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#math
<slxAlex> ??
<arctanx> hold on
<arctanx> you should get quite a lot more
<slxAlex> yes
<slxAlex> I have to use k(pi)2 to get them
<arctanx> I've never seen that
<slxAlex> well you just add that
<slxAlex> while K is any integer
<slxAlex> so you add (or take) multiples of 2(pi)
<arctanx> oh right that's what you mean
<arctanx> got you
<slxAlex> until you reach your limits
<arctanx> so what are all your values?
<slxAlex> doing it now :D
<arctanx> hold on a second
<arctanx> actually, never mind.
<slxAlex> arctanx: I end up getting pi, 6(pi)/8
<slxAlex> and then add k(pi)2 to those
<slxAlex> that right?
<arctanx> nope
<arctanx> for such a small domain I would suggest forgetting about 2kpi for the moment
<slxAlex> ok
<slxAlex> have you got any answers for this?
<arctanx> start at -9pi/4 and go up from there. where is the first instance of -1/sqrt(2)?
<arctanx> I haven't worked it out to the end but I know what angles you should be getting
<slxAlex> arctanx: I don't get this
<slxAlex> am I right with 2x - (pi)/4 = 7(pi)/4, 5(pi)/4 ??
<arctanx> yes
<slxAlex> so then can I add (pi)/4 to both sides?
<arctanx> you need to keep going into negative angles
<slxAlex> arctanx: the way I've been taught that's all I should get.
<slxAlex> then I have to use k(pi)2
<arctanx> logical enough
<arctanx> except that this time you've found the two highest possible angles
<arctanx> you need to be subtracting 2*k*pi instead of adding
<slxAlex> well
<slxAlex> that's the same as making K negative
<arctanx> yes it is, if you'd prefer to think about it that way
<slxAlex> and arctanx for some reason the book says I should get 0 as one of my answers, how is that possible?
<arctanx> because your eventual answer will be in the form x=
<arctanx> you're still solving for 2x - pi/4
<slxAlex> ahhhh yes
<arctanx> so ignore your book's answers for now
<slxAlex> of course
<slxAlex> arctanx: how many answers should I get?
<arctanx> 5.
<slxAlex> done
<slxAlex> 7(pi)/4, 5(pi)/4, 3(pi)/2, (pi)/2, -(pi)/2
<arctanx> problem there
<slxAlex> ?
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