\begin{align}\sin(x)&=\sum_{k=0}^{\infty}\frac{D^k(\sin(x))_{x=0}}{k!}\cdot x^k
\\&=\sin(0)+\cos(0)x+\frac{-\sin(0)}{2!}\cdot x^2+\frac{-\cos(0)}{3!}\cdot x^3+
\frac{\sin(0)}{4!}\cdot x^4+\frac{\cos(0)}{5!}\cdot x^5+o(x^5)
\\&=\underbrace{0+x+0\cdot x^2-\frac{x^3}{3!}+0\cdot x^4+\frac{x^5}{5!}}_{\text{polynomial expansion up to the $5$-th term }}+o(x^5).
\end{align}$$
</s>