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?
 A: 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.
A: $x^2-y^2=(x+y)(x-y)$, we have
$$\sin^2A-\sin^2B\\
=(\sin A+\sin B)(\sin A-\sin B)\\
=2\sin\frac{A+B}2\cos\frac{A-B}2\cdot2\cos\frac{A+B}2\sin\frac{A-B}2\\
=2\sin\frac{A+B}2\cos\frac{A+B}2\cdot2\sin\frac{A-B}2\cos\frac{A-B}2\\
=\sin(A+B)\sin(A-B)$$
A: You wrote:
$$
\begin{align}
& \phantom{={}}\sin^2(A) - \sin^2(B) \overset{(1)}{=} \sin(A + B)\sin(A -B) \\[8pt]
& = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt]
& \overset{(2)}{=} (\sin(A)+\sin(B))(\cos(B)+\cos(A))(\sin(A)-\sin(B))(\cos(B)-\cos(A)) \\[8pt]
& \overset{(3)}{=} (\sin(A)+\sin(B))^2(\cos(B)-\cos(A))^2
\end{align}
$$
I've put numbers above things that are dubious or wrong.
The problem with the $(1)$ is that that is something you want to prove, not something you already know.
The problem with $(2)$ is that if you expand that product of four sums, you get $16$ terms, and if you expand the thing before it, you get a sum of only $8$ terms.  You haven't shown that they're equal.
The problem with $(3)$ is that if you multiply $\cos B+\cos A$ by $\cos B-\cos A$, you get $\cos^2 B-\cos^2 A$, not $(\cos B-\cos A)^2$, and a similar thing applies to the sines.
I think some others have posted some things you can try instead.
A: HINT:
Expand the right-hand side using the addition identities. You'll end up with:
$$\sin^2A\cos^2B - \sin^2B\cos^2A$$
Then use the basic trigonometric identity.
A: $2\sin(x)\sin(y) = \cos(x-y)-\cos(x+y)$
Let $ x = A-B , y = A+B$  
Therfore you have , RHS 
$$0.5(\cos(2B)-\cos(2A))$$
Now $\cos(2X) = 1-2\sin^2(X)$
Use this identity to put in the previously written question . You will get it equal to LHS
