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 $\sin(x) = x-\frac{x^3}{3!}$?
 A: 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:
$$
\begin{align}\sin(x)&=\sum_{k=0}^{\infty}\frac{D^k(\sin(x))_{x=0}}{k!}\cdot x^k
\\&=\sin(0)+\cos(0)x+\frac{-\sin(0)}{2!}\cdot x^2+\frac{-\cos(0)}{3!}\cdot x^3+
\frac{\sin(0)}{4!}\cdot x^4+\frac{\cos(0)}{5!}\cdot x^5+o(x^5)
\\&=\underbrace{0+x+0\cdot x^2-\frac{x^3}{3!}+0\cdot x^4+\frac{x^5}{5!}}_{\text{polynomial expansion up to the $5$-th term }}+o(x^5).
\end{align}$$
A: If you know Euler's formula
$e^{ix} = \cos(x)+i\sin(x)$,
the power series
$e^x
= \sum_{n=0}^{\infty} \dfrac{x^n}{n!}$,
and don't mind using
complex expressions 
to get real results,
then
$$e^{ix}-e^{-ix}
=(\cos(x)+i\sin(x))-(\cos(-x)+i\sin(-x))
=2i\sin(x)
$$
so
$$\sin(x)
=-i(e^{ix}-e^{-ix})/2
,$$
and
$$e^{ix}-e^{-ix}
=\sum_{n=0}^{\infty} \dfrac{(ix)^n-(-ix)^n}{n!}
=\sum_{n=0}^{\infty} ((i^n)-(-i)^n)\dfrac{x^n}{n!}
$$
and,
with period 4,
$$(i^n)-(-i)^n
=1-1, i-(-i), -1-(-1), -i-(i)
=0, 2i, 0, -2i
$$
so
$-i/2$ times this is
$0, 1, 0, -1$.
The power series is therefore
$$x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\dfrac{x^7}{5040}...
=\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}
.$$
A: Terms are the members of a summation (whatever the formula) and they are enumerated in the order of their appearance (left to right, starting from first)


*

*$x$ is the first term,

*$-\dfrac{x^3}{3!}$ is the second term,

*$\dfrac{x^5}{5!}$ is the third term.
When speaking of a polynomial, a term is said of the $n^{th}$ degree depending on the power of the variable. So here there are no terms of even degree and


*

*$x$ is the first degree term,

*$-\dfrac{x^3}{3!}$ is the third degree term,

*$\dfrac{x^5}{5!}$ is the fifth degree term.
Now a Taylor expansion is written up to a remainder term, with as many terms as you like. The word order is used and equals the highest degree. So you can say


*

*$\sin(x)=x+r_1(x)$ is the first order expansion,

*$\sin(x)=x-\dfrac{x^3}{3!}+r_3(x)$ is the third order expansion,

*$\sin(x)=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+r_5(x)$ is the fifth order expansion.

Similarly, for the cosine you would have
First term $1$, second term $-\dfrac{x^2}2$, third term $\dfrac{x^4}{4!}$, and zero$^{th}$ order development/expansion $1$, second order $1-\dfrac{x^2}2$, fourth order $1-\dfrac{x^2}2+\dfrac{x^4}{4!}$.
