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I've seen many trigonometric identities but here is one that I encountered for which I did not find a reference.

In case you wonder where this came from, I was investigating certain $q$-series in this paper on Eulerian polynomials.

Question. Is this true? Can you provide a reference or a proof? $$\sum_{k=-n}^n\frac{2k+1}{n+k+1}\binom{2n}{n-k}\sec((2k+1)\theta)=\sec((2n+1)\theta).$$

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The sum from k=-n to k=n-1 equals zero by pairing off the end terms and by using the sign difference together with the evenness of the function sec.

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  • $\begingroup$ Ahh, too easy. Thanks all the same. $\endgroup$
    – T. Amdeberhan
    Commented Jun 3, 2017 at 17:31

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