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#math
<slxAlex> yep
<slxAlex> -pi < x < pi (> are greater than or = to signs)
<slxAlex> so what answers will I get for sin^-1(-1/sqrt(2))
<arctanx> you mean -pi <= x <= pi ?
<slxAlex> yes
<slxAlex> sorry, didn't think of writing it like that
<arctanx> what you should do is first get it to the form sin(2x - pi/4) = - 1/sqrt(2), which you've done
<slxAlex> yep
<arctanx> now you know that -pi <= x <= pi
<slxAlex> mmhmmm
<slxAlex> so it is valid in all quadrants (one turn)?
<arctanx> if you had an equation sin x = 1/sqrt(2), where -pi <= x <= pi, could you solve that?
<slxAlex> sin(x) = pi/4
<arctanx> right
<slxAlex> and ...
<arctanx> just checking. now you don't have sin x, you have sin (2x - pi/4)
<slxAlex> ohh....
<arctanx> so what you do is you work out your domain for 2x - pi/4
<slxAlex> so there is only going to be one answer atm?
<arctanx> I suspect there will be more, but wait and see on that one
<slxAlex> arctanx: so I don't do the ambiguous case yet?
<arctanx> take the inequality: pi <= x <= pi. What will be the domain for 2x - pi/4?
<arctanx> I don't know what you mean by ambiguous case
<slxAlex> 2pi - pi/4
<slxAlex> to find the other angle it could be?
<arctanx> no, you're thinking about this wrong
<slxAlex> ohhh
<slxAlex> is the domain just that? -pi <= x <= pi
<arctanx> yes
<slxAlex> ok
<arctanx> now you've been given a domain for x, but to solve the sin part of your problem, you need to know the domain for 2x - pi/4 instead
<slxAlex> ohhh
<slxAlex> -2pi - pi/4 <= 2x - pi/4 <= 2pi - pi/4
<slxAlex> arctanx: that right?
<arctanx> that's right
<slxAlex> so what next?
<arctanx> you can simplify those outer ones to -9pi/4 and 7pi/4 if it makes it easier counting quadrants
<arctanx> well on the right hand side of your equation you have -1/sqrt(2). so you look at your unit circle and and see where you get -1/sqrt(2) between -9pi/4 and 7pi/4
<slxAlex> here's my problem
<slxAlex> it ain't on the unit circle diag
<arctanx> it might have a rationalised denominator
<tomfitzyuk> If lim_{x -> inf} f'(x) = 0, show lim_{x -> inf) f(x) exists. I've wrote the definitions out but can't seem to figure out how to show it.
<arctanx> i.e. top and bottom multiplied by sqrt(2), giving you sqrt(2)/2
<tomfitzyuk> lim_{x->inf} lim_{h->0} f(x
<slxAlex> arctanx: that is sooooo annoying
<tomfitzyuk> lim_{x->inf} lim_{h->0} (f(x+h) - f(x)) / h = 0
<slxAlex> ok
<arctanx> slxAlex: so what angles within the domain you worked out for 2x - pi/4 give you -1/sqrt(2)?
<slxAlex> 7pi/4 and 5pi/4
<slxAlex> and now I use k(pi)2
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