AEA QuestionWatch

Announcements
This discussion is closed.
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.
0
14 years ago
#2
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
0
14 years ago
#3
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)
0
Thread starter 14 years ago
#4
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.
0
14 years ago
#5
(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. 0
14 years ago
#6
Which pure units do you need for AEA maths, I thought it would only be up to P3?
0
14 years ago
#7
(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.
0
14 years ago
#8
Ja, cheers dude
0
X
new posts Back
to top
Latest
My Feed

Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

University open days

• University of Bristol
Wed, 23 Oct '19
• University of Exeter
Undergraduate Open Day - Penryn Campus Undergraduate
Wed, 23 Oct '19
• University of Nottingham
Mini Open Day Undergraduate
Wed, 23 Oct '19

Poll

Join the discussion

Have you made up your mind on your five uni choices?

Yes I know where I'm applying (135)
62.79%
No I haven't decided yet (46)
21.4%
Yes but I might change my mind (34)
15.81%