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Math Help - Double Angle Formula (Sin 2A, Cos 2A...)

  1. #1
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    Double Angle Formula (Sin 2A, Cos 2A...)

    Hi,

    I'm fairly new to more complex trigonometry, and i've found myself stumped on a certain question which can be found in exercise 6C of the Cambridgbe Advanced Mathematics course textbook. The question reads as follows...

    If Sin A = 2/3, and A is obtuse, find the exact values of Cos A, Sin 2A and tan 2A.

    I know how to use the double angle formula to calculate Sin 2A and tan 2A, but i'm stuck on trying to find Cos A which i need to put into the double angle formula to find the last two parts...

    Any ideas? Thanks
    Last edited by lrwatton93; September 18th 2011 at 06:39 AM.
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  2. #2
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    Re: Double Angle Formula (Sin 2A, Cos 2A...)

    Quote Originally Posted by lrwatton93 View Post
    Hi,

    I'm fairly new to more complex trigonometry, and i've found myself stumped on a certain question which can be found in exercise 6C of the Cambridgbe Advanced Mathematics course textbook. The question reads as follows...

    If Sin A = 2/3, and A is obtuse, find the exact values of Cos A, Sin 2A and tan 2A.

    I know how to use the double angle formula to calculate Sin 2A and tan 2A, but i'm stuck on trying to find Cos A which i need to put into the double angle formula to find the last two parts...

    Any ideas? Thanks
    familiar with this identity?

    \cos^2{A} + \sin^2{A} = 1

    solving for \cos{A} ...

    \cos{A} = \pm \sqrt{1 - \sin^2{A}}

    since A is obtuse, the value of \cos{A} would be negative (why?)
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  3. #3
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    Re: Double Angle Formula (Sin 2A, Cos 2A...)

    Would i start off using the Pythagorean identity:

    (Sin A)^2 + (Cos A)^2 = 1

    let Sin A = 2/3

    (2/3)^2 + (Cos A)^2 = 1

    take (2/3)^2 from each side...

    (Cos A)^2 = 1 - 4/9

    because Sin A is Obtuse, the square root of the RHS must be negative (Deduced from Sin / Cos graphs)

    Cos A = - [SQRT]5 / 3

    so does that work?
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  4. #4
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    Re: Double Angle Formula (Sin 2A, Cos 2A...)

    that's all, folks.
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