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I got this equation in a research paper but I couldn't understand how it was calculated:

$$\sum_{n=0}^{N-1}\text{cos}\left[ \frac{(\pi k/2)\left( 2n+1 \right)}{2N} \right]=\frac 12\frac{\text{sin}\left( \pi k/2 \right)}{\text{sin} \left( \frac{\pi k/2}{2N} \right)}$$

Could you please help me proving if that equality is correct or no?

Sajjad
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  • On the left-hand side, $k$ is an indexing variable for the sum. On the right-hand side, $k$ is a physical variable in the expression. This makes me think there is something wrong with the equation..? – Frank W Aug 29 '22 at 06:19
  • @FrankW Sorry, I edited it. – Sajjad Aug 29 '22 at 06:22
  • You will make it easier to understand and prove if you write $\theta$ for $\frac{\pi k}{4N}$. – ancient mathematician Aug 29 '22 at 06:41
  • @ancientmathematician I am sorry, it is $\pi k/2N$, I agree with you about that simplification, but will that equation becomes correct? – Sajjad Aug 29 '22 at 06:43
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    See "How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?". – Blue Aug 29 '22 at 06:44
  • @Blue it seems different about mine above results ! – Sajjad Aug 29 '22 at 06:50
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    @Sajjad: The results are equivalent. To match the other question better, substitute $n\to k$, $N\to n$, $k\to p$, and rewrite a little; this is what you want to show: $$\sum_{k=0}^{n-1}\cos\left(\frac{p\pi}{4n}+\frac{p\pi}{2n}k \right)=\frac{\sin\dfrac{p\pi}2}{2\sin\dfrac{p\pi}{4n}}$$ Taking $d=\dfrac{p\pi}{2n}$ and $a=\dfrac{p\pi}{4n}=\dfrac{d}2$, the other question says that the result should be $$\frac{\sin(nd/2)}{\sin(d/2)}\cos(a+(n-1)d/2)=\frac{\sin(na)}{\sin a}\cos(na)=\frac{2\sin(na)\cos(na)}{2\sin a}=\frac{\sin2na}{2\sin a}$$ Substituting $a$ gives the desired result here. – Blue Aug 29 '22 at 07:17
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    @Blue thank you for your explanation, please put it as an answer to close the question. – Sajjad Aug 29 '22 at 11:43

1 Answers1

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This is a special case of a formula given in the question "How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?" and proven in the answers.

$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \left( \frac{ 2 a + (n-1)\cdot d}{2}\right) \tag{$\star$}$$

To match that notation, substitute $n\to k$, $N\to n$, $k\to p$ into the current relation, and rewrite a little. This is what the question is trying to show: $$\sum_{k=0}^{n-1}\cos\left(\frac{p\pi}{4n}+\frac{p\pi}{2n}k \right)=\frac{\sin\dfrac{p\pi}2}{2\sin\dfrac{p\pi}{4n}}$$ Taking $d=\dfrac{p\pi}{2n}$ and $a=\dfrac{p\pi}{4n}=\dfrac{d}2$, relation $(\star)$ says that the sum should be $$\frac{\sin(nd/2)}{\sin(d/2)}\cos(a+(n-1)d/2)=\frac{\sin(na)}{\sin a}\cos(na)=\frac{2\sin(na)\cos(na)}{2\sin a}=\frac{\sin2na}{2\sin a}$$ Substituting $a$ gives the desired result. $\square$

Blue
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