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$$\frac{\cos(2x)+\cos(2y)}{\sin(x)+\cos(y)} = 2\cos(y)-2\sin(x)$$

The question asks to prove the identity. I tried simplifying the first half, thought maybe I could expand and simplify with the double angle formulas.

Changed it to $$\cos(x)^2 - sin(x)^2 + cos(y)^2 - sin(y)^2$$ and tried a few thing like that, but I'm stuck at that point. Am I even on the right track here, or way off?

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2 Answers

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We have, using the identity $\cos^2{x}+\sin^2{x}=1$: $$ \cos{2x} = \cos^2{x}-\sin^2{x} = 2\cos^2{x}-1 = 1-2\sin^2{x}. $$ Then $$ \cos{2x}+\cos{2y} =1-2\sin^2{x} + 2\cos^2{y}-1 = 2(\cos^2{y}-\sin^2{x}) \\ = 2(\cos{y}-\sin{x})(\cos{y}+\sin{x}), $$ and dividing gives the required identity.

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The two give expressions are equal if and only if $\sin x+\cos y\ne0$ and $$ \cos2x+\cos2y=2(\sin x+\cos y)(\cos y-\sin x) $$ The right hand side can be rewritten as $$ 2\cos^2y-2\sin^2x $$ and we can recall the half-angle formulas $$ \cos^2y=\frac{1+\cos2y}{2},\qquad \sin^2x=\frac{1-\cos2x}{2} $$ so the right hand side becomes $$ 2\frac{1+\cos2y}{2}-2\frac{1-\cos2x}{2} $$

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