hey all, knowing that this forum is used for hw a lot, ill join the trend

how do you prove |sinx-siny|<|x-y| using the mean value theorem? (this is taken from option 12 of the ib math hl syllabus.)

thanks alot! sexual favors to whoever can get it first, i guess...

quote:

Originally posted by PhaseFour thanks alot! sexual favors to whoever can get it first, i guess...

Half now, the other half after I give you the answer.

pics or stfu

I think its -3

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quote:

Originally posted by PhaseFour hey all, knowing that this forum is used for hw a lot, ill join the trend how do you prove |sinx-siny|<|x-y| using the mean value theorem? (this is taken from option 12 of the ib math hl syllabus.) thanks alot! sexual favors to whoever can get it first, i guess...

Since I don't want to write out that whole thing, here's something simpler.

|sin a| < |a|
|sin a| < |sin a - sin 0|

sin a - sin 0
______________ = f' (c) = cos c
a - 0

|sin a| = |sin a - sin 0| = |cos c| |a - 0| = |cos c| |a|

|sin a| = |cos c| |a| < (1) |a| = |a|

now just expand that to what you have, it'll be longer but basically same concept, if you don't get it you shouldn't be taking calculus

Nice thread

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quote:

Originally posted by Aristronica Since I don't want to write out that whole thing, here's something simpler. |sin a| < |a| |sin a| < |sin a - sin 0| sin a - sin 0 ______________ = f' (c) = cos c a - 0 |sin a| = |sin a - sin 0| = |cos c| |a - 0| = |cos c| |a| |sin a| = |cos c| |a| < (1) |a| = |a| now just expand that to what you have, it'll be longer but basically same concept, if you don't get it you shouldn't be taking calculus

thanks alot! i get it, but they didnt ask it on the exam (which was buttrape all the way)

speaking of buttrape, i now owe u one favor

this could be done much more easily using the formula:

sin x - sin y = 2 cos[(x+y)/2]sin[(x-y)/2]

then

|sin x - sin y| = 2|cos[(x+y)/2]||sin[(x-y)/2]|

since |cos z| ≤ 1 ∀z∈R and |sin u| ≤ |u| ∀u∈R

we immediately get

|sin x - sin y| ≤ 2|(x-y)/2| = |x-y|

equality is achieved when x=y

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Last edited by Noisician on May-09-2004 at 02:16

quote:

Originally posted by Noisician this could be done much more easily using the formula: sin x - sin y = 2 cos[(x+y)/2]sin[(x-y)/2] then |sin x - sin y| = 2|cos[(x+y)/2]||sin[(x-y)/2]| since |cos z| ≤ 1 ∀z∈R and |sin u| ≤ |u| ∀u∈R we immediately get |sin x - sin y| ≤ 2|(x-y)/2| = |x-y| equality is achieved when x=y

dude...what??? wow...

quote:

Originally posted by PhaseFour thanks alot! i get it, but they didnt ask it on the exam (which was buttrape all the way) speaking of buttrape, i now owe u one favor

Ok!!! *bends over and drops pants*