To figure out the derivatives of trig functions we need:

Two extremely important limits (derived below):
$$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1
\qquad \mbox{and} \qquad \lim_{x \to 0}\frac{1-\cos(x)}{x}=0;$$

The addition-of-angle formulas for sine and cosine:
\begin{eqnarray*}\sin(A+B) &=& \sin(A)\cos(B) + \cos(A)\sin(B), \cr
\cos(A+B) & = & \cos(A)\cos(B) - \sin(A)\sin(B),
\end{eqnarray*}
which we covered earlier (review).

We just showed that the first of the limits on the last line is 1, and
the second limit is $\displaystyle{\frac{\sin(0)}{1+\cos(0)}=\frac{0}{2}=0}$,
so the product is 0.