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FP3 complex no. Q

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 :smile:

Reply 1

AlphaNumeric

10

\cos A = \frac{e^{iA}+e^{-iA}}{2}

\sin A = \frac{e^{iA}-e^{-iA}}{2i}

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.

Reply 2

JohnSPals

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 :smile:

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

So 2cosx = e^ix + e^-ix

Similarly 2isinx = e^ix - 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 :tongue:

)

Reply 3

Clearvision

OP

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?

Reply 4

JohnSPals

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 :tongue:

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!

Reply 5

AlphaNumeric

10

\cos^{2}x \sin^{4}x = (\frac{e^{ix}+e^{-ix}}{2})^{2}(\frac{e^{ix}-e^{-ix}}{2i})^{4}

Binomial expansion on that (boring and painful) and then collect together terms as

\frac{e^{iAx}+e^{-iAx}}{2} = \cos Ax

Reply 6

Clearvision

OP

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 :frown:

Reply 7

AlphaNumeric

10

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.

Reply 8

Clearvision

OP

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'

But this didn't seem to give the right answer... i had 1/64cos^6x instead of 1/32

FP3 complex no. Q

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