tan ((11pi)/8) = tan ((3pi)/8 + (8pi)/8) = tan ((3pi)/8 + pi) = tan ((3pi)/8)
Find tan ((3pi)/8) = tan t--> tan 2t = tan ((3pi)/4) = - 1
Use trig identity;
tan 2t = (2tan t)/(1 - tan^2 t)
-1 = (2tan t)/(1 - tan^2 t)
Cross multiply:
tan^2 t - 1 = 2tan t
tan^2 t - 2tan t - 1 = 0
Solve this quadratic equation for tan t:
D = d^2 = b^2 - 4ac = 4 + 4 = 8 --> d = +- 2sqrt2
There are 2 real roots:
tan t = -b/(2a) +- d/(2a) = 1 +- sqrt2
tan t = 1 + sqrt2 and tan t = 1 - sqrt2
Since tan t = tan ((3pi)/8) is positive, then
tan ((11pi)/8) = tan ((3pi)/8) = 1 + sqrt2
