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Trig Identities for Refraction

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem is essentially (I've rephrased it, but this is what it is asking)
    show that [ (2n1cosa) / (n2cosa + n1cosb) ]2

    is equal to (sin2a sin2b) / (sin2(a+b)cos2(a-b))

    where these are for refraction through materials and satisfy n1sina=n2sinb



    2. The attempt at a solution
    I just took pictures of my work because it would be easier than typing it all out. As far as I can tell I didn't do anything wrong up till then
    first:
    http://img526.imageshack.us/img526/6325/0221091218.jpg [Broken]
    second:
    http://img10.imageshack.us/img10/7136/0221091220.jpg [Broken]

    the lines are numbered so you can refer to them easily if you need to


    tl;dr:
    show how [2cosasinb]2 = sin2a sin2b


    Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 20, 2009 #2

    Mark44

    Staff: Mentor

    Just looking at your bottom line, I think you have made a mistake somewhere.
    Your last line is to show that 2cosa*sinb = sin2a*sin2b, which is not an identity.

    Starting from the right side:
    sin2a*sin2b = 2sina*cosa*2sinb*cosb = 4sina*cosa*sinb*cosb, which is different from the left side.

    You don't show how you got your first expression to [ (2cosasinb) / (sin2a / 2 + sin2b / 2) ]^2, so it's possible you made a mistake getting to that point.
     
  4. Feb 20, 2009 #3
    ok thanks, I'll double check my work again, and if it doesn't work out, post my complete work to see
     
  5. Feb 21, 2009 #4
    Ok, I actually just made a mistake typing the question in here. I fixed it now, and posted pictures of my work so you can see how I got there. The last line (now correct in first post) should have been

    show how
    [2cosasinb]2 = sin2a sin2b

    Thanks for your help :)
     
  6. Feb 23, 2009 #5
    bump for help. is this an identity?

    [2cosasinb]2 = sin2a sin2b
     
  7. Feb 23, 2009 #6

    Mark44

    Staff: Mentor

    I think it might be. The equation is true for a = pi/6 and b = pi/3. I don't see how to prove that it's true for arbitrary a and b, but let me look into it...
     
  8. Feb 23, 2009 #7

    Mark44

    Staff: Mentor

    Not an identity. In my earlier post I neglected to square a number, so it's not true for a = pi/6 and b = pi/3.
     
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