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Math Help - Itsa me again T.T

  1. #1
    Junior Member Freaky-Person's Avatar
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    Itsa me again T.T

    Yeah... I'll get professional help with this tomorrow if I can...

    Until then... I love you guys XD

    The question said to solve the following using the double angle formulas. but I have like no clue what the double angle formula possibly has to do with any of this..

    \sin (\frac{3\pi}{8})
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  2. #2
    Super Member

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    Hello, Freaky-Person!

    Solve the following using the double-angle formulas: . \sin\left(\frac{3\pi}{8}\right)
    One of the formulas is: . \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad\sin\theta \;=\;\pm\sqrt{\frac{1-\cos2\theta}{2}}


    Let \theta \,=\,\frac{3\pi}{8}
    Then we have: . \sin\left(\frac{3\pi}{8}\right) \;=\;\pm\sqrt{\frac{1 - \cos\frac{3\pi}{4}}{2}} \;=\;\pm\sqrt{\frac{1 - \left(\text{-}\frac{\sqrt{2}}{2}\right)}{2}} \;=\;\pm\sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}

    . . = \;\pm\sqrt{\frac{2+\sqrt{2}}{4}} \;=\;\pm\frac{\sqrt{2+\sqrt{2}}}{2}


    Since \frac{3\pi}{8} is in Quadrant 1, where sine is positive: . \sin\left(\frac{3\pi}{8}\right) \;=\;\frac{\sqrt{2+\sqrt{2}}}{2}

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