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Determine Sign of Half-Angle Identities

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Use the figure to evaluate the function that f(x)=sin(x)

    f(θ/2)
    2. Relevant equations

    n/a

    3. The attempt at a solution

    x2+y2=1

    x= -2/7

    (-2/7)2+y2=1

    y=√(6)/7
    sin(θ/2)= +/- √(1-cos(θ)/2) (the whole function is over 2 inside of the square root)

    =+/- √(1-√(6)/7/2)

    =+/- √(7-√(6)/14)

    I keep coming out with the wrong sign. How do you determine the sign in a Half- Angle identity?
     

    Attached Files:

  2. jcsd
  3. Mar 14, 2013 #2

    eumyang

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    ^ Not sure how you got that. Also, the y value has to be negative. Do you know why?

    Is it supposed to be theta or alpha? Your diagram has the angle labeled alpha. Assuming theta, and assuming you know what quadrant it is in, you can guess what quadrant θ/2 is in. You also should know that certain trig functions are positive in certain quadrants.
     
  4. Mar 14, 2013 #3
    I used the Pythagorean theorem to find the (square root) 6/7 but it should be negative because of the fact that it is the third quadrant. What I don't understand is how to determine whether the whole function is (+) or (-). Can you help me?
     
  5. Mar 14, 2013 #4
    it is supposed to alpha.
     
  6. Mar 14, 2013 #5
    Why should it be negative? I understand that there are certain trig functions that are (+) and (-) in certain quadrants, but why is the the whole function negative?
     
  7. Mar 14, 2013 #6

    eumyang

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    I'm afraid you made a mistake somewhere. Even with the negative, I am not getting
    [tex]y = -\frac{\sqrt{6}}{7}[/tex]
     
  8. Mar 14, 2013 #7
    The answer to this question is 3/(square root) 14. That is what my homework is telling me. How do you come to that answer?
     
  9. Mar 14, 2013 #8
    I believe I did make a mistake somewhere, just not sure where.
     
  10. Mar 14, 2013 #9

    eumyang

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    I was not answering the overall question. I was pointing out that you got the wrong y-coordinate of the point in the beginning of your work:

    Another thing:
    I forgot to mention that this is also wrong. On the unit circle, does cos θ equal the y-coordinate?
     
  11. Mar 14, 2013 #10
    Cos of theta equals the x- coordinate. I realize that I got it wrong, but can you explain to me where i went wrong and how the homework came to that answer?
     
  12. Mar 15, 2013 #11

    eumyang

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    It would be better if you try it again yourself.

    Instead of plugging in the y-coordinate (which you still haven't fixed), plug in the x-coordinate, (-2/7):
    [itex]\sin \left ( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}[/itex]
    [itex]\sin \left ( \frac{\theta}{2} \right) = \pm \sqrt{\frac{1 - (-2/7)}{2}}[/itex]

    This will simplify to [itex]\frac{3}{\sqrt{14}}[/itex]. Try it.
     
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