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Math Help - midpoint issue

  1. #1
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    midpoint issue

    Hi;
    Can someone just tell me the midpoint of

    -pi/6 and 5pi/6

    Thanks.
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  2. #2
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    Re: midpoint issue

    Quote Originally Posted by anthonye View Post
    Hi;
    Can someone just tell me the midpoint of

    -pi/6 and 5pi/6

    Thanks.
    Their midpoint is the same as their mean in this case so add them and divide by 2. You should get \frac{\pi}{3}
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  3. #3
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    Re: midpoint issue

    Yeah thanks thats what I get. the web gets pi/2 thought I was right.
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    Re: midpoint issue

    so in this case pi/3 and 5pi/7 I get 7pi/12.
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    Re: midpoint issue

    Quote Originally Posted by anthonye View Post
    so in this case pi/3 and 5pi/7 I get 7pi/12.
    I get \dfrac{11\pi}{21}. I'm not sure how you got to twelfths from twenty-firsts (21 being the LCD of 3 and 7)

    Spoiler:
    \dfrac{1}{2}\left(\frac{\pi}{3} \cdot \frac{7}{7} + \frac{5\pi}{7} \cdot \dfrac{3}{3} \right) = \dfrac{1}{2}\left(\frac{7\pi}{21} + \frac{15\pi}{21}\right) = \dfrac{1}{2} \cdot \dfrac{22\pi}{21} = \dfrac{11\pi}{21}



    More generally lets say you have a line between a and b and you want to find the midpoint (we'll call this c). By definition the midpoint (c) is equidistant from either end (a and b). Thus b - c = c - a.

    With a little bit of manipulation: b+a = 2c and c = \dfrac{b+a}{2}. This is the same as the (arithmetic) mean of a and b.
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  6. #6
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    Re: midpoint issue

    Quote Originally Posted by anthonye View Post
    Yeah thanks thats what I get. the web gets pi/2 thought I was right.
    Seems to be a sign mistake because: \frac{+\frac \pi6+\frac56 \pi}2=\frac \pi2

    Quote Originally Posted by anthonye View Post
    so in this case pi/3 and 5pi/7 I get 7pi/12.
    How did you get this result?

    \frac{\frac \pi3 + \frac57 \pi}2=\frac{\frac7{21} \pi+\frac{15}{21} \pi}2

    which will never yields your result.
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  7. #7
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    Re: midpoint issue

    Oops sorry wrote it wrong it should be

    pi/3 and 5pi/6 gives 7pi/6 then dived by 2 for midpoint 7pi/12.
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  8. #8
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    Re: midpoint issue

    Quote Originally Posted by anthonye View Post
    Oops sorry wrote it wrong it should be

    pi/3 and 5pi/6 gives 7pi/6 then dived by 2 for midpoint 7pi/12.
    Yes, that's right
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  9. #9
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    Re: midpoint issue

    Thanks.
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  10. #10
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    Re: midpoint issue

    This is all to do with setting up the axis for a sing graph when my midpoint is pi/3.

    this is the first point after the origin should'nt it be pi/6 to match the phase shift which is -pi/6.
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  11. #11
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    Re: midpoint issue

    I am right about the phase shift.

    y=3sin2(x + pi/6).

    Thanks.
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  12. #12
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    Re: midpoint issue

    Quote Originally Posted by anthonye View Post
    Yeah thanks thats what I get. the web gets pi/2 thought I was right.
    \pi/2 is the midpoint of + \pi/6 and 5\pi/6.
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