Joan Thompson

2022-01-15

If ${a}_{k}\phantom{\rule{0.222em}{0ex}}=\mathrm{tan}\left(\sqrt{2}+\frac{k\pi }{2011}\right)$ , then evaluate $\frac{{a}_{1}+{a}_{2}+\dots +{a}_{2011}}{{a}_{1}{a}_{2}\dots {a}_{2011}}$

usumbiix

Using Sum of tangent functions where arguments are in specific arithmetic series,
$\mathrm{tan}\left(2m+1\right)x=\frac{\left(\begin{array}{c}2m+1\\ 1\end{array}\right)t-\left(\begin{array}{c}2m+1\\ 3\end{array}\right){t}^{3}+\dots +{\left(-1\right)}^{m}\left(\begin{array}{c}2m+1\\ 2m+1\end{array}\right){t}^{2m+1}}{\left(\begin{array}{c}2m+1\\ 0\end{array}\right)-\left(\begin{array}{c}2m+1\\ 2\end{array}\right){t}^{2}+\dots ++{\left(-1\right)}^{m}\left(\begin{array}{c}2m+1\\ 2m\end{array}\right){t}^{2m}}$
So, if $\mathrm{tan}\left(2m+1\right)x=\mathrm{tan}y$
$\left(2m+1\right)x=k\pi +y$ where k is any integer

So, the roots of
$\mathrm{tan}y=\frac{\left(\begin{array}{c}2m+1\\ 1\end{array}\right)t-\left(\begin{array}{c}2m+1\\ 3\end{array}\right){t}^{3}+\dots +{\left(-1\right)}^{m}\left(\begin{array}{c}2m+1\\ 2m+1\end{array}\right){t}^{2m+1}}{\left(\begin{array}{c}2m+1\\ 0\end{array}\right)-\left(\begin{array}{c}2m+1\\ 2\end{array}\right){t}^{2}+\dots +{\left(-1\right)}^{m}\left(\begin{array}{c}2m+1\\ 2m\end{array}\right){t}^{2m}}$
$⇔{\left(-1\right)}^{m}{t}^{2m+1}-\mathrm{tan}y{\left(-1\right)}^{m}\left(2m+1\right){t}^{2m}+\dots +\mathrm{tan}y=0$
are

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