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

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?

  • $\begingroup$ Try cross-multiplying with the denominator as a start. $\endgroup$ – Henning Makholm Mar 30 '15 at 0:40
  • $\begingroup$ That involves assuming it's an equality in the first place... $\endgroup$ – Chappers Mar 30 '15 at 0:40
  • $\begingroup$ Try using the other identities for cos(2x) (and cos(2y)). $\endgroup$ – user222031 Mar 30 '15 at 0:41
  • $\begingroup$ @Chappers Two ratios are equal if and only if the cross multiplications are equal. $\endgroup$ – egreg Apr 4 '15 at 13:16
  • $\begingroup$ @egreg I'm just cautious because I constantly see students start with $a=b$ and work down to $0=0$ at the end, rather than starting with something that is known to be mathematically coherent like $a-b$ and showing it's equal to zero, or $a/b$ and showing it's equal to $1$. $\endgroup$ – Chappers Apr 4 '15 at 14:11

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.


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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