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Trying to prove a trig identity

  1. Aug 6, 2013 #1
    1. The problem statement, all variables and given/known data
    prove that (sinA +sin3A + sin5A)/(cosA + cos3A + cos5A) = tan3A


    2. Relevant equations
    sinP + sinQ = 2sin((P+Q)/2)cos((P-Q)/2)
    cosP + cosQ = 2cos((P+Q)/2)cos((P-Q)/2)


    3. The attempt at a solution
    (sin3A + sinA) + sin5A = 2sin2AcosA + 2sin((5/2)A)cos((5/2)A)
    (cos3A + cosA) + cos5A = 2cos2AcosA + 2cos((5/2)A + 45)cos((5/2)A - 45)
    It started to feel like a bit of a cul de sac at this point so I tried pursuing variations on this theme by starting with (sin5A + sin3A) + sinA etc, but these seemed just as fruitless.
     
  2. jcsd
  3. Aug 6, 2013 #2
    Because the right hand side is a function of 3A, it could be useful to combine A and 5A on the left hand side.
     
  4. Aug 6, 2013 #3

    verty

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    Homework Helper

    I haven't checked what Voko suggested but you could also try this: let p = 3A, q = 2A.
     
  5. Aug 6, 2013 #4
    voko's suggestion is nice. I was able to solve the problem without pen and paper using his hint. I would suggest Appleton to try that. :)
     
  6. Aug 7, 2013 #5
    Thanks for all your suggestions, voko's suggestion led me to the proof, however, I'm still unable to eliminate the pen and paper.
     
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