tan-pi-9-tan-4pi-9-tan-7pi-9-

Question Number 96260 by bemath last updated on 31/May/20

\tan (\frac{π}{9}) + \tan (\frac{4 π}{9}) + \tan (\frac{7 π}{9}) = ?

Answered by john santu last updated on 31/May/20

set x = \frac{π}{9}, \frac{4 π}{9}, \frac{7 π}{9}

3 x = \frac{π}{3}, \frac{4 π}{3}, \frac{7 π}{3}

\tan 3 x = \tan \frac{π}{3} = \tan \frac{4 π}{3} = \tan \frac{7 π}{3} = \sqrt{3}

let q = \tan x \Rightarrow \tan 3 x = \frac{3 q - q^{3}}{1 - 3 q^{2}}

\Rightarrow \sqrt{3} - 3 \sqrt{3} q^{2} = 3 q - q^{3}

q^{3} - 3 \sqrt{3} q^{2} - 3 q + \sqrt{3} = 0

h a v e r o o t s a r e \tan \frac{π}{9}, \tan \frac{4 π}{9}, \tan \frac{7 π}{9}

by {Vieta}^{'} s rule

\tan \frac{π}{9} + \tan \frac{4 π}{9} + \tan \frac{7 π}{9} = 3 \sqrt{3}