Write down each of the terms in the expansion of $\sin(x^2)-(\sin(x))^2$. Write down each of the terms in the expansion of $\sin(x^2)-(\sin(x))^2$. Taylor's Theorem applies at the point $a=0$ and with $n=4$.

Got no idea how to proceed. My lecture notes have one example that I barely understand. I'd really appreciate a semi-detailed overview of what I should be doing. I can go to class and ask questions, but I need to have some idea of what I am looking at in order to ask intelligent questions.
 A: $$
\sin x = x - \frac{x^3}6 + \frac{x^5}{120} - \cdots
$$
So
$$
\sin (x^2) = x^2 - \frac{x^6}6+\frac{x^{10}}{120} - \cdots
$$
and
$$
(\sin x)^2 = x^2 - \frac{x^4}3 + \frac{2x^6}{45} -\cdots.
$$
To get the last series just look at
$$
\left(x - \frac{x^3}6 + \frac{x^5}{120} - \cdots\right)\left(x - \frac{x^3}6 + \frac{x^5}{120} - \cdots\right).
$$
Multiply the $x$ from one by the $x$ from the other to get $x^2$.  Then multiply $x$ from the first by $-x^3/6$ from the second to get $-x^4/6$, and $x$ from the second by $-x^3/6$ from the first, and add those to get $-x^4/3$.  Then $x$ from the first by $x^5/120$ from the second, plus $-x^3/6$ from the first by $-x^3/6$ from the second, plus $x^5/120$ from the first by $x$ from the second, adding up to $2x^6/45$.  And so on. 
So the first non-zero term in the expansion of $\sin(x^2)-(\sin x)^2$ is the fourth-degree term.
A: Since:
$$\sin x =\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$$
we have:
$$ \sin(x^2) = \sum_{n\geq 0}\frac{(-1)^n x^{4n+2}}{(2n+1)!}.\tag{1}$$
On the other hand, since $\sin^2 x = \frac{1-\cos(2x)}{2}$ and
$$\cos x = \sum_{n\geq 0}\frac{(-1)^n x^{2n}}{(2n)!},$$
we have:
$$ \sin^2 x = \sum_{n\geq 1}\frac{(-1)^{n+1} 4^n x^{2n}}{(2n)!}\tag{2}$$
and the result follows from combining $(1)$ and $(2)$. In a neighbourhood of the origin, we have:
$$ \sin x^2-\sin^2 x = \frac{x^4}{3}+o(x^5).\tag{3}$$
A: There are quite a few good resources online on how to do these problems. Check out:
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-38-taylors-series/
and http://en.wikipedia.org/wiki/Taylor_series
The idea is that given say n derivatives at a given point a, for some function f, one can find a polynomial that close to that point is a "good" (taylor remainder theorem says how good) approximation to the function and gives a better approximation with higher degree terms.
A: we will keep track of upto powers of $x^4.$  $$\sin(x) = x - \frac 16 x^3 + \cdots, (\sin x)^2 = x^2 - \frac 13 x^4 + \cdots, \sin(x^2) = x^2+ 0x^4 + \cdots $$ therefore $$\sin (x^2) - (\sin x)^2 = \frac 13 x^4 + \cdots $$
