Question
prove the identity
sin2x-sin2y/sin2x+sin2y=tan(x-y)/tan(x+y)
sin2x-sin2y/sin2x+sin2y=tan(x-y)/tan(x+y)
Answers
Steve
This one is easier if you work on the right side.
(sin(x-y)/cos(x-y)) / (sin(x+y)/cos(x+y))
(sinxcosy-cosxsiny)/(cosxcosy+sinxsiny)
-------------------------------------
(sinxcosy+cosxsiny)/(cosxcosy-sinxsiny)
Expand those products, and you will be able to factor out some (sin^2+cos^2) and watch it become just like the left side.
Working on the left side first requires you to insert some canceling factors which are not obvious.
(sin(x-y)/cos(x-y)) / (sin(x+y)/cos(x+y))
(sinxcosy-cosxsiny)/(cosxcosy+sinxsiny)
-------------------------------------
(sinxcosy+cosxsiny)/(cosxcosy-sinxsiny)
Expand those products, and you will be able to factor out some (sin^2+cos^2) and watch it become just like the left side.
Working on the left side first requires you to insert some canceling factors which are not obvious.