<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] 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 $1-\dfrac{x^2}2+\dfrac{x^4}{4!}$.
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