How do you find the exact value of #sec((5pi)/12)#?

1 Answer
Feb 29, 2016

#2/(sqrt(2 - sqrt3)#

Explanation:

#sec ((5pi)/12) = 1/cos #. Find #cos ((5pi)/12)#
On the trig unit circle,
#cos ((5pi)/12) = cos ((6pi)/12 - pi/12) = cos (pi/2 - (pi)/12)#=
#= sin (pi/12)#. (complementary arcs)
Find #sin (pi/12)# by using trig identity:
#cos (pi/6) = sqrt3/3 = 1 - 2sin^2 (pi/12)#
#sin^2 (pi/12) = 1 -sqrt3/2 = (2 - sqrt3)/4#
#sin (pi/12) = sqrt(2 - sqrt3)/2 #--> (sin (pi/12) is positive.
We have:
#cos ((5pi)/12) = sin (pi/12) = sqrt(2 - sqrt3)/2#
Finally, flip the value of cos.
#sec ((5pi)/12) = 2/(sqrt(2 - sqrt3)#