Hi;

Can someone just tell me the midpoint of

-pi/6 and 5pi/6

Thanks.

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- December 14th 2011, 05:58 AManthonyemidpoint issue
Hi;

Can someone just tell me the midpoint of

-pi/6 and 5pi/6

Thanks. - December 14th 2011, 06:02 AMe^(i*pi)Re: midpoint issue
- December 14th 2011, 06:04 AManthonyeRe: midpoint issue
Yeah thanks thats what I get. the web gets pi/2 thought I was right.

- December 14th 2011, 06:07 AManthonyeRe: midpoint issue
so in this case pi/3 and 5pi/7 I get 7pi/12.

- December 14th 2011, 06:20 AMe^(i*pi)Re: midpoint issue
I get . I'm not sure how you got to twelfths from twenty-firsts (21 being the LCD of 3 and 7)

__Spoiler__:

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

With a little bit of manipulation: and . This is the same as the (arithmetic) mean of a and b. - December 14th 2011, 06:25 AMearbothRe: midpoint issue
- December 14th 2011, 06:29 AManthonyeRe: 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. - December 14th 2011, 06:32 AMe^(i*pi)Re: midpoint issue
- December 14th 2011, 06:35 AManthonyeRe: midpoint issue
Thanks.

- December 14th 2011, 08:38 AManthonyeRe: 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. - December 17th 2011, 08:20 AManthonyeRe: midpoint issue
I am right about the phase shift.

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

Thanks. - December 21st 2011, 03:32 AMHallsofIvyRe: midpoint issue