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Math Help - Half-Angle Formula Help

  1. #1
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    Nov 2009
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    Half-Angle Formula Help

    Hey Guys,

    I'm trying to solve the following problem:

    Sin(∆)= 3/5 0 < ∆ < pi/2

    Sin(∆/2) = +/- Sqrt[(1-cos∆)/2]

    How would I go about that?

    I started with
    5^2 - 3^2 = sqrt(16) = +/- 4

    cos(∆) = 4/5

    So --
    Sin(∆/2) = sqrt[(1-4/5)/2]
    Sqrt[1/5/2] = 1/Sqrt(10) = Sqrt(10)/10

    But the book says that this is incorrect. Can someone please assist?

    Thank you,
    Jason V.
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  2. #2
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    skeeter's Avatar
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    North Texas
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    Quote Originally Posted by JasonV View Post
    Hey Guys,

    I'm trying to solve the following problem:

    Sin(∆)= 3/5 0 < ∆ < pi/2

    Sin(∆/2) = +/- Sqrt[(1-cos∆)/2]

    How would I go about that?

    I started with
    5^2 - 3^2 = sqrt(16) = +/- 4

    cos(∆) = 4/5

    So --
    Sin(∆/2) = sqrt[(1-4/5)/2]
    Sqrt[1/5/2] = 1/Sqrt(10) = Sqrt(10)/10

    But the book says that this is incorrect. Can someone please assist?

    Thank you,
    Jason V.
    you are correct ... books have been wrong before.
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  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
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    Quote Originally Posted by skeeter View Post
    you are correct ... books have been wrong before.
    Thanks man.
    Last edited by JasonV; November 4th 2009 at 05:37 PM.
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