Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$ This is my attempt:
$$
\begin{align}
& \phantom{={}}\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B) \\[8pt]
& = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt]
& = (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8pt]
&  = (\sin(A)+\sin(B))^2(\cos(B)-\cos(A))^2
\end{align}
$$
But now I can't get rid of the cosines. How can I get rid of them?

You got off to a good start:
$$
\sin(A+B)\sin(A-B) = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B)-\cos(A)\sin(B))
$$
This is of the form $(x+y)(x-y)$ so
$$
\sin(A+B)\sin(A-B) = \sin^2(A)\cos^2(B)-\cos^2(A)\sin^2(B)
$$
Eliminate the cosines (since $\sin^2(x)+\cos^2(x)=1$, so $\cos^2(x)=1-\sin^2(x)$) and expand:
$$
\begin{align}
\sin(A+B)\sin(A-B) &= \sin^2(A)(1-\sin^2(B))-(1-\sin^2(A))\sin^2(B)\\
  &= \sin^2(A)-\sin^2(A)\sin^2(B)-\sin^2(B)+\sin^2(A)\sin^2(B)\\
   &=\sin^2(A)-\sin^2(B)
\end{align}
$$
as required.
