Scott
Ripley
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| Posted on Thursday, 10
July, 2003 - 09:46 am: |
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I am given that zn+1/zn=2cosnq, for
n a positive integer, and z=cisq. I
am supposed to show that 4cos3q = cos3q+ 3cosq.
In part (a) of this question I was to show that
(z+1/z)3=(z3+1/z3)+(3z+3/z) which I showed
easily. I'm supposed to use the result from (a)
to show that 4cos3q = cos3q+3cos q but I don't know how.
I know that 4cos3q = 2(2cosnq)
so that is in my mind but from there I get stuck.
I try substituting cisq into the large
equation above but I can't find any things to
cancel out to ultimately get just cos.
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Andre
Rzym
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| Posted on Thursday, 10
July, 2003 - 10:10 am: |
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You are almost there. Firstly, you
mean
z=cos(q)+isin(q) and we are to prove 4cos3q = ...
Now focus on each term of your identity:
(z+1/z)3=(z3+1/z3)+(3z+3/z)
Now we know
(z+1/z)=2cos(q)
So
8cos3(q)=(z3+1/z3)+(3z+3/z)
Can you carry on from here?
Andre
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Scott
Ripley
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| Posted on Thursday, 10
July, 2003 - 10:37 am: |
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I understood how you got 8cos3(q)=(z3+ 1/z3)+(3z+3/z) but from there I'm just stuck
again. I tried substituting cisq into
(z3+1/z3)+(3z+3/z) but once again nothing.
I thought that if 8cos3(q)=(z3+1/z3)+ (3z+3/z) then if I divided it by two it would
give me 4cos3(q) which is what I
require, but then I don't know how dividing all the
cos and sin parts by 2 would do any good. Is there
something I'm missing here?
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Andre
Rzym
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| Posted on Thursday, 10
July, 2003 - 11:02 am: |
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Now focus on the (3z+3/z) term.
We can rewrite this as 3(z+1/z). But from your very
first statement, what does z+1/z equal?
This leaves the z3+1/z3 term. For this, note that
z3=cos(3q)+isin(3q). What, therefore,
does 1/z3 equal? What, therefore, does z3+ 1/z3 equal?
Andre
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Scott
Ripley
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| Posted on Thursday, 10
July, 2003 - 12:44 pm: |
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Oh i got it
now, thanks a bunch
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