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Trigonometry reflection problem

  1. Jun 20, 2011 #1
    Hi,

    I was about solving a refrection problem and I'm just one step away from the solution. I stuck at some simple problem I guess.

    I have:

    [tex] \alpha=60°+\frac1 3 arcsin(\frac1 {1,33} sin(\alpha)) [/tex]

    and I want to find the alpha but I have troubles to solve this.

    I guess it's not that hard, but I'm failing right now :blushing:

    Thanks for your help in advance
     
  2. jcsd
  3. Jun 20, 2011 #2

    I like Serena

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    Homework Helper

    Hi Lindsayyyy! :smile:

    The problem you have here has afaik no algebraic solution.
    If that is what you're looking for, you won't find it.

    What you can do, of course, is make an approximation.
    That's easy with for instance http://wolframalpha.com.

    Cheers! :smile:
     
  4. Jun 20, 2011 #3
    Hhhmm...I gave it a try, and I solved by hand pretty easily.

    This was my approach:

    Multiply both side of the equation by 3, to eliminate the 3 in the denominator.
    Move the 180 from the righ to the left
    take the sine of both sides of the equation
    you should have

    sin(3a - 180) = sin( arcsin( (1/1.33) sin(a) ) )

    but the sin(p-180) = -sin(p)

    -sin(3a) = sin( arcsin( (1/1.33) sin(a) ) )

    and the sin ( arcsin(x) ) = x

    -sin(3a) = (1/1.33) sin(a)

    and sin(3a) = 3sin(a) - 4(sin(a))^3

    -3sin(a) + 4(sin(a))^3 = (1/1.33)sin(a)

    put everything on one side of the equation and you have a cubic equation without a quadratic or constant term:

    4(sin(a))^3 - (3 + 1/1.33)sin(a) = 0

    Now, we simply say

    X = sin(a)
    A = 4
    C = -(3 + 1/1.33)

    and we have

    AX^3 + CX = 0

    or, if we factor one X out:

    X ( AX^2 + C ) = 0

    So, one solution is X1 = zero, due to the single X...now, we need to solve the roots of the quadratic equation with no B term:

    X2 = +sqrt ( -C/A ) = 0.9685
    X3 = -sqrt ( -C/A ) = -0.9685

    So, from X = sin(a)

    a = arcsin(X)
    a1 = arcsin(0) = 0 degrees
    a2 = arcsin(0.9685) = 75.578 degrees
    a3 = arcsin(-0.9685) = -75.578 degrees

    and that's it.
     
  5. Jun 20, 2011 #4
    You still need to check if the solutions work.

    You created an extra solution when you replaced

    3a - 180 = arcsin( (1/1.33) sin(a) ) with

    sin(3a - 180) = sin(arcsin( (1/1.33) sin(a) ))
     
  6. Jun 20, 2011 #5

    I like Serena

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    Homework Helper

    Awww! You're right.
    I made a mistake in my calculation and came out with sine's with different arguments that could not be made equal.
    Good catch! :smile:

    Yes. That is right too.......

    There is an extra condition that -90 < 3a - 180 ≤ 90.
    That is, 30 < a ≤ 90.


    Cheers! :smile:
     
  7. Jun 21, 2011 #6
    Thank you very much for help. Nice approach @ gsal, this helped me a lot.:approve:
     
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