"If ak=tan⁡(2+kπ2011) , then evaluate a1+a2+…+a2011a1a2…a2011" was answered by our experts

Joan Thompson

Answered question

2022-01-15

a_{k} = \tan (\sqrt{2} + \frac{k π}{2011})

, then evaluate

\frac{a_{1} + a_{2} + \dots + a_{2011}}{a_{1} a_{2} \dots a_{2011}}

usumbiix

Beginner2022-01-16Added 33 answers

Using Sum of tangent functions where arguments are in specific arithmetic series,

\tan (2 m + 1) x = \frac{(\begin{matrix} 2 m + 1 \\ 1 \end{matrix}) t - (\begin{matrix} 2 m + 1 \\ 3 \end{matrix}) t^{3} + \dots + {(- 1)}^{m} (\begin{matrix} 2 m + 1 \\ 2 m + 1 \end{matrix}) t^{2 m + 1}}{(\begin{matrix} 2 m + 1 \\ 0 \end{matrix}) - (\begin{matrix} 2 m + 1 \\ 2 \end{matrix}) t^{2} + \dots + + {(- 1)}^{m} (\begin{matrix} 2 m + 1 \\ 2 m \end{matrix}) t^{2 m}}

So, if

\tan (2 m + 1) x = \tan y

(2 m + 1) x = k π + y

where k is any integer

x = \frac{k π + y}{2 m + 1}; k = 1, 2, \dots, 2 m + 1

So, the roots of

\tan y = \frac{(\begin{matrix} 2 m + 1 \\ 1 \end{matrix}) t - (\begin{matrix} 2 m + 1 \\ 3 \end{matrix}) t^{3} + \dots + {(- 1)}^{m} (\begin{matrix} 2 m + 1 \\ 2 m + 1 \end{matrix}) t^{2 m + 1}}{(\begin{matrix} 2 m + 1 \\ 0 \end{matrix}) - (\begin{matrix} 2 m + 1 \\ 2 \end{matrix}) t^{2} + \dots + {(- 1)}^{m} (\begin{matrix} 2 m + 1 \\ 2 m \end{matrix}) t^{2 m}}

\Leftrightarrow {(- 1)}^{m} t^{2 m + 1} - \tan y {(- 1)}^{m} (2 m + 1) t^{2 m} + \dots + \tan y = 0

are

\tan x; x = \frac{k π + y}{2 m + 1}, k = 1, 2, \dots, 2 m + 1

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