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Math Help - about multiple angles of trigonometry

  1. #1
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    about multiple angles of trigonometry

    my answer is different from the provided answer.
    can someone please help me to check where i gone wrong??

    the question is:

    given that sin X= - 5/13, cos Y = - 4/5 and that X and Y are in the same quadrant, find the value of sin(X/2).

    my solution :
    cos X = 2 cos^2 (X/2) -1
    cos (X/2) = square root [(1/13)/2]
    cos (X/2)= 1/(square root 26)
    cos (X/2)= (square root 26)/26

    then, sin X = 2(sin X/2)(cos X/2)
    -5/13=2(sin X/2)[(square root 26)/26]
    -5/13 x [26/(square root 26)] = 2(sin X/2)
    [-10/(square root 26)] x 1/2 = sin X/2
    sin (X/2) = -5/(square root 26)
    rationalise
    sin (X/2) = [(-5square root 26)/26]

    but the provided answer is (5square root 26)/26 .....

    which means mine is negative but the provided answer is positive...

    can anyone tell me where i gone wrong???
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  2. #2
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by wintersoltice View Post
    my answer is different from the provided answer.
    can someone please help me to check where i gone wrong??

    the question is:

    given that sin X= - 5/13, cos Y = - 4/5 and that X and Y are in the same quadrant, find the value of sin(X/2).

    my solution :
    cos X = 2 cos^2 (X/2) -1
    cos (X/2) = square root [(1/13)/2]
    cos (X/2)= 1/(square root 26)
    cos (X/2)= (square root 26)/26

    then, sin X = 2(sin X/2)(cos X/2)
    -5/13=2(sin X/2)[(square root 26)/26]
    -5/13 x [26/(square root 26)] = 2(sin X/2)
    [-10/(square root 26)] x 1/2 = sin X/2
    sin (X/2) = -5/(square root 26)
    rationalise
    sin (X/2) = [(-5square root 26)/26]

    but the provided answer is (5square root 26)/26 .....

    which means mine is negative but the provided answer is positive...

    can anyone tell me where i gone wrong???
    it was said that X and Y are in the same quadrant. sin X is negative, so X must be in III or IV quadrant.. cos Y is also negative, thus, Y is in II or III..

    therefore, X and Y must be in III..
    note that if you divide an angle in the III quadrant, the resulting angle will be on the II quadrant..

    thus, your cos (X/2) must be negative..
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  3. #3
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    does that mean my answer is correct???
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  4. #4
    MHF Contributor kalagota's Avatar
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    Taguig City, Philippines
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    Quote Originally Posted by wintersoltice View Post
    does that mean my answer is correct???
    no.. the answer is the positive one..

    look at your calculations.

    sin x = 2 sin (x/2) cos (x/2)

    sin x is negative, cos (x/2) is also negative, then sin (x/2) must be positive..

    besides, since x/2 is in the second quadrant, it should be logical that sin (x/2) must be positive..
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