C3 January 2013 25/01/2013

It gave me the correct solutions but some are saying that the method is not right

hmm im not sure, I think their method is wrong because
1 +cotx doesnt = cosecx
because the identity is
1 +cot^2x =cosec^2x
its when cot and cosec are squared


isnt it?

yes, $1+cot^2x = cosec^2x$
but, we are saying that $1+cotx = cosecx$ (without the squared)

yes, $1+cot^2x = cosec^2x$
but, we are saying that $1+cotx = cosecx$ (without the squared)

did u look at the question

hmm im not sure, I think their method is wrong because
1 +cotx doesnt = cosecx
because the identity is
1 +cot^2x =cosec^2x
its when cot and cosec are squared


isnt it?

I think your solutions are wrong to part c, what did you get for b?

the identity is right? havee you solved that question yet the trig, I am sure I did it correct because if it was wrong I would have not got the correct solutions

hmmm i found the 2003 and 2005 ones easy, I cant reember jan 2012 P

looking at the specimen it looks pretty OK with some of the end questions alittle more challenging (as expected)

I think your solutions are wrong to part c, what did you get for b?

I'm confused by your comment about the 2003 paper. C3 was first examined in May/June 2005.

you are right I did make a mistake on the solution but then like how would you do it?

Yeah, I'm just agreeing :P

for b i proved 2sin(1/2 θ)[cos(1/2θ) + sin(1/2 θ)]

forc, equate 2sin(1/2 θ)[cos(1/2θ) + sin(1/2 θ)] to 0
therefore, that will get you two solutions
2sin(1/2 θ) = 0, cos(1/2θ) + sin(1/2 θ) = 0
θ = 0 because sin(0) = 0, sorry i'm not sure how to get the second solution.

what grade boundary would u give for the specimen paper?

I go 93 in this. All you have to do is understand the material and being fiddly with the identities.

can anyone explain a technique :- with cos2x there are 3 types of ways/formulas when we prove an equation how do we know which one to chose?

Well if the equation is in terms of cos^2x

the you would use cos2x = 2cos^2(x) -1

if in terms of sin^2x then cos2x = 1 - 2sin^2(x)

oh, sorry.
how would you solve cos(1/2θ) + sin(1/2 θ) = 0? for 0<θ<2π