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hw help :D
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| 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... |
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| Boomer187 |
| 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. |
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| torontotrance |
| pics or stfu |
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| Aristronica |
| 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 |
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| DigitalMP |
| my ideal partner is female, mixes records, and a geek. |
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| PhaseFour |
| 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 :) |
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| 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 |
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| Aristronica |
| 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... |
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| Aristronica |
| 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* :eyes: |
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