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Math Help - half angle trouble

  1. #1
    Junior Member
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    half angle trouble

    I am trying to use the half angle formula to solve these problems:
    Sin(-7pi/12)
    Sin(7pi/12)
    cos(13pi/12)
    cos(3pi/8)
    here is what I did cos(3pi/8) = 1/2(1+cis13pi/4)= sqrt (2)+sqrt(3)/2 but the book shows a different answer.
    for Sin(-7pi/12) I got sqrt(2)-sqrt(3)/2 I must be doing something wrong and I must have not understood what I read in the text. This area of the text seems more difficult than the rest of the material we have covered. Please explain the half angle formula so even I can understand it.
    Thank you for your paitience!
    Keith Stevens
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  2. #2
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    Heloo, Keith!

    The formulas are:

    . . \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{\frac{1-\cos2\theta}{2}}

    . . \cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}\quad\Rightarrow\quad\cos\theta \:=\:\pm\sqrt{\frac{1+\cos2\theta}{2}}

    and we must choose the sign carefully.



    Use the half angle formulas:

    (1)\;\;\sin\left(\text{-}\frac{7\pi}{12}\right)\quad(2)\;\sin\!\left(\frac  {7\pi}{12}\right)\quad(3)\;\cos\!\left(\frac{13\pi  }{12}\right)\quad(4)\;\cos\!\left(\frac{3\pi}{8}\r  ight)

    (1)\;\text{-}\frac{7\pi}{12} is in Quadrant 3, where sine is negative.
    \sin\left(\text{-}\frac{7\pi}{12}\right)\:=\:-\sqrt{\frac{1 - \cos\left(\text{-}\frac{7\pi}{6}\right)}{2}} \:=\: -\sqrt{\frac{1 - \left(\text{-}\frac{\sqrt{3}}{2}\right)}{2}} \:=\:-\sqrt{\frac{2 + \sqrt{3}}{4}}\:=\:-\frac{\sqrt{2 + \sqrt{3}}}{2}


    (2)\;\;\frac{7\pi}{12} is in Quadrant 2, where sine is positive.
    \sin\left(\frac{7\pi}{12}\right)\:=\:\sqrt{\frac{1 - \cos\left(\frac{7\pi}{6}\right)}{2}} \:=\: \sqrt{\frac{1 - \left(\text{-}\frac{\sqrt{3}}{2}\right)}{2}} \:=\:\sqrt{\frac{2 + \sqrt{3}}{4}}\:=\:\frac{\sqrt{2 + \sqrt{3}}}{2}


    (3)\;\;\frac{13\pi}{12} is in Quadrant 3, where cosine is negative.
    \cos\left(\frac{13\pi}{12}\right)\:=\:-\sqrt{\frac{1 + \cos\left(\frac{13\pi}{6}\right)}{2}} \:=\: -\sqrt{\frac{1 + \left(\frac{\sqrt{3}}{2}\right)}{2}} \:=\:-\sqrt{\frac{2 + \sqrt{3}}{4}}\:=\:-\frac{\sqrt{2 + \sqrt{3}}}{2}


    (4)\;\;\frac{3\pi}{8} is in Quadrant 1, where cosine is positive.
    \cos\left(\frac{3\pi}{8}\right)\:=\:\sqrt{\frac{1 + \cos\left(\frac{3\pi}{4}\right)}{2}} \:=\: \sqrt{\frac{1 + \left(\text{-}\frac{\sqrt{2}}{2}\right)}{2}} \:=\:\sqrt{\frac{2 - \sqrt{2}}{4}}\:=\:\frac{\sqrt{2 - \sqrt{2}}}{2}

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