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Math Help - Unit Circle: Trig Functions of Real Numbers

  1. #1
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    Unit Circle: Trig Functions of Real Numbers

    The question asks to find the sine of 5pi/7. My teacher says to add/subtract an n value of pi to get a value on the unit circle; and then I find the sine. I've been at this problem for awhile and there has to be a better way than guessing how many pi's it will take to get a value on the circle.

    What is the answer?

    I tried to add 2 pi.
    5pi/7 + 14pi/7 = 19pi/7 (Not on the unit circle!)

    Then 4 pi.
    5pi/7 + 28pi/7 = 33pi/7

    The idea is that its supposed to reduce to a value on the unit circle. . . But I tried some other numbers, and I can't get it!
    Last edited by tinu; December 1st 2009 at 04:11 PM.
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  2. #2
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    Quote Originally Posted by tinu View Post
    The question asks to find the sine of 5pi/7. My teacher says to add/subtract an n value of pi to get a value on the unit circle; and then I find the sine. I've been at this problem for awhile and there has to be a better way than guessing how many pi's it will take to get a value on the circle.

    What is the answer?

    I tried to add 2 pi.
    5pi/7 + 14pi/7 = 19pi/7 (Not on the unit circle!)

    Then 4 pi.
    5pi/7 + 28pi/7 = 33pi/7

    The idea is that its supposed to reduce to a value on the unit circle. . . But I tried some other numbers, and I can't get it!
    The angle you're given isn't an integral multiple of any of the special angles.

    \left(\frac{5\pi}{7}\right)\left(\frac{180^{\circ}  }{\pi}\right)=\frac{900^{\circ}}{7}\approx 129^{\circ}
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  3. #3
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    Oh wow. I feel silly. Thank you.
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  4. #4
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    sin(5pi/7)
    =sin(7pi-2pi/7)
    =sin(7pi/7-2pi/7)
    =sin(pi-2pi/7)
    =sin(pi)cos(2pi/7)-cos(pi)sin(2pi/7)]
    =0-(-1)sin(2pi/7)
    =sin(2pi/7)

    as for the unit circle that ur teacher mentioned it is just a convention to remember what sign is taken by sin, cos and tan in which quadrant. the value neednot satisfy any circle's equation.

    s | A
    2 | 1
    ---------------
    T | C
    3 | 4

    for ur question we found out that sin(5pi/7)=sin(pi-2pi/7) which means that it lies on the quadrant 2 where sin is always positive . and thats wat we found out by expanding it using sine(a+b)=sinacosb+cosasinb.was this helpful????please comment.
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  5. #5
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    Thats a new way to look at it. I'll keep what you said in mind as we go onto later chapters. I think what we are doing is a bit simpler. I am pretty sure I get it though.
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