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Tan=cos/sin help

  1. Nov 14, 2005 #1
    sin(a+b)/sin(a+c)=[ sin(2b)/sin(2c)]^(1/2)
    then prove tan^2a=tanbtanc
    I have reached till {tan(a)cos(b)+ sin(b)} * {sin(c)cos(c)}^(1/2)=
    {tan(a)cos(c)+sin(c)}* { sin(b) cos(b)}^(1/2)
     
  2. jcsd
  3. Nov 15, 2005 #2

    Can you assume that a+b+c = 180 degrees ?

    I think there is something missing in your question. How elese can you prove the first equality ?

    Or is it given ? In that case, you can find a connection between the angles by manipulating this first equality.

    marlon
     
  4. Nov 15, 2005 #3
    you have all the info you need. i squared everything, and then solved for tan^2a. its messy, but it does work out. when you square it all, the tana terms cancel. and then you just have to factor and divide ect.
     
  5. Nov 15, 2005 #4
    no there is no relation between the angles.the relation given is not going to help much. I think I will be getting a method for that sum.however thank u
     
  6. Nov 15, 2005 #5
    it works out. i said that already. just keep trucking. like i said, square everything and start gathering terms. its just a lot of math, but it does work.
     
  7. Nov 17, 2005 #6
    try using some of these identities

    Tan=cos/sin

    Sin(a+b)=
    SinaCosb+CosaSinb

    Cos(a+b)
    CosaCosb-SinaSinb

    ya thats my knowledge of trig
     
  8. Nov 17, 2005 #7
    the OP already used those identities. the only other identy needed is knowing tanx= sinx/cosx. otherwise its all just rearranging the equation.
     
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