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-- hw help :D

hw help :D

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...

Re: hw help :D

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

42

Re: hw help :D

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

my ideal partner is female, mixes records, and a geek.

Re: Re: hw help :D

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

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...

Re: Re: Re: hw help :D

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*

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