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What I'd like to find is an identity for $$\sum_{k=0}^{n-1}\csc\left(w+ k \frac{\pi}{n}\right)\csc\left(x+ k \frac{\pi}{n}\right)\csc\left(y+ k \frac{\pi}{n}\right)\csc\left(z+ k \frac{\pi}{n}\right)$$

Further I wonder if a method can be extended to $$\sum_{k=0}^{n-1}\csc\left(x+ k \frac{\pi}{n}\right)\csc\left(y+ k \frac{\pi}{n}\right)\csc\left(z+ k \frac{\pi}{n}\right)$$

Here it is shown that,

$$\sum_{k=0}^{n-1}\csc\left(x+\frac{k\pi}n\right)\csc\left(y+\frac{k\pi}n\right)=\frac{n\sin n(x-y)}{\sin(x-y)\sin nx\sin ny}$$

So I thought I could extend this idea using the trigonometry product to sums identity

giving: $$\cos \left(\theta +\frac{\pi k}{n}\right) \cos \left(\frac{\pi k}{n}+\tau \right) \cos \left(\frac{\pi k}{n}+\phi \right)=\frac{1}{4} \left(\cos \left(\theta +\frac{\pi k}{n}+\tau -\phi \right)+\cos \left(\theta +\frac{\pi k}{n}-\tau +\phi \right)+\cos \left(-\theta +\frac{\pi k}{n}+\tau +\phi \right)+\cos \left(\theta +\frac{3 \pi k}{n}+\tau +\phi \right)\right)$$ and $$\cos \left(\theta +\frac{\pi k}{n}\right) \cos \left(\frac{\pi k}{n}+\tau \right) \cos \left(\frac{\pi k}{n}+\omega \right) \cos \left(\frac{\pi k}{n}+\phi \right)=\frac{1}{8} \left(\cos (\theta +\tau -\omega -\phi )+\cos (\theta -\tau -\omega +\phi )+\cos (\theta -\tau +\omega -\phi )+\cos \left(\theta +\frac{2 \pi k}{n}+\tau -\omega +\phi \right)+\cos \left(\theta +\frac{2 \pi k}{n}+\tau +\omega -\phi \right)+\cos \left(\theta +\frac{2 \pi k}{n}-\tau +\omega +\phi \right)+\cos \left(-\theta +\frac{2 \pi k}{n}+\tau +\omega +\phi \right)+\cos \left(\theta +\frac{4 \pi k}{n}+\tau +\omega +\phi \right)\right)$$

and then taking the logarithmic derivative and making use of Chebyshev polynomials however I cannot quite manipulate this to work.

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1 Answer 1

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Edition of 28.07.23

Let $C=\cos\dfrac k n\pi,\quad S=\sin\dfrac k n\pi,\quad T=\tan\dfrac k n\pi,\quad$ then $$Q_k(p,q)=\csc\left(p+\frac k n\pi\right)\csc\left(q+\frac k n\pi\right) =\dfrac1{\left(C\sin p+S\cos p\right)\left(C\sin q+S\cos q\right)}$$ $$=\dfrac1{C^2\cos p\cos q(T+\tan p)(T+\tan q)},$$ $$Q_k(p,q)=\dfrac{(1+T^2)\sec p \sec q}{ \left(T+\tan p\right)\left(T+\tan q\right)},\tag1$$ $$Q_k(p,q)=\sec p \sec q\left(1+\dfrac E{T+\tan p}+\dfrac F{T+\tan q}\right),$$ where $$E=\lim_{T\to -\tan p}(T+\tan p)Q_k(p,q)\cos p\cos q=\dfrac{1+\tan^2 p}{\tan q-\tan p},$$ $$F=\lim_{T\to -\tan q}(T+\tan q)Q_k(p,q)\cos p\cos q =\dfrac{1+\tan^2 q}{\tan p-\tan q}.$$ Therefore, $$Q_k(p,q)=\dfrac{\sec p\sec q}{\tan q-\tan p}\left(\tan q-\tan p+\dfrac{1+\tan^2 p}{T+\tan p}-\dfrac {1+\tan^2 q}{T+\tan q}\right).$$ $$Q_k(p,q)=\dfrac{h(q)-h(p)}{\sin(q-p)},\tag2$$ where $$h(p)=\tan p-\dfrac{1+\tan^2 p}{T+\tan p} =\dfrac{T\tan p-1}{T+\tan p},$$ $$\dfrac{h(p)h(q)-\tan p\tan q}{\tan p \tan q+1} =\dfrac1{\tan p \tan q+1}\left(\dfrac{T\tan p-1}{T+\tan p}\,\dfrac{T\tan q-1}{T+\tan q} - \tan p \tan q\right)$$ $$=\dfrac1{\tan p \tan q+1}\left(\dfrac{T\tan p-1}{T+\tan p}\,\dfrac{T\tan q-1}{T+\tan q}+1 - (\tan p \tan q+1)\right)$$ $$=\dfrac{(T\tan p-1)(T\tan q-1)+(T+\tan p)(T+\tan q)}{(\tan p \tan q+1)(T+\tan p)(T+\tan q)}-1$$ $$=\dfrac{T^2(1+\tan p \tan q)+(1+\tan p\tan q)}{(\tan p \tan q+1)(T+\tan p)(T+\tan q)}-1 = \dfrac{1+T^2}{(T+\tan p)(T+\tan q)}-1$$ $$=Q_k(p,q)\cos p\cos q-1,$$ i.e. $$h(p)h(q)=(Q_k(p,q)\cos p\cos q-1)(1+\tan p\tan q)+\tan p \tan q,$$ $$h(p)h(q)=Q_k(p,q)\cos(p-q)-1,\tag3$$

From OP, $$\sum\limits_{k=0}^{n-1} Q_k(p,q) = \dfrac{n\sin n(p-q)}{\sin(p-q)\sin np\sin nq}.\tag4$$

Then, $$\sum\limits_{k=0}^{n-1} h(p)h(q) = \dfrac{n\sin n(p-q)\cot(p-q)}{\sin np\sin nq}-1.\tag5$$

Therefore, $$\begin{align} &\sigma =\sum\limits_{k=0}^{n-1} \csc\left(w+\frac k n\pi\right)\csc\left(x+\frac k n\pi\right) \csc\left(y+\frac k n\pi\right)\csc\left(z+\frac k n\pi\right)\\[4pt] &=\sum\limits_{k=0}^{n-1}Q_k(w,x) Q_k(y,z) =\dfrac1{\sin(x-w)\sin(z-y)}\sum\limits_{k=0}^{n-1}(h(x)-h(w))(h(z)-h(y))\\[4pt] &=\dfrac1{\sin(x-w)\sin(z-y)}\left(\sum\limits_{k=0}^{n-1} h(x)h(z)-\sum\limits_{k=0}^{n-1} h(x)h(y)-\sum\limits_{k=0}^{n-1} h(w)h(z)+\sum\limits_{k=0}^{n-1} h(w)h(y)\right), \end{align}$$ $$\color{green}{\mathbf{\begin{align} &\sigma=\dfrac{n}{\sin(x-w)\sin(z-y)} \left(\dfrac{\sin n(x-z)\cot(x-z)}{\sin nx\sin nz} -\dfrac{\sin n(x-y)\cot(x-y)}{\sin nx\sin ny}\right.\\[4pt] &\left.-\dfrac{\sin n(w-z)\cot(w-z)}{\sin nw\sin nz} +\dfrac{\sin n(w-y)\cot(w-y)}{\sin nw\sin ny}\right). \end{align}}}$$

Old Edition

$$\begin{align} &P_k=\csc\left(w+\frac k n\pi\right)\csc\left(x+\frac k n\pi\right) \csc\left(y+\frac k n\pi\right)\csc\left(z+\frac k n\pi\right)\\[4pt] &=\dfrac1{\left(C\sin w+S\cos w\right)\left(C\sin x+S\cos x\right) \left(C\sin y+S\cos y\right)\left(C\sin z+S\cos z\right)},\\[4pt] &=\dfrac{\sec w\sec x\sec y\sec z\,(1+T^2)^2}{(T+\tan w)(T+\tan x)(T+\tan y)(T+\tan z)},\\[4pt] &=\sec w\sec x\sec y\sec z\left(1+\dfrac A{T+\tan w}+\dfrac B{T+\tan x}+\dfrac C{T+\tan y}+\dfrac D{T+\tan z}\right),\\[4pt] \end{align}$$ where $$\begin{align} &A=\cos w\cos x\cos y\cos z\lim\limits_{T\to -\tan w} (T+\tan w)P_k\\[4pt] &=\dfrac{\cos x\cos y\cos z}{\cos^3w(\tan x-\tan w)(\tan y-\tan w)(\tan z-\tan w)},\\[4pt] &B=\cos w\cos x\cos y\cos z\lim\limits_{T\to -\tan x} (T+\tan x)P_k\\[4pt] &=\dfrac{\cos w\cos y\cos z}{\cos^3x(\tan w-\tan x)(\tan y-\tan x)(\tan z-\tan x))},\\[4pt] &C=\cos w\cos x\cos y\cos z\lim\limits_{T\to -\tan y} (T+\tan y)P_k\\[4pt] &=\dfrac{\cos w\cos x\cos z}{\cos^3y(\tan w-\tan y)(\tan x-\tan y)(\tan z-\tan y))},\\[4pt] &D=\cos w\cos x\cos y\cos z\lim\limits_{T\to -\tan z} (T+\tan z)P_k\\[4pt] &=\dfrac{\cos w\cos x\cos y}{\cos^3z(\tan w-\tan z)(\tan x-\tan z)(\tan y-\tan z))},\\[4pt] \end{align}$$

My first idea was to use $\;Q_k(w,x),\; Q_k(x,y)\;$ etc. for definition of the terms $\;\dfrac1{T+\tan w},\; \dfrac1{T+\tan x},\; \dfrac1{T+\tan y},\; \dfrac1{T+\tan z}.\;$

But the rank of the obtained system is insufficient for this task.

Current idea is to express $\;P_k\;$ via the set of $\;Q_k.$ But I did not have success yet.

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    $\begingroup$ thank you for sharing your idea. $\endgroup$
    – onepound
    Jul 25, 2023 at 19:07
  • $\begingroup$ @onepound Full edition. $\endgroup$ Jul 27, 2023 at 22:01
  • 1
    $\begingroup$ much appreciated, thanks $\endgroup$
    – onepound
    Jul 28, 2023 at 7:50

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