[ASK] Prove

Apr 2018
163
38
Sukoharjo
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
Feb 2007
11,681
4,225
New York, USA
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?
 
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Jun 2013
1,112
590
Lebanon
\(\displaystyle \cos (2x)+\cos (2y)=2\cos (x+y)\cos (x-y)\)

\(\displaystyle \sin (2x)-\sin (2y)=2\cos (x+y)\sin (x-y)\)
 
Apr 2018
163
38
Sukoharjo
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.
 
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