Simplifying trig expression I was working through some trig exercises when I stumbled upon the following problem:
Prove that: $ \cos(A+B) \cdot \cos(A-B)=\cos^2A- \sin^2B$.
I started out by expanding it such that 
$$ \cos(A+B) \cdot \cos(A-B)=(\cos A \cos B-\sin A \sin B) \cdot (\cos A \cos B+ \sin A \sin B),$$
which simplifies to: 
$$ \cos^2 A \cos^2 B- \sin^2 A \sin^2 B .$$ 
However, I don't know how to proceed from here. Does anyone have any suggestions on how to continue.

Here is a detailed answer.Let's rock!
$$
\require{cancel}\begin{align}
\cos\left(A-B\right)\cdot\cos\left(A+B\right)&=\left(\cos A\cos B-\sin A\sin B\right)\left(\cos A\cos B+\sin A\sin B\right)\\
&=\cos^2A\cos^2B-\sin^2A\sin^2B\\
&=\cos^2A\left(1-\sin^2B\right)-\sin^2A\sin^2B\\
&=\cos^2A-\sin^2B\cos^2A-\sin^2A\sin^2B\\
&=\cos^2A-\sin^2B\cancelto{1}{\left(\cos^2A+\sin^2A\right)}\\
&=\cos^2A-\sin^2B
\end{align}
$$
I hope this helps.
