Showing $ n=\sum_{k=1}^{(n+1)/2}\sin{\frac{2\pi k}{n+2}}\sin{\frac{\pi(n-2(k-1))}{n+2}}\sec^2{\frac{\pi(n-2(k-1))}{2n+4}}$ for natural $n$

Question

Good afternoon,

I am a little confused and intrigued by this finite summation formula I came up with.

If $n$ is a natural number then $$ n=\sum_{k=1}^{(n+1)/2}\sin{\frac{2\pi k}{n+2}}\sin{\frac{\pi(n-2(k-1))}{n+2}}\sec^2{\frac{\pi(n-2(k-1))}{2n+4}} $$ This is kind of obtuse to look at so here are some examples:

$n=1$: $$ 1=\sin{\frac{2\pi}{3}}\sin{\frac{\pi}{3}}\sec^2{\frac{\pi}{6}} $$ $n=3$: $$ 3=\sin{\frac{2\pi}{5}}\sin{\frac{3\pi}{5}}\sec^2{\frac{3\pi}{10}}+\sin{\frac{4\pi}{5}}\sin{\frac{\pi}{5}}\sec^2{\frac{2\pi}{10}} $$ $n=5$: $$ 5=\sin{\frac{2\pi}{7}}\sin{\frac{5\pi}{7}}\sec^2{\frac{5\pi}{14}}+\sin{\frac{4\pi}{7}}\sin{\frac{3\pi}{7}}\sec^2{\frac{3\pi}{14}}+ \sin{\frac{6\pi}{7}}\sin{\frac{\pi}{7}}\sec^2{\frac{\pi}{14}} $$ Does this look familiar to anybody? I have no idea what to compare this to? Is this new or even vaguely interesting? I managed to prove that this is true for odd values of n. It seems to hold for even $n$ too (if you sum over $\left \lfloor\frac{m+1}{2}\right \rfloor$) but I don't know how to prove it. Thanks for any help!

Answer 1

We have, $$\theta=\frac {\pi(n+2-2k)}{2n+4}$$ Then $\frac {2k\pi}{n+2}=\pi-2\theta $

Thus, the series is equivalent to: $$S=\sum_{k=1}^{\frac {n+1}{2}} \sin (\pi-2\theta)\sin 2\theta \sec^2 \theta $$ Using $\sin 2\theta=2\cos \theta \sin \theta$ and $\sin (\pi-2\theta)=\sin 2\theta$, we get: $$S=\sum 4\sin^2 \theta =\sum 4 \cos^2\left(\frac {k\pi}{n+2}\right)$$ Now note the identity: $$4\cos^2 x=2\cos 2x+2$$ Hence we get: $$S=\sum_{k=1}^{\left \lfloor \frac {n+1}{2} \right \rfloor} \left(2+2\cos \frac {2k\pi}{n+2}\right) {\tag 1}$$ It is already a well-known result that $$\sum_{k=1}^{n} \cos(a+(k-1)d)=\frac {\cos\left(a+\frac {(n-1)d}{2}\right) \sin \frac {nd}{2}}{\sin \frac d2} {\tag 2}$$ Thus, using $(2)$ in $(1)$ and putting $n=2m$, we arrive at the result for even $n$, as shown. Note that if $n=2m$ then $\left \lfloor \frac {n+1}{2} \right \rfloor=m$. Hence: $$S=\left(\sum_{k=1}^m 2\right)+2\left(\sum_{k=1}^m \cos \frac {2k\pi}{2m+2}\right)$$ The first term obviously evaluates to $2m=n$, and if you note that the second term is a summation of $\cos$ in AP, and hence use $(2)$, you'll see that it evaluates to $0$.

Thus $$S=n+0=n$$

This completes the proof of the even case.