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Math Help - Finding an exact value using half angle.

  1. #1
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    Finding an exact value using half angle.

    Hello, all.

    I need to find the value of cos(\frac{17\pi}{12}) using the half angle formula for cos.

    Here's what I've done.

    I found that the reference angle for cos(\frac{17\pi}{12}) is cos(\frac{\pi}{12}) in quadrant II, making the sign for the formula negative.

    So I've got:

    -\sqrt{ \frac{1+cos(\frac{\pi}{6})}{2} }

    I evaluate this to:

    -\frac{ \sqrt{ 2+\sqrt{3} } }{2}

    This is incorrect, but I'm not sure how to arrive at the correct answer, which is:

    \frac{\sqrt{2-\sqrt{3}}}{2}

    Thanks for the help!
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  2. #2
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    I get

    \cos \frac{17\pi}{12} = \cos\left(\frac{1}{2} \frac{17\pi}{6}\right) \pm\sqrt{ \frac{1+\cos\frac{17\pi}{6}}{2} }

     =\pm\sqrt{ \frac{1+\cos(\frac{12\pi}{6}+\frac{5\pi}{6})}{2} }<br />

     =\pm\sqrt{ \frac{1+\cos\frac{5\pi}{6}}{2} }<br />

     =\pm\sqrt{ \frac{1-\cos\frac{\pi}{6}}{2} }<br />

     =\pm\sqrt{ \frac{1-\frac{\sqrt{3}}{2}}{2} }<br />

    simplify.
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  3. #3
    Junior Member
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    Since the reference angle is negative, shouldn't we use the ( - ) sign in front of the equation and then place the reference angle inside the equation with a ( + ) sign?

    Obviously, my thinking is wrong, but I was hoping someone could explain.
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