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Math Help - Half-Angle formulas

  1. #1
    Newbie
    Joined
    Apr 2010
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    7

    Half-Angle formulas

    find the exact value of sin(5pi/12)

    here's what I have, can you tell me if I'm doing it correctly and if I've simplified. I'm more than a little confused.

    [sin(5pi/6)]/2 = +/- √[(1-cosx)/2]

    = +/- √{[1-cos(5pi/6)]/2}

    = +/- √ ({1-[-√(3)/2]}/2)

    = +/- √ {[1+√(3)/2]/2}

    And that's as far as I got. I'm not sure if I'm doing it correctly or if i am, i have no idea how to simplify further. :/

    Thanks ahead of time, all help is very much appreciated.
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Can't quite tell exactly what you're doing.

    Here's a way...

    \sin(5\pi/12)=\sin(\pi/4 + \pi/6)

    = \sin(\pi/4)\cos(\pi/6)+\cos(\pi/4)\sin(\pi/6)

    = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2}

    = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}

    = \frac{(1 + \sqrt{3})}{2\sqrt{2}}

    = \frac{\sqrt{2}(1 + \sqrt{3})}{4}
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