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My book's throwing equations out of nowhere

  • Thread starter VietDao29
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VietDao29
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My book's throwing equations out of nowhere!!!

On proving one theorem, my book says:
[tex]1 + 2\sum_{k = 1} ^ {n} \cos k \alpha = \frac{\sin \left( n + \frac{1}{2} \right) \alpha}{\sin \frac{\alpha}{2}}[/tex]
I have no idea where that formula comes from :confused:, and I tried in vain proving it, but I failed. :cry: Can anybody gives me a hint?
Thank you,
Any help will be appreciated,
Viet Dao,
 
Last edited:

Answers and Replies

Hurkyl
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How did you try? If I remember, it's pretty easy.
 
Tide
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Think de Moivre! :)
 
lurflurf
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[tex]2\sin\left(\frac{a}{2}\right)\cos(k a)=\sin\left(k a+\frac{a}{2}\right) \ - \ \sin\left(k a-\frac{a}{2}\right)[/tex]
so
[tex]\cos(k a)=\frac{\sin\left(k a+\frac{a}{2}\right) \ - \ \sin\left(k a-\frac{a}{2}\right)}{2\sin\left(\frac{a}{2}\right)}[/tex]
 
VietDao29
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Thanks, After hours of struggling, I realised that it could be proved by induction. Not very hard, though. :smile:
And I also tried lurflurf's way. Thanks.
Thanks a lot,
But... I still don't know how to prove it using de Moivre's formula...
[tex](\cos x + i\sin x) ^ n = \cos (nx) + i \sin (nx)[/tex]
How can I do from there?
Viet Dao,
 
Last edited by a moderator:
Hurkyl
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Note that cos(nz) = Re[ cis(z)^n ]

( cis(z) := cos z + i sin z = exp(z) )
 
VietDao29
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Thanks a lot, guys. :smile:
Viet Dao,
 

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