FurtherMaths2020
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#1
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why is it done this way? I know cos(x) = cos(-x) and -sin(x) = sin(-x)

and that z1/z2 = z1 - z2 = cos(x1 -x2) + isin(x1-x2) = cos(11x) = isin(11x) - the same answer as solution bank.

they've not even done: (z1/z2)(z*2/z*2)
Last edited by FurtherMaths2020; 1 year ago
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RDKGames
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#2
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(Original post by FurtherMaths2020)
z1 - z2 = cos(x1 -x2) + isin(x1-x2)
This is not true. Where is this coming from?

Anyway, there is a typo on the second line. Obviously it should read

\dfrac{(\cos 2x + i\sin 2x)(\cos 9x + i\sin 9x)}{(\cos 9x - i \sin 9x)(\cos 9x + i\sin 9x)}
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FurtherMaths2020
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#3
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(Original post by RDKGames)
This is not true. Where is this coming from?

Anyway, there is a typo on the second line. Obviously it should read

\dfrac{(\cos 2x + i\sin 2x)(\cos 9x + i\sin 9x)}{(\cos 9x - i \sin 9x)(\cos 9x + i\sin 9x)}
the textbook
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FurtherMaths2020
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#4
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(Original post by RDKGames)

\dfrac{(\cos 2x + i\sin 2x)(\cos 9x + i\sin 9x)}{(\cos 9x - i \sin 9x)(\cos 9x + i\sin 9x)}
but if they did it like that, that following work wouldn't be the same. have a look at the answer again.
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mqb2766
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(Original post by FurtherMaths2020)
but if they did it like that, that following work wouldn't be the same. have a look at the answer again.
RDKGames is right. Its a typo. You're multipying by the denom conjugate on the top and bottom. The bottom is unity magnitude (as its a unit vector). The top becomes
cos(11x) + isin(11x)
by addition formulae or more simply by de Moivre (if you've done it).
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FurtherMaths2020
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#6
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(Original post by mqb2766)
RDKGames is right. Its a typo. You're multipying by the denom conjugate on the top and bottom. The bottom is unity magnitude (as its a unit vector). The top becomes
cos(11x) + isin(11x)
by addition formulae or more simply by de Moivre (if you've done it).
look at their answer - they've not multiplied through using the conjugate
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mqb2766
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(Original post by FurtherMaths2020)
look at their answer - they've not multiplied through using the conjugate
They have, its a typo.
Line 3 is a scalar (magnitude - squared) on the denom. They've multiplied by the conjugate.

We are talking about Q17?
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RDKGames
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(Original post by FurtherMaths2020)
look at their answer - they've not multiplied through using the conjugate
Yes they have!

cos(9x) + isin(9x) is the conjugate of cos(9x) - isin(9x)

Image
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NotNotBatman
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#9
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(Original post by FurtherMaths2020)
look at their answer - they've not multiplied through using the conjugate
It's a typo. Just multiply top and bottom by the complex conjugate of the bottom.

expand out and it works. where do you think cos^2+sin^2 has come from?
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RDKGames
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(Original post by FurtherMaths2020)
the textbook
Yes what they show is the result for division, but you said z1 - z2 instead of z1/z2 therefore I said your claim is incorrect.
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FurtherMaths2020
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#11
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(Original post by RDKGames)
Yes they have!

cos(9x) + isin(9x) is the conjugate of cos(9x) - isin(9x)

Image
sorry, I see that they have. anyway, a per the textbook, I used the method they taught.
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FurtherMaths2020
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#12
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(Original post by FurtherMaths2020)
why is it done this way? I know cos(x) = cos(-x) and -sin(x) = sin(-x)

and that z1/z2 = z1 - z2 = cos(x1 -x2) + isin(x1-x2) = cos(11x) = isin(11x) - the same answer as solution bank.

they've not even done: (z1/z2)(z*2/z*2)
(Original post by RDKGames)
Yes what they show is the result for division, but you said z1 - z2 instead of z1/z2 therefore I said your claim is incorrect.
I said z1/z2 = z1 - z2
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FurtherMaths2020
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#13
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(Original post by mqb2766)
They have, its a typo.
Line 3 is a scalar (magnitude - squared) on the denom. They've multiplied by the conjugate.

We are talking about Q17?
ok, I see that. anyway, why haven't they done it this way: z1/z2 = z1 - z2 = cos(x1 -x2) + isin(x1-x2) = cos(11x) = isin(11x)
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wbanner2001
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#14
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These new spec. Edexcel textbooks have hundreds of mistakes in them across the 8 Maths and F Maths textbooks I have used (I do options FS1 and FM1). The teachers hate all the errors as well
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FurtherMaths2020
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#15
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(Original post by NotNotBatman)
It's a typo. Just multiply top and bottom by the complex conjugate of the bottom.

expand out and it works. where do you think cos^2+sin^2 has come from?
ok, I see it now
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RDKGames
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#16
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(Original post by FurtherMaths2020)
I said z1/z2 = z1 - z2
Yeah, hence automatically saying that division is the same as subtraction is incorrect.

Anyway, hopefully you understand how having z1 - z2 is incorrect in that line.
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FurtherMaths2020
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#17
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(Original post by RDKGames)
Yeah, hence automatically saying that division is the same as subtraction is incorrect.

Anyway, hopefully you understand how having z1 - z2 is incorrect in that line.
yeah, I think I do. thanks
Last edited by FurtherMaths2020; 1 year ago
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RDKGames
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#18
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(Original post by FurtherMaths2020)
yeah, I think I do
You think?

Well ok, let me put it this way;

It's like me saying \dfrac{10}{2} = 10 - 2

It's just a simple nonsensical error, nothing more to it. Just be careful in the future with what you write, that's the take-home message.
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FurtherMaths2020
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#19
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(Original post by RDKGames)
You think?

Well ok, let me put it this way;

It's like me saying \dfrac{10}{2} = 10 - 2

It's just a simple nonsensical error, nothing more to it. Just be careful in the future with what you write, that's the take-home message.
sorry, I was confused with arg(z1/z2) = arg(z1) - arg(z2)
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mqb2766
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(Original post by FurtherMaths2020)
sorry, I was confused with arg(z1/z2) = arg(z1) - arg(z2)
Because the question is based on arguments (imaginary exponentials, unity magnitude), this actually "works" on this question.
Last edited by mqb2766; 1 year ago
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