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Gaz031
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#1
Report Thread starter 14 years ago
#1
I'll include the whole question, as this might be meant to give ideas for the part I'm having trouble with.

(a)
Simplify:
(i)sin7xcosx + sinxcos7x
(ii)sin7xcosx - sinxcos7x.
I get sin8x and sin6x respectively.

(b) Find expressions in terms of r, for P and Q such that:
2sinxcos(2r-1)x = sinPx - sinQx.
I used the fact that subtracting (ii) from (i) gives 2cos7xsinx = sin(7x+x) - sin(7x-x)
So: 2cos(2rx-x)sinx = sin(x+2rx-x) - sin(2rx - x - x) = sin(2rx) - sin x(2r-2)
Hence P = 2r, Q=2(r-1)

(c) Here's the part i'm having difficulty with:
Prove for positive integers n,
sin2nx = 2sinx[sum to n from r=1][cos[x(2r-1)]

At first i tried using part b. and letting r=n. Giving:
2sinxcos(2n-1)x = sin2nx - sinx(2n-2)
2sinxcos(2n-1)x + sin(2n-2)x = sin2nx
sin2nx = sinxcos(2n-1)x + sin2x(n-1)
Which seems to give me a redundant term.
Thanks for any help.
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dvs
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b) 2sinxcos(2r-1)x = sin(2r)x - sin[2(r-1)]x

c)
Use b):
r=1 => 2sinxcosx = sin2x - 0
r=2 => 2sinxcos3x = sin4x - sin2x [Note that they're the same but with opposite signs]
r=3 => 2sinxcos5x = sin6x - sin4x [Same applies to sin4x here]
...
r=n => 2sinxcos(2n-1)x = sin(2nx) - sin[2(n-1)]x
Add these up to get:
2sinxcosx+2sinxcos3x+2sinxcos5x+ ...+2sinxcos(2n-1)x = sin(2nx) [All the other terms cancel out!]
2sinx[cosx+cos3x+cos5x+...+cos(2n-1)x] = sin(2nx)
QED
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JamesF
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For part c, notice that
[sum to n from r=1] {2sinxcos(2r-1)x} = [sum to n from r=1] {sinPx - sinQx}
Of which all but the first SinQx and last SinPx terms cancel out. Expand the first few terms and the last few if you want to be more clear. Anyway, so you are left with
[sum to n from r=1] {2sinxcos(2r-1)x} = -sin(0) + sin(2nx) = sin(2nx)
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Gaz031
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Thanks very much for the help both of you. It's clear now.
So basically the summing was similar to that used in P4 to find sums of a function with identity, where the terms cancel ?

I haven't focused on pure as I had only applied exams this January. I'll try to get my P1-P4 knowledge back up to speed with AEA papers before starting P5/P6.
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dvs
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(Original post by Gaz031)
So basically the summing was similar to that used in P4 to find sums of a function with identity, where the terms cancel ?
Yup.

Good luck.
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~Raphael~
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Which pure units do you need for AEA maths, I thought it would only be up to P3?
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JamesF
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(Original post by ~Raphael~)
Which pure units do you need for AEA maths, I thought it would only be up to P3?
Yea, pretty much. That last part didnt actually need any special knowledge, just expand a few terms and see almost everything cancels.
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~Raphael~
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#8
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Ja, cheers dude
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