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Maths C3 Trig help please

Ok heres the question paper:

http://www.scribd.com/doc/16015592/Edexcel-GCE-core-3-mathematics-C3-666501-advanced-subsidiary-jan-2006-question-paper

Heres the markscheme;

http://www.wimbledoncollege.org.uk/TheKnowledge/Maths/Y13Maths/13PastPapers/C3%20Jan%2006%20MS.pdf

I just don't understand 7b whatsoever, can 't get my head around it
any help would be very much appreciated!
Original post by guitarmike456
Ok heres the question paper:

http://www.scribd.com/doc/16015592/Edexcel-GCE-core-3-mathematics-C3-666501-advanced-subsidiary-jan-2006-question-paper

Heres the markscheme;

http://www.wimbledoncollege.org.uk/TheKnowledge/Maths/Y13Maths/13PastPapers/C3%20Jan%2006%20MS.pdf

I just don't understand 7b whatsoever, can 't get my head around it
any help would be very much appreciated!

There isn't too much to understand. It just wants you to rearrange the first expression to get sin2x=cos2x USING the results from part (a). Post some working and we'll tell you if you make a mistake.
(edited 13 years ago)
Reply 2
Original post by Farhan.Hanif93
There isn't too much to understand. It just wants you to rearrange the first expression to get sin2x=cos2x USING the results from part (a). Post some working and we'll tell you if you make a mistake.


I really don't follow, how can you get sin2x=cos2x from part A
Original post by guitarmike456
I really don't follow, how can you get sin2x=cos2x from part A

We're trying to go from:
(1) cosx(cos2xsinx+cosx)=12\cos x \left(\dfrac{\cos 2x}{\sin x + \cos x}\right) = \dfrac{1}{2}

to

(2) cos2x=sin2x\cos 2x = \sin 2x

From part (a), you know that cos2xcosx+sinxcosxsinx\frac{\cos 2x}{\cos x + \sin x} \equiv \cos x - \sin x and this term occurs in what we are given. So replace the fraction in (1) with cosx - sinx for a start to get:

cosx(cosxsinx)=12\cos x (\cos x - \sin x) = \frac{1}{2}.

Now expand out the LHS. Does it look familiar? (take a look at your answer to part (b))
(edited 13 years ago)
Reply 4
AH i have it
i feel foolish thanks!

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