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Math Help - Location of half angle

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    Location of half angle

    Sin u = 5/13 and pi/2<u<pi.

    Find Sin (u/2); How do you determine the sign (positive or negative) for the value?

    Using that information find cos (u/2): I have sqrt(26)/26. Is this positive or negative? How do I know?
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    Quote Originally Posted by IDontunderstand View Post
    Sin u = 5/13 and pi/2<u<pi.

    Find Sin (u/2); How do you determine the sign (positive or negative) for the value?
    Which quadrant will it be in?

    for \cos u consider pythagorean triplet 5,12,13.
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    Quote Originally Posted by IDontunderstand View Post
    Sin u = 5/13 and pi/2<u<pi.

    Find Sin (u/2); How do you determine the sign (positive or negative) for the value?

    Using that information find cos (u/2): I have sqrt(26)/26. Is this positive or negative? How do I know?
    \cos\left(u\right)=\frac{-12}{13}

    \sin\left(\frac{u}{2}\right)=\sqrt{\frac{1-\cos(u)}{2}}
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    Quote Originally Posted by IDontunderstand View Post
    Sin u = 5/13 and pi/2<u<pi.

    Find Sin (u/2); How do you determine the sign (positive or negative) for the value?

    Using that information find cos (u/2): I have sqrt(26)/26. Is this positive or negative? How do I know?
    if ...

    \displaystyle \frac{\pi}{2} < u < \pi

    then ...

    \displaystyle \frac{\pi}{4} < \frac{u}{2} < \frac{\pi}{2} ... quad I , correct?
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    sin u =5/13 and pi/2<u<pi

    so find cos u = -12/13 as it is in 2nd quadrant.

    use cos u= 2  cos^2 u/2 -1 to evaluate cos u/2.

    you take only positive value as it is in 1st quadrant.

    using that you can find sin u/2 also.



    since u/2(pi/4<u/2<pi/2 ) is in 1st quadrant

    both cos u/2 and sin u/2 are positive.
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    @saravananbs
    Why not learn to post in symbols? You can use LaTeX tags
    You do understand that this is not a homework service?
    [tex] \sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}} [/tex]
    gives  \sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}}
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    Quote Originally Posted by Plato View Post
    @saravananbs
    Why not learn to post in symbols? You can use LaTeX tags
    You do understand that this is not a homework service?
    [tex] \sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}} [/tex]
    gives  \sin\left(\frac{u}{2}\right)=\sqrt{\dfrac{1-\cos(u)}{2}}
    yes i try to use latex.

    i have given only the approach to the problem.
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