My book's throwing equations out of nowhere

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The discussion centers on a challenging equation from a book, specifically the formula relating to cosine sums and sine functions. The original poster, Viet Dao, struggles to understand and prove the equation, prompting responses from others who suggest using de Moivre's theorem and induction as possible methods for proof. One participant clarifies the relationship between cosine and sine using complex exponentials, which aids in understanding the proof process. Viet Dao expresses gratitude for the hints but still seeks clarity on applying de Moivre's theorem effectively. The conversation highlights collaborative problem-solving in mathematical proofs.
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My book's throwing equations out of nowhere!

On proving one theorem, my book says:
1 + 2\sum_{k = 1} ^ {n} \cos k \alpha = \frac{\sin \left( n + \frac{1}{2} \right) \alpha}{\sin \frac{\alpha}{2}}
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:
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How did you try? If I remember, it's pretty easy.
 
Think de Moivre! :)
 
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)
so
\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)}
 
Thanks, After hours of struggling, I realized 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...
(\cos x + i\sin x) ^ n = \cos (nx) + i \sin (nx)
How can I do from there?
Viet Dao,
 
Last edited by a moderator:
Note that cos(nz) = Re[ cis(z)^n ]

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