#### Monoxdifly

Prove that $$\displaystyle \frac{cos2x+cos2y}{sin2x-sin2y}=\frac1{tan(x-y)}$$. Can someone provide me some hints? I tried to manipulate the right-hand expression but got back to square one.

#### Jhevon

MHF Helper
Remember that $$\displaystyle \tan \theta = \frac {\sin \theta}{\cos \theta}$$. So...

Start on the right hand side by writing $$\displaystyle \frac 1{\tan (x - y)}$$ as $$\displaystyle \frac {\cos (x - y)}{\sin (x - y)}$$, then apply the addition formula for each. Where can you get with that?

• topsquark

#### Idea

$$\displaystyle \cos (2x)+\cos (2y)=2\cos (x+y)\cos (x-y)$$

$$\displaystyle \sin (2x)-\sin (2y)=2\cos (x+y)\sin (x-y)$$

• topsquark and Jhevon

#### Monoxdifly

Remember that $$\displaystyle \tan \theta = \frac {\sin \theta}{\cos \theta}$$. So...

Start on the right hand side by writing $$\displaystyle \frac 1{\tan (x - y)}$$ as $$\displaystyle \frac {\cos (x - y)}{\sin (x - y)}$$, then apply the addition formula for each. Where can you get with that?
I tried that and ended up going back to the right-hand side form.

$$\displaystyle \cos (2x)+\cos (2y)=2\cos (x+y)\cos (x-y)$$

$$\displaystyle \sin (2x)-\sin (2y)=2\cos (x+y)\sin (x-y)$$
This one proved to be more useful, but to each their own.