[MAMDS1993]

v = tan(pi/4-t) I think! I don't remember anymore though to be honest!

[willamgg1993]

That's what I got!! And then to integrate that again I made it sin(pi/4-t)/(cos(pi/4-t), did the compound angle identities to get the top was the differential of the bottom so took log of the bottom and found c for the distance equation. That last bit baffled me though. I wrote loads of **** :L

[MAMDS1993]

yeah you should've gotten lncos(pi/4-t) + c can't remember the c value now...but yeah basically it was what you had to get! Coz integral of tanx is lnsecx and then because of the negative x it becomes -lnsecx which is lncosx!

[willamgg1993]

I hate it when you think back to a paper (STATS ) and realise you had either made a stupid mistake or were like 2 steps from getting the show that

[willamgg1993]

(Original post by MAMDS1993)
The integration last qn was honestly not too bad coz u get s=lncos(pi/4-t) and then for the expression u get s=lncos(pi/4-t) - ln(sqrt(2)/2) which when u simplify becomes s=lncos(pi/4-t)+1/2ln(2) then u had to compare it with v using (cos^2(x)) = 1/(tan^2x+1) = 1/(sqrt(v^2+1)) so s becomes s = ln((v^2+1)^(-1/2) - ln(1/2)^(-1/2) then simplify that you use laws of logs you get the show that thing!

I got s=ln(cost+sint) and just put both our answers in the GDC and it's exactly the same function as s=lncos(pi/4-t)+1/2ln(2)

Freaky **** goes on in the world of maths :L

[MAMDS1993]

Yeah haha OMG! I can see how it would be the same thing though when you simplify the natural log and then you simplify using cos(A+B)=cosAcosB-SinASinB.

[mathz]

anyone have or know where i can find copies of the 2012 higher level papers?

[mathz]

(Original post by gummy_bear10)
they probably won't be released to the IB store until November. Having said that, it's likely your teacher held back a few copies... although I don't think they're meant to give them to you.

hi. im not a student,just wanted to have a look. can find most other papers on the maths exam forum but not the ib papers.