How do you evaluate #tan (pi/12)# using the half angle formula?

1 Answer
Jun 21, 2016

#2 - sqrt3#

Explanation:

Use the trig identity:
#tan 2a = (2tan a)/(1 - tan^2 a)#
Trig table -->
#tan 2a = tan (pi/6) = 1/sqrt3#
Call tan (pi/12) = t, we get:
#1/sqrt3 = (2t)/(1 - t^2).# Cross multiply -->
#(1 - t^2) = 2sqrt3 t#
#t^2 + 2sqr3t - 1 = 0#
Solve this quadratic equation for t.
#D = d^2 - 4ac = 4(3) + 4 = 16# --> #d = +- 4#
There are 2 real roots:
#t = tan (pi/12) = -b/(2a) +- d/(2a) = -(2sqrt3)/2 +- 4/2 = - sqrt3 +- 2#.
Since #tan (pi/12)# is positive, therefor,
#tan (pi/12) = 2 - sqrt3#
Check by calculator.
#tan (pi/12) = tan 15^@ = 0.27#
#2 - sqrt3 = 2 - 1.73 = 0.27#. OK