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Math Help - [SOLVED] double argument properties

  1. #1
    Sevens
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    [SOLVED] double argument properties

    Need some urgent help solving double argument properties for Calc/Trig course. Anybody out there that can help?

    If cos A=3/5 and 0<A<90, find sin 2A and cos 1/2 A

    Any help on this?
    Last edited by Sevens; December 20th 2006 at 05:11 PM. Reason: added problem
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  2. #2
    Super Member

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    Hello, Sevens!

    You're expected to know the necessary identities:

    . . \sin2A\:=\:2\sin A\cos A

    . . \cos\frac{A}{2}\:=\:\pm\sqrt{\frac{1 + \cos A}{2}}


    If \cos A = \frac{3}{5} and 0^o < A < 90^o, find [1]\;\sin2A and [2]\;\cos\frac{A}{2}

    [1] Since \cos A = \frac{3}{5}, we have a 3-4-5 right triangle. . Hence: . \sin A = \frac{4}{5}

    Therefore: . \sin2A \;=\;2\sin A\cos A \;=\;2\left(\frac{4}{5}\right)\left(\frac{3}{5}\ri  ght)\;=\;\boxed{\frac{24}{25}}


    [2] \cos\frac{A}{2} \;=\;\sqrt{\frac{1 + \cos A}{2}} \;=\;\sqrt{\frac{1 + \frac{3}{5}}{2}} \;=\;\sqrt{\frac{8}{10}}\;=\;\sqrt{\frac{4}{5}} \;=\;\frac{2}{\sqrt{5}}\;=\;\boxed{\frac{2\sqrt{5}  }{5}}

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