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Could you give me some guidance on proving the following trig identity?

$$\cos(A + B)\cos(A - B) = -(\sin A + \cos B)(\sin A - \cos B).$$

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You just have to expand by using the compound angle results.

$\begin{align} LHS&=cos(A+B)cos(A-B)\\ &=(cosAcosB-sinAsinB)(cosAcosB+sinAsinB) \\ &=(1-sin^2A)cos^2B-sin^2A(1-cos^2B)\\ &=cos^2B-sin^2A\\ &=-(sinA+cosB)(sinA-cosB) \\ &=RHS \end{align}$

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You have $\cos(A+B)=\cos A \cos B - \sin A \sin B$ and $\cos(A-B)=\cos A \cos B + \sin A \sin B$, these are standard formulas.

Multiply them together and simplify as needed.

Them multiply out the RHS, and see what you get.

I won't finish it for you, because the exercise is worth doing by hand.

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Consider $\cos(x\mp y)=\cos x\cos y\pm\sin x\sin y$, add them together, you'll get $$\cos(x+y)+\cos(x-y)=2\cos x\cos y.$$ Now let $x=A+B,\,y=A-B$, so $2\cos(A+B)\cos(A-B)=\cos(2A)+\cos(2B)=2\cos^2A-1+1-2\sin^2B$,
hence we get the desired result as $\cos^2A-\sin^2B=(\cos A+\sin B)(\cos A-\sin B)$.

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