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Sin(2sin-1(x/3)) stuck at last step

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Simplify this expression.

    sin(2* arcsin(x/3))

    2. Relevant equations

    3. The attempt at a solution

    Let t = arcsin(x/3), we get sin(2t) which is double angle
    sin(2T) = 2 sinT cosT

    I did cosT first

    T = arcsin x/3, and we know inverse of arcsin is sin T = x/3
    this gives us the hypotenuse 3, and opposite side = x (3sin delta = x)
    using the identity of cos^2 + sin^2 = 1, and solve for cos, I get cosT = sqrt(9-x^2)

    This is corrected.

    Now, solve for sinT.

    sin(arcsin(x/3)), is the same as just x/3, by definition I think.
    Well, the answer the book gave was 2x/9 * cosT
    I get the cosT part right, but now the x/9

    I don't understand how he produced the number 9.

    Thank you. I have an exam so please pardon me posting so many questions these two days.
  2. jcsd
  3. Apr 7, 2010 #2
    You said that you used [tex]sin^2+cos^2=1[/tex] to solve for the cos T. It should be [tex]\cos T= \sqrt{1-sin^2 T} [/tex]
    [tex]\cos T=\sqrt {1-\frac{x^2}{3^2}}[/tex]. Find a common denominator under the radical and a 3 pops out in the denominator.
    Last edited: Apr 7, 2010
  4. Apr 7, 2010 #3
    I still don't get it.
    Right, the identity is that, so in this case, we have sin^2 + cos^2 = 3^2
    so cost is right (what i wrote)
    i still can't understand how we get 1/9 outside for sinT
  5. Apr 7, 2010 #4
    The identity doesn't have [tex] 3^2[/tex], it has 1 [tex] \sin^2 T + \cos ^2 T=1[/tex] , not 9.

    You are mixing up the the identity with the triangle.

    So far you have [tex] \frac{2x}{3}\sqrt{1-\frac{x^2}{9}[/tex]
    Then you find the common denominator under the radical:

    Last edited: Apr 7, 2010
  6. Apr 7, 2010 #5
    Brilliant. Thank you for pointing this out.
    My math professor only repeats what the book prints, and my background in Math is a bit weak.
    Thanks for all the help!
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