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.

$$
\sin x = x - \frac{x^3}6 + \frac{x^5}{120} - \cdots
$$
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.
