# midpoint issue

• Dec 14th 2011, 05:58 AM
anthonye
midpoint issue
Hi;
Can someone just tell me the midpoint of

-pi/6 and 5pi/6

Thanks.
• Dec 14th 2011, 06:02 AM
e^(i*pi)
Re: midpoint issue
Quote:

Originally Posted by anthonye
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 $\displaystyle \frac{\pi}{3}$
• Dec 14th 2011, 06:04 AM
anthonye
Re: midpoint issue
Yeah thanks thats what I get. the web gets pi/2 thought I was right.
• Dec 14th 2011, 06:07 AM
anthonye
Re: midpoint issue
so in this case pi/3 and 5pi/7 I get 7pi/12.
• Dec 14th 2011, 06:20 AM
e^(i*pi)
Re: midpoint issue
Quote:

Originally Posted by anthonye
so in this case pi/3 and 5pi/7 I get 7pi/12.

I get $\displaystyle \dfrac{11\pi}{21}$. I'm not sure how you got to twelfths from twenty-firsts (21 being the LCD of 3 and 7)

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

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

With a little bit of manipulation: $\displaystyle b+a = 2c$ and $\displaystyle c = \dfrac{b+a}{2}$. This is the same as the (arithmetic) mean of a and b.
• Dec 14th 2011, 06:25 AM
earboth
Re: midpoint issue
Quote:

Originally Posted by anthonye
Yeah thanks thats what I get. the web gets pi/2 thought I was right.

Seems to be a sign mistake because: $\displaystyle \frac{+\frac \pi6+\frac56 \pi}2=\frac \pi2$

Quote:

Originally Posted by anthonye
so in this case pi/3 and 5pi/7 I get 7pi/12.

(Sadsmile)How did you get this result?

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

which will never yields your result.
• Dec 14th 2011, 06:29 AM
anthonye
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.
• Dec 14th 2011, 06:32 AM
e^(i*pi)
Re: midpoint issue
Quote:

Originally Posted by anthonye
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
• Dec 14th 2011, 06:35 AM
anthonye
Re: midpoint issue
Thanks.
• Dec 14th 2011, 08:38 AM
anthonye
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.
• Dec 17th 2011, 08:20 AM
anthonye
Re: midpoint issue
I am right about the phase shift.

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

Thanks.
• Dec 21st 2011, 03:32 AM
HallsofIvy
Re: midpoint issue
Quote:

Originally Posted by anthonye
Yeah thanks thats what I get. the web gets pi/2 thought I was right.

$\displaystyle \pi/2$ is the midpoint of +$\displaystyle \pi/6$ and $\displaystyle 5\pi/6$.