Question 386784
Find the exact value of {{{tan((3pi)/16)}}}.
Recall the identity {{{tan(2x) = (2tanx)/(1-tan^2 (x))}}}.
Then from this come
(1)-----> {{{tan((3pi)/8) = (2tan((3pi)/16))/(1-tan^2(3pi/16))}}}, and
(2)-----> {{{tan((3pi)/4) = (2tan((3pi)/8))/(1-tan^2(3pi/8))}}}. Let {{{alpha = tan((3pi)/8)}}}
From (2) we get {{{-1 = (2alpha)/(1-alpha^2)}}}.
From this we get {{{alpha^2 - 2alpha - 1 = 0}}}.  This has solutions
{{{alpha = (2 +- sqrt(8))/2 = 1 +- sqrt(2)}}}, using the quadratic formula. 
Eliminate {{{1-sqrt(2)}}}, because {{{alpha = tan((3pi)/8)}}} >0.

Then substitute this value into (1):
{{{1+sqrt(2) = (2tan((3pi)/16))/(1-tan^2(3pi/16)) = (2beta)/(1-beta^2)}}}, letting {{{beta =  tan((3pi)/16)}}}.
Hence the quadratic equation {{{(1+sqrt(2))beta^2 + 2beta - (1+sqrt(2)) = 0}}}.
The solutions to this equation are {{{beta = (-1 +- sqrt(4+2sqrt(2)))/(1+sqrt(2))}}}.  Again eliminate the negative {{{beta}}} value.  Hence {{{beta = ( sqrt(4+2sqrt(2)) - 1)/(1+sqrt(2))}}}, the exact value of {{{tan((3pi)/16)}}}.