Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$ $$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$
I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.
The way I did it:
$$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$
 A: Maybe you didn't see you can use the formula $(x+y)(x-y)=x^2-y^2$. Other than that, I don't see any other simpler method.
A: You can use the addition theorem which states that 
$$\cos(\alpha+\beta)=\cos(\alpha)\cdot \cos(\beta) -\sin(\alpha)\sin(\beta)$$
$$\cos(\alpha + \beta) \cdot \cos(\alpha -\beta)= ( \cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha) \cdot \cos(-\beta)-\sin(\alpha)\cdot \sin(-\beta))$$
As $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$ this is equal to
$$(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta))$$
With the third binom you get 
$$\cos^2(\alpha) \cos^2(\beta) -\sin^2(\alpha)\sin^2(\beta)$$
As $\cos^2(\beta)+\sin^2(\beta)=1$ we have
$$\cos^2(\alpha)(1-\sin^2(\beta))-(1-\cos^2(\alpha))\sin^2(\beta)$$
Multplying it out gives you 
$$\cos^2(\alpha) -\sin^2(\beta)\cos^2(\alpha)-\sin^2(\beta)+\cos^2(\alpha)\sin^2 (\beta)$$
And this is 
$$\cos^2(\alpha)-\sin^2(\beta)$$
Another way:
$$2\cos(\alpha)\cdot \cos(\beta)=\cos(\alpha+\beta) + \cos(\alpha-\beta)$$
Using this in your formula gives us
$$\cos(A+B)\cdot \cos(A-B)=\frac{1}{2}\left( \cos(2A) + \cos(-2B)\right)=\frac{1}{2} \left(\cos(2 A)+ \cos(2 B)\right)$$
As $\cos(2 A)=1-2\sin^2(A)$ and $\cos(2 B) = 1-2 \sin^2 (B)$ this is equal to
$$1-\sin^2(A) -\sin^2 (B)=\cos^2(A)-\sin^2(B)$$
A: Another way of looking at it:
$$\begin{align*}
\cos(A+B)\cos(A-B)  &= \frac14\left(e^{i(A+B)}+\frac{1}{e^{i(A+B)}}\right)\left(e^{i(A-B)}+\frac{1}{e^{i(A-B)}}\right) \\
&= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\
&= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2 - 2 + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\
&= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2\right] + \frac14\left[ - 2 + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\
&= \left(\frac{e^{iA} + e^{-iA}}{2}\right)^2 - \left( \frac{e^{iB}-e^{-iB}}{2i}\right)^2 \\
&= \cos^2 A - \sin^2 B.
\end{align*}
$$
A: Here is an interesting different way: Let
$$x = \cos(A+B)\cos(A-B) \\
y = \sin(A+B)\sin(A-B)$$
Then,
$$x+y = \cos(A+B-A+B) = \cos(2B) \\
x-y = \cos(A+B+A-B) = \cos(2A)$$
You can add these to get $2x$ or subtract them to get $2y$. Then expand using the double-angle formulas. This gives you two trig product formulas at the same time.
A: For a slightly different solution,
$2\cos^2 A-2\sin^2 B$
$=2\cos^2 A-1+1-2\sin^2 B$
$=\cos 2A+\cos 2B$
$=2\cos (A+B) \cos (A-B)$
and halve each side.
