$$\cos(x+y)\cos(x-y)=\cos^2(x)-\sin^2(y)$$
I use the sum and difference formula to reduce the right hand side of the equation to $$[\cos(x)\cos(y)-\sin(x)\sin(y)][\cos(x)\cos(y)+\sin(x)\sin(y)]$$ then foil to end up with $$\cos^2(x)\cos^2(y)-\sin^2(x)\sin^2(y)$$ and I can't see what to do from here any help? I know it's really simple but I can't figure out how to get it.
$\endgroup$ 31 Answer
$\begingroup$HINT: Notice,
$$LHS=\cos(x+y)\cos(x-y)=(\cos x\cos y-\sin x\sin y)(\cos x\cos y+\sin x\sin y)$$ $$=(\cos x\cos y)^2-(\sin x\sin y)^2$$ $$=\cos^2 x\cos^2 y-\sin^2 x\sin^2 y$$ $$=\cos^2 x(1-\sin^2 y)-(1-\cos^2 x)\sin^2 y$$ $$=\cos^2 x-\cos^2 x\sin^2 y-\sin^2 y+\cos^2 x\sin^2 y$$ $$=\cos^2 x-\sin^2 y$$
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