# Trigonometric Ratios

1. Jun 21, 2008

### ritwik06

1. The problem statement, all variables and given/known data
Given:
sin x + sin y= a
cos x + cos y=b
find tan(x+y)
Prove it to be equal to"
$$\frac{2ab}{b^{2}-a^{2}}$$

3. 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!

Last edited: Jun 21, 2008
2. Jun 21, 2008

### konthelion

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

For the part,
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.

Last edited: Jun 21, 2008
3. Jun 21, 2008

### ritwik06

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