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Trigonometric Ratios

  1. Jun 21, 2008 #1
    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"
    [tex]\frac{2ab}{b^{2}-a^{2}}[/tex]


    3. The attempt at a solution

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

    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. jcsd
  3. Jun 21, 2008 #2
    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) [tex]2 sin \left( \frac{2x+2y}{2} \right) cos \left( \frac{2x+2y}{2} \right) = 2sin(x+y)cos(x-y)[/tex]

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

    Edited.
     
    Last edited: Jun 21, 2008
  4. Jun 21, 2008 #3
    Thanks a lot friend. I hav solved it now. Thanks once again.
     
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