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Math Help - having trouble with a 1/2 angle

  1. #1
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    having trouble with a 1/2 angle

    having trouble with a 1/2 angle-half_angle.jpg

    Ive run into a point I cant seem to figure out. Where do I go from where I stopped at cos^2(x/2)= 6/14??
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    Quote Originally Posted by ninobrn99 View Post
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    Ive run into a point I cant seem to figure out. Where do I go from where I stopped at cos^2(x/2)= 6/14??
    \displaystyle \cos^2\left(\frac{x}{2}\right) = \frac{3}{7}

    \displaystyle \cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{3}{7}}

    since x is in quad II , which value works?
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    okay, Im not crazy then. I came up with that b4, but it wasn't one of my answer choices. Since x is in quad 2, then cos is neg and sin is posi.

    Thank you.
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    Quote Originally Posted by ninobrn99 View Post
    okay, Im not crazy then. I came up with that b4, but it wasn't one of my answer choices. Since x is in quad 2, then cos is neg and sin is posi.

    Thank you.
    for x it is ... you need to remember that the question is asking about \cos\left(\frac{x}{2}\right) ... care to rethink that conclusion?
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    Ive been staring at this problem and cant figure out what to do from there. I would think to set cos(1/2x)=-(sqrt3/7) but that doesnt work out either.
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  6. #6
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    if 90 < x < 180 , then 45 < \frac{x}{2} < 90 ... so \cos\left(\frac{x}{2}\right) = ?
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  7. #7
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    Got it! Ugh, I've been going at this solo this past weekend trying to get ahead and was finally shown today what you meant. Thank you again for the help!
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