Question #48a63

1 Answer
Nghi N.
Feb 8, 2017
  • (1 + sqrt2)

Explanation:

Call tan (5pi)/8 = tan t
Trig table and unit circle give:
#tan 2t = tan ((10pi)/8) = tan ((5pi)/4) = tan (pi/4 + pi) = #
#tan (pi/4) = 1#
Use trig identity: #tan 2t = (2tan t)/(1 - tan^2 t)#
In this case, we have:
#(2tan t)/(1 - tan^2 t) = 1#
#2tan t = 1 - tan^2 t#
#tan^2 t + 2tan t - 1 = 0#
Solve this quadratic equation for tan t, using the improved quadratic formula (Socratic Search):
#D = d^2 = b^2 - 4ac = 4 + 4 = 8# --> #d = +- 2sqrt2#
There are 2 real roots:
#tan t = -b/(2a) +- d/(2a) = - 2/2 +- (2sqrt2)/2 = - 1 +- sqrt2#
Since #tan ((5pi)/8)# is negative (Quadrant II), there for;
#tan t = tan ((5pi)/8) = - (1 + sqrt2)#
Check by calculator:
#tan ((5pi)/8) = tan (112.5) = - 2.414#
#- ( 1 + sqrt2) = - 2.414# . OK

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