[Clarity Incognito]

(Original post by Pheylan)
post the CF please

Good spot pheylan hahahaha, I thought e^(-t) was part of it but it wasn't, it was just the e^(-t) was factorised !!!!!! Thanks guys!! oh can you answer the cos2x and sin2x thing btw?

[Pheylan]

[little_miss_sunshine_xx]

(Original post by refref)
Paper A first question

Are those arguements right? Book says the argument should always be -pi<x<pi but one of them is 17pi/10 and another 13pi/10

I noticed that too! Must be a printing error.

The nerves are beginning to get to me now!!!!! :O

[Pheylan]

(Original post by Clarity Incognito)
oh can you answer the cos2x and sin2x thing btw?

yeah you do, in general if the RHS is sin(ax) or cos(ax), the PI will be of the form psin(ax) + qcos(ax)

[Clarity Incognito]

(Original post by Pheylan)
yeah you do, in general if the RHS is sin(ax) or cos(ax), the PI will be of the form psin(ax) + qcos(ax)

Ok thanks, I haven't done any past papers before today so I haven't seen any in that form, have you, or have you not looked at the older past papers? Also, I saw ages ago in a thread that it might be sin2x + cos2x + sinx + cosx?? Sorry so many questions. I'll rep when I can.

(Original post by henil3110)
...

I'm in but I can't just yet.

[refref]

Also in the exam style paper in the back of the book, they do the same thing in 5b!

WHAT THE ****?

[Pheylan]

(Original post by Clarity Incognito)
have you not looked at the older past papers?

nope, but i've done all the exam questions from the textbook

(Original post by Clarity Incognito)
I saw ages ago in a thread that it might be sin2x + cos2x + sinx + cosx??

i don't think so

[Fiasco]

(Original post by Clarity Incognito)
To what term and can it be left in fractional form??

Probably something like the cubic term and in fractional form, though this is not a good one to differentiate. Best way would be to do what neomilan said below.

(Original post by neomilan)
that would just be expanding e^x but replacing the x in the expansion with x^2 right? and the multiplying by 1/sinx expansion?

Yeah, that's the easiest way.

[henil3110]

(Original post by refref)
Also in the exam style paper in the back of the book, they do the same thing in 5b!

WHAT THE ****?

I know right! I think just put the values between -Pi and Pi...I did the June 2009 paper, and they've done it correctly there.

[Clarity Incognito]

(Original post by Fiasco)
Probably something like the cubic term and in fractional form, though this is not a good one to differentiate. Best way would be to do what neomilan said below.

That's what I did, I could have given you the answer immediately to the eighth power but it was in fractional form, which is non standard for a series questions so I thought there might be a sneaky trick

[henil3110]

(Original post by Clarity Incognito)
Ok thanks, I haven't done any past papers before today so I haven't seen any in that form, have you, or have you not looked at the older past papers? Also, I saw ages ago in a thread that it might be sin2x + cos2x + sinx + cosx?? Sorry so many questions. I'll rep when I can.

I'm in but I can't just yet.

Ok! There's always tomorrow!

[rbnphlp]

as kosy 91 asked can anyone see how to show z^n+z^-n=2costheta using z=re^{i\theta}

[dan9]

i dont think im under as much pressure as you guys thankfully i dont have to take this exam that serious as further maths is not apart of my uni offer, be nice to try and get an A* though

[Fiasco]

(Original post by henil3110)
I know right! I think just put the values between -Pi and Pi...I did the June 2009 paper, and they've done it correctly there.

All the values are correct, the graph repeats. If you were to draw the points on an argand diagram you would see that it is all the same ****. However, usually if a range is given (which it sometimes isn't) you should answer in that range.

I'm sure it was an error in the answers, or originally the question might not have included a range.

[Clarity Incognito]

(Original post by rbnphlp)
as kosy 91 asked can anyone see how to show z^n+z^-n=2costheta using z=re^{i\theta}

Are you sure you don't mean 2cosntheta??

[dan9]

(Original post by rbnphlp)
as kosy 91 asked can anyone see how to show z^n+z^-n=2costheta using z=re^{i\theta}

yh you just rewrite it as (cos nO + i sin nO) + (cos nO - i sin nO)

[rbnphlp]

(Original post by Clarity Incognito)
Are you sure you don't mean 2cosntheta??

yh soorry thats it..I know cosnx=e{inx}+e^{-inx}/2 , but what about the modulous r^n and r^-n?

[henil3110]

(Original post by rbnphlp)
as kosy 91 asked can anyone see how to show z^n+z^-n=2costheta using z=re^{i\theta}

The Sample paper has that question. The marking scheme says:

[Fiasco]

If any one wants a hard differential equation question:



STEP I Q6 2010

[Clarity Incognito]

(Original post by rbnphlp)
yh soorry thats it..I know cosnx=e{inx}+e^{-inx}/2 , but what about the modulous r^n and r^-n?

You're going to have to eventually convert into cosines and sines (which cancel) like they've done the above post, I'm not sure what alternative you were thinking of, perhaps you could show me with an example? Seeing something new is always good the night before.