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# FP3 complex no. Q watch

1. From the OCR paper in Jan 06...

*I substitued x for theta to make it easier to type

By expressing cosx and sinx in terms of e^ix and e^-ix, or otherwise, show that

cos^2x sin^4x = 1/32(cos6x - 2cos4x - cos2x + 2)

I have no clue how to do it going about using the e^ix form so i figured i'd make the bit on the LHS in terms of cos using the i dentity cos^2x + sin^2x = 1 Then work out each term using (2cosx)^n = (z+1/z)^n But itdoesn't seem i have got anywhere, expanding cos^6x i got a 1/64 cos6x term which doesn't fit with the given answer...

Any help appreciated

2. Put that into the left hand side, with A = 2x and A = 4x. Put the expansions into the RHS too, and then mess about with the LHS till it equals the right.
3. (Original post by Clearvision)
From the OCR paper in Jan 06...

*I substitued x for theta to make it easier to type

By expressing cosx and sinx in terms of e^ix and e^-ix, or otherwise, show that

cos^2x sin^4x = 1/32(cos6x - 2cos4x - cos2x + 2)

I have no clue how to do it going about using the e^ix form so i figured i'd make the bit on the LHS in terms of cos using the i dentity cos^2x + sin^2x = 1 Then work out each term using (2cosx)^n = (z+1/z)^n But itdoesn't seem i have got anywhere, expanding cos^6x i got a 1/64 cos6x term which doesn't fit with the given answer...

Any help appreciated
eix = cosx + isinx
e-ix = cos(-x) + isin(-x) = cosx - isinx
[Since cos and sin are even and odd functions respectively]

So 2cosx = eix + e-ix

Similarly 2isinx = eix - e-ix

I'll leave you to have anther go suing that, get back to me if you can't get the answer from there (i.e. I can't be bothered to do any more for now )
4. Urgh thanks, but it's not going too well.

Could you complete please if it's not too much hassle. Or even better if anyone has the markscheme could i have that?
5. (Original post by Clearvision)
Urgh thanks, but it's not going too well.

Could you complete please if it's not too much hassle. Or even better if anyone has the markscheme could i have that?
Sure

2cosx = eix + e-ix
Similarly 2isinx = eix - e-ix

We want to show that cos^2x sin^4x = 1/32(cos6x - 2cos4x - cos2x + 2)...

---

(2cosx)^2 = 16(cos^2x) so (cosx)^2 = 1/16[eix + e-ix]^2

Similarly, since i^4 = 1,

(2isinx)^4 = 16(sinx)^4 so (sinx)^4 = 1/16[eix - e-ix]^4

Multiply the two bits in bold together... Ugh That loks nasty, but I think that's how you get it (after a lot of messing about)

(cosx)^2(sinx)^4 = 1/(16^2) [ [e2ix - 1)^2 . (eix - e-ix)^2]
= 1/(16^2) [ [e4ix - 2e2ix + 1] [e2ix - 2 + e-2ix]]
= ... This is the stupidest question in teh world and whoever set it should hang himself

I'm sure I remember this ebing easier at A-level than I'm making it out to be, so maybe I've done something wrong... DOn't think so though!

6. Binomial expansion on that (boring and painful) and then collect together terms as
7. That's pretty much what i'm doing but GAH it's so frustrating.

This was on OCR's Jan 06 FP3 paper :-/

I Have the exam tomorrow
8. It does seem an unusually pointless question, even for A Level, because it involves such a big expansion. The odd thing is that A Level doesn't expect you to think up an "or otherwise" method very often (a nifty trick is more what a STEP question would expect) but following their method gives a 10 term expansion.
9. It has a part ii) where you have to integrate it (0 - pi/3) but the first part is worth 6 marks and it seems to take much longer than 6 mins.

My 'otherwise' method was 'make the bit on the LHS in terms of cos using the i dentity cos^2x + sin^2x = 1 Then work out each term using (2cosx)^n = (z+1/z)^n'

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