Closed form for $\prod_{k = 1}^{n} \sqrt{1 - \cos(a_k - b_k)}$?

Question

When calculating $\prod_{k = 1}^{n} | e^{-i b_k \xi_k} - e^{-i a_k \xi_k} |$, I found out that (I'll omit the index for simplicity) $$ | e^{-i b} - e^{-i a} | = \sqrt{\left(-\cos(a) + \cos(b)\right)^2 + \left(\sin(a)- \sin(b)\right)^2} = \sqrt{2} \sqrt{1 - \cos(a - b)} = 2 \left| \sin\left(\frac{a - b}{2}\right) \right|. $$ for $a,b \in \mathbb{R}$. I wondered if the is a closed form (no product) of $$ \prod_{k = 1}^{n} \sqrt{1 - \cos(a_k - b_k)}, $$ where $b_k > a_k$ are real numbers.

Answer 1

In accordance with the OP, $$P\left(\vec a,\vec b\right)=\prod\limits_{k=1}^n\sqrt{1-\cos(a_k-b_k)} = 2^{\large^n\!/\!_2} \left|\prod\limits_{k=1}^n\sin\dfrac{a_k-b_k}2\right|.$$

To avoid infinity productions, the logarithmic function can be used: $$P\left(\vec a,\vec b\right)= \begin{cases} 0,\quad\text{if}\quad \bigcup\limits_{k=1}^n\left(\dfrac{a_k-b_k}{2\pi}\in\mathbb N\right),\\ 2^{\large^n\!/\!_2}\exp\left(\sum\limits_{k=1}^n\ln\left|\sin\dfrac{a_k-b_k}2\right|\right),\quad\text{otherwise.} \end{cases}$$