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<arctanx> look at 3pi/2 on your unit circle
<arctanx> it's straight down, i.e. sin is -1
<slxAlex> mmm
<arctanx> My opinion is that you should forget about this 2kpi thing for a moment and try my way of finding your other solutions
<slxAlex> ok
<slxAlex> soo what's wrong there?
<arctanx> ok you know that anticlockwise rotation means positive angles, and clockwise rotation means negative angles, right?
<slxAlex> yep
<slxAlex> ahhh
<arctanx> your minimum value for your domain is -9pi/4. so trace around a unit circle with your finger clockwise until you reach -9pi/4
<arctanx> now, is -9pi/4 a solution?
<slxAlex> yeeeppp
<slxAlex> yes
<arctanx> good, write that down
<slxAlex> ok
<arctanx> now keep tracing your finger around in an anticlockwise direction until you reach 7pi/4, which is the maximum end of your domain, taking note of any cases of sin angle = -1/sqrt(2) along the way
<slxAlex> I already know that one is an answer
<slxAlex> ahhh
<slxAlex> I see
<arctanx> what angles have you found this time?
<slxAlex> hang on just a tic
<slxAlex> 7(pi)/4, 5(pi)/4, 3(pi)/2, (pi)/2, -(pi)/2, -3(pi)/2, -9(pi)/4
<slxAlex> they are the ones in between 7(pi)/4 and -9(pi)/4
<slxAlex> is that right arctanx?
<arctanx> nope. give me a second to work out what you've done
<slxAlex> fuck
<sexcopter> hi there, I'm having some bother with a problem sheet on general relativity (http://www.maths.bris.ac.uk/~macpd/gen_rel/prob3.pdf). question 2 i just can't seem to get anywhere, and question 4, i'm not sure how you apply a lorentz transform to an equation. any ideas?
<arctanx> slxAlex: please tell me why 3(pi)/2 is a solution
<arctanx> slxAlex: you should get -9pi/4, -3pi/4, -pi/4, 5pi/4, 7pi/4
<slxAlex> ok...
<arctanx> if you look at each of those angles on the unit circle, you will see that they fall on one of two places, both of which will result in a sin of -1/sqrt(2)
<slxAlex> yes
<slxAlex> arctanx: forgive me I'm getting a bit tired and restless now
<slxAlex> I do understand though
<arctanx> so you understand what you did wrong with your answers?
<slxAlex> yes
<arctanx> okay
<arctanx> the next part is easy, you worked out that 2x - pi/4 = -9pi/4, -3pi/4, ...
<arctanx> so now you work it out like any other equation, by adding pi/4 to everything, then dividing by 2
<arctanx> what values of x do you then get?
<slxAlex> arctanx: 0, (pi), -(pi), -(pi)/4, 3(pi)/4
<arctanx> slxAlex: right. does your book agree?
<slxAlex> yes
<slxAlex> now just quickly arctanx
<arctanx> mm?
<slxAlex> just to make sure I've got this
<slxAlex> 2Cos^2(t) = 1
<slxAlex> Cos^2(t) = 1/2
<slxAlex> Cos(t) = 1/sqrt(2)
<slxAlex> t = (pi)/4, 7(pi)/4
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