Expert: Paul Klarreich - 8/13/2007

Question
i had hard time trying to solve this identity,,,
its get loop every time i got further,, i am stuck with it

could you please show me how to deal with it

(sin 2x + sin 2y) tan(x-y) = cos 2x - cos2y

Answer
Hi, Saleh,

I know this is late, but I have been on vacation and didn't see it until today.

There might be a small error in your identity, because things work out like this: (unless I blew a sign somewhere)

First, find these two identities at:

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

1. sin u + sin v = 2 sin((u + v)/2) cos(u - v)/2)

2. cos u - cos v = - 2 sin((u + v)/2) sin((u - v)/2)

Apply them to your example:

sin 2x + sin 2y =  2 sin(x + y) cos(x - y)
cos 2x - cos 2y = -2 sin(x + y) sin(x - y)

Substitute that into your left side and right sides, along with the well-known quotient identity.

(sin 2x + sin 2y) tan(x-y) = cos 2x - cos2y

2 sin(x + y) cos(x - y) tan(x-y) = -2 sin(x + y) sin(x - y)

2 sin(x + y) cos(x - y) sin(x-y)/cos(x-y) = -2 sin(x + y) sin(x - y)

2 sin(x + y) sin(x - y)  = -2 sin(x + y) sin(x - y)

So this is your identity EXCEPT for that minus sign, which I don't see a way to get rid of.