Prove $\cos(A + B) \cos(A − B) = \cos^2A − \sin^2B$ I have a question regarding trigonometric identities. The question I am currently struggling to understand is:

Prove that $$\cos(A + B) \cos(A − B) = \cos^2A − \sin^2B$$

When approaching this problem I know that there is
$$\cos2A = \cos^2A - \sin^2A$$
but how would I apply this here, or is it completely wrong way of approaching it, or should I try
$$\cos(A+B)=\cos A \cos B - \sin A \sin B$$
Many Thanks.
 A: \begin{align*}
\cos(x+y)\cdot\cos(x-y)&=
(\cos x\cos y-\sin x\sin y)(\cos x\cos y+\sin x\sin y)\\
&=\cos^2 x\cos^2 y-\sin^2 x\sin^2 y\\
&=\cos^2 x(1-\sin^2 y)-(1-\cos^2 x)\sin^2 y\\
&=\cos^2 x-\cos^2 x\sin^2 y-\sin^2 y+\cos^2 x\sin^2y\\
&=\cos^2 x-\sin^2y
\end{align*}
A: Note
\begin{align}
\cos^2A − \sin^2B =&
\frac12(1+ \cos 2A )-\frac12(1-\cos2B)\\
=& \frac12( \cos 2A+\cos2B)
=\cos(A + B) \cos(A − B) 
\end{align}
A: Hint:
The only identities you should need to prove your claim are:

*

*$\cos(x+y)=\cos x\cos y - \sin x \sin y$

*$\sin(x+y)=\sin x\cos y + \cos x\sin y$

*$\sin(-x)=-\sin(x)$

*$\cos(-x)=\cos(x)$
A: If
$$\cos(a+b)\cos(a-b)=\cos^2(a)-\sin^2(b)$$ holds, then it also holds with $\dfrac\pi2-a$ instead of $a$ and
$$\sin(a-b)\sin(a+b)=\sin^2(a)-\sin^2(b).$$
Then by subtraction,
$$\cos(a+b+a-b)=\cos(2a)=\cos^2(a)-\sin^2(a).$$
A: You need to use the addition/subtraction formulas
$$
\cos(A+B)=\cos A\cos B-\sin A\sin B\\
\cos(A-B)=\cos A\cos B+\sin A\sin B
$$
Therefore the product is
$$
\cos(A+B)\cos(A-B)=\cos^2A\cos^2B-\sin^2A\sin^2B
$$
The identity you're due to prove only has $\cos^2A$ and $\sin^2B$: then change $\cos^2B=1-\sin^2B$ and $\sin^2B=1-\cos^2B$. Plug these in and see what happens.
A: Using the formulas you mentioned, you can derive the following convenient forms (they are useful when doing integration):

*

*$\cos A\cos B=\frac 1 2(\cos(A+B)+\cos(A-B))$

*$\cos(2A)=\cos^2 A-\sin^2 A=2\cos^2 A-1=1-2\sin^2 A$
Applying the above formulas, one easily sees that $$\cos(A+B)\cos(A-B)=\frac 12(\cos(2A)+\cos(2B))$$
$$=\frac 12(2\cos^2 A-1+1-2\sin^2 B)=\cos^2 A-\sin^2 B.$$
