Identities for combinatorial sums involving trigonometric functions

preprint2022arXivOpen access

Abstract

Let Am,n(a)=∑j=0m(−4)j(m+j2j)∑k=0n−1sin⁡(a+2kπ/n)cos⁡2j(a+2kπ/n) A_{m,n}(a)=\sum_{j=0}^m (-4)^j {m+j\choose 2j}\sum_{k=0}^{n-1} \sin(a+2kπ/n) \cos^{2j}(a+2kπ/n) and Bm,n(a)=∑j=0m(−4)j(m+j+12j+1)∑k=0n−1sin⁡(a+2kπ/n)cos⁡2j+1(a+2kπ/n), B_{m,n}(a)=\sum_{j=0}^m (-4)^j {m+j+1\choose 2j+1}\sum_{k=0}^{n-1} \sin(a+2kπ/n) \cos^{2j+1}(a+2kπ/n), where m≥0m\geq 0 and n≥1n\geq 1 are integers and aa is a real number. We present two proofs for the following results: (i) If $2m+1 \equiv 0 \, (\mbox{mod} \, n)$, then Am,n(a)=(−1)mnsin⁡((2m+1)a). A_{m,n}(a)=(-1)^m n \sin((2m+1)a). (ii) If $2m+1 \not\equiv 0 \, (\mbox{mod} \, n)$, then Am,n(a)=0A_{m,n}(a)=0. (iii) If $2(m+1) \equiv 0 \, (\mbox{mod} \, n)$, then Bm,n(a)=(−1)mn2sin⁡(2(m+1)a). B_{m,n}(a)=(-1)^m \frac{n}{2} \sin(2(m+1)a). (iv) If $2(m+1) \not\equiv 0 \, (\mbox{mod} \, n)$, then Bm,n(a)=0B_{m,n}(a)=0.

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