Solve Trigonometric Ratios: 2ab/(b^2 - a^2)

[ritwik06]

Homework Statement

Given:
sin x + sin y= a
cos x + cos y=b
find tan(x+y)
Prove it to be equal to"
\frac{2ab}{b^{2}-a^{2}}

The Attempt at a Solution

I get stuck after this:
2ab= 2 sin (x+y)+sin 2x+sin 2y
b^{2}-a^{2}= cos 2x + cos 2y + 2 cos (x+y)

Division of the two never results in tan (x+y).
Plase help me express tan (x+y) in terms of a,b!

 

[konthelion]

Alright, I checked these two corrections and they look correct.

2ab= 2 sin (x+y)+sin 2x+sin 2y
b^{2}-a^{2}= cos 2x + cos 2y + 2 cos (x+y)

For the part,

sin(2x)+sin(2y)

you can use the sum and difference formulas for sines and cosines you will get
(3) 2 sin \left( \frac{2x+2y}{2} \right) cos \left( \frac{2x+2y}{2} \right) = 2sin(x+y)cos(x-y)

For the part, cos(2x)+cos(2y)
(4) 2 cos \left( \frac{2x+2y}{2} \right) cos \left( \frac{2x-2y}{2} \right)=2cos(x+y)cos(x-y)

Edited.

 

[ritwik06]

Since, you know that
(1) sin x + sin y= a
(2) cos x + cos y=b

Then by the sum and difference formulas, you will get
(3) 2 sin \left( \frac{x+y}{2} \right) cos \left( \frac{x+y}{2} \right) = a
(4) 2 cos \left( \frac{x+y}{2} \right) cos \left( \frac{x-y}{2} \right) = b

No, wait scratch this. This won't help.

Thanks a lot friend. I hav solved it now. Thanks once again.