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.
 A: You're on the right track! From where you left off:
$\cos^2A\cos^2B-\sin^2A\sin^2B = \cos^2A(1-\sin^2B)-\sin^2A\sin^2B$
$ = \cos^2A - \sin^2B(\cos^2A+\sin^2A)=\cos^2A-\sin^2B$
Note: using \sin and \cos results in prettier TeX formatting.
A: The identities
$$\cos(\theta) = \frac{e^{i \theta}+e^{- i \theta}}{2}$$
$$\sin(\theta) = \frac{e^{i \theta}-e^{- i \theta}}{2i}$$
can reduce a trigonometric identity to a identity of polynomials. Let's see how this works in your example:
$$\cos(A+B) \cos(A-B)=\cos(A)^2-\sin(B)^2$$
is rewritten into:
$$\frac{e^{i (A+B)}+e^{- i (A+B)}}{2} \frac{e^{i (A-B)}+e^{- i (A-B)}}{2}=\left(\frac{e^{i A}+e^{- i A}}{2}\right)^2-\left(\frac{e^{i B}-e^{- i B}}{2i}\right)^2$$
now we change it into a rational function of $X = e^{i A}$ and $Y = e^{i B}$:
$$\frac{(X Y + \frac{1}{X Y})(\frac{X}{Y} + \frac{Y}{X})}{4} = \frac{(X + \frac{1}{X})^2 + (Y - \frac{1}{Y})^2}{4}$$
and you can simply multiply out both sides to see that they are both $\frac{1}{4}\left(X^2 + \frac{1}{X^2} + Y^2 + \frac{1}{Y^2}\right)$ which proves the trigonometric equality.
A: 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.
