De Moivre's Theorem (1 Viewer)

Pedro123 · Oct 13, 2020

Not sure how to solve the following:
$(1+cos(a)+isin(a))^k+(1+cos(a)-isin(a))^k = (2^k*2)*cos(ka/2)cos^k(a/2)$
I can get it down to proving:
$1 + cos(a) + cos(2a) + .... cos(ka) = 2^k*cos(ka/2)cos^k(a/2)$
But don't know where to go from there. I'm almost certain you need to use:
$2cos(A)cos(B) = cos(A-B) + cos(A+B)$
But I'm not sure how I'm supposed to use that. Doing it with induction also is a bit of a pain in the inductive step then.
For reference, this is Cambridge Maths Year 12 Extension 2 3A Question 17. I found this much harder than the enrichment problems, which leads me to believe I'm fucking this up somehow, but I don't know how.

Thanks

B1andB2 · Oct 13, 2020

Have you tried getting rid of the one by using double angle?

Eg first bracket
(1 + cos^a/2 -sin^a/2 + 2i Sin a/2 cos a/2)^k
And then simplify 1- sin^ a/2 to be cos^ a/2

so you get 2cos^2 (a/2) + ...

idk just a thought

YonOra · Oct 13, 2020

It's times like this where I'm happy i didn't pursue Ext 2 maths

idkkdi · Oct 13, 2020

Pedro123 said:
Not sure how to solve the following:
$(1+cos(a)+isin(a))^k+(1+cos(a)-isin(a))^k = (2^k*2)*cos(ka/2)cos^k(a/2)$
I can get it down to proving:
$1 + cos(a) + cos(2a) + .... cos(ka) = 2^k*cos(ka/2)cos^k(a/2)$
But don't know where to go from there. I'm almost certain you need to use:
$2cos(A)cos(B) = cos(A-B) + cos(A+B)$
But I'm not sure how I'm supposed to use that. Doing it with induction also is a bit of a pain in the inductive step then.
For reference, this is Cambridge Maths Year 12 Extension 2 3A Question 17. I found this much harder than the enrichment problems, which leads me to believe I'm fucking this up somehow, but I don't know how.

Thanks

someone's learning ahead.

mani88 · Oct 13, 2020

Pedro123 said:
Not sure how to solve the following:
$(1+cos(a)+isin(a))^k+(1+cos(a)-isin(a))^k = (2^k*2)*cos(ka/2)cos^k(a/2)$
I can get it down to proving:
$1 + cos(a) + cos(2a) + .... cos(ka) = 2^k*cos(ka/2)cos^k(a/2)$
But don't know where to go from there. I'm almost certain you need to use:
$2cos(A)cos(B) = cos(A-B) + cos(A+B)$
But I'm not sure how I'm supposed to use that. Doing it with induction also is a bit of a pain in the inductive step then.
For reference, this is Cambridge Maths Year 12 Extension 2 3A Question 17. I found this much harder than the enrichment problems, which leads me to believe I'm fucking this up somehow, but I don't know how.

Thanks

Use Double angle formulas to get rid of the 1 in the LHS and then start from there

B1andB2 · Oct 13, 2020

This is what I’ve been doing

I don’t know how to finish off the last line, but most of it is there
Sorry it’s so messy

also just realised I forgot to write brackets to show it’s all to the power of k whoops just imagine I did

Trebla · Oct 13, 2020

Pedro123 said:
Not sure how to solve the following:
$(1+cos(a)+isin(a))^k+(1+cos(a)-isin(a))^k = (2^k*2)*cos(ka/2)cos^k(a/2)$
I can get it down to proving:
$1 + cos(a) + cos(2a) + .... cos(ka) = 2^k*cos(ka/2)cos^k(a/2)$
But don't know where to go from there. I'm almost certain you need to use:
$2cos(A)cos(B) = cos(A-B) + cos(A+B)$
But I'm not sure how I'm supposed to use that. Doing it with induction also is a bit of a pain in the inductive step then.
For reference, this is Cambridge Maths Year 12 Extension 2 3A Question 17. I found this much harder than the enrichment problems, which leads me to believe I'm fucking this up somehow, but I don't know how.

Thanks

As a hint
$1+\cos\theta-i\sin\theta$
$= 2\cos^2\frac{\theta}{2}-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}$
$= 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2} - i\sin\frac{\theta}{2}\right)$
$= 2\cos\frac{\theta}{2}\left(\cos\left(-\frac{\theta}{2}\right) + i\sin\left(-\frac{\theta}{2}\right)\right)$

mani88 · Oct 13, 2020

Pedro123 · Oct 13, 2020

Thanks guys - turns out I just pulled a bit of a dumb

Out of curiosity, does anyone have any ideas to prove the sum of cosine expression I thought I needed?

B1andB2 · Oct 13, 2020

mani88 said:
$LHS&=(1+\cos\alpha+i\sin\alpha)^k+(1+\cos\alpha-i\sin\alpha)^k\\ =\left(2\cos^2\frac{\alpha}{2}+2i\sin\frac{\alpha}{2}\cos\frac{\alpha}{2})^k+(2\cos^2\frac{\alpha}{2}-2i\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\right)^k\\ =2^k\cos^k\frac{\alpha}{2}\left(\cos\frac{k\alpha}{2}+i\sin\frac{k\alpha}{2}\right)+2^k\cos^k\frac{\alpha}{2}\left(\cos\frac{-k\alpha}{2}+i\sin\frac{-k\alpha}{2}\right)\\ =2^{k+1}\cos^k\frac{\alpha}{2}\left(2\cos\frac{k\alpha}{2}\right)\\$

Where did the k+1 come from?

mani88 · Oct 13, 2020

B1andB2 said:
Where did the k+1 come from?

i cant even see what i typed, its just coming as a small square

mani88 · Oct 13, 2020

Yep I see it now
sorry that was a bit incorrect, i just fixed it up

B1andB2 · Oct 13, 2020

mani88 said:
Yep I see it now
sorry that was a bit incorrect, i just fixed it up

How did you get that k+1

mani88 · Oct 13, 2020

The extra 2 comes for $2\cos\frac{k\alpha}{2}$

Trebla · Oct 13, 2020

Pedro123 said:
Out of curiosity, does anyone have any ideas to prove the sum of cosine expression I thought I needed?

Just expand the RHS using compound angles and it works out. You actually don't even need to memorise this because it's in the reference sheet.

mani88 · Oct 13, 2020

B1andB2 said:
How did you get that k+1

This is the full solution

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De Moivre's Theorem (1 Viewer)

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