How do you use the half angle formula to determine the exact values of the sine, cosine, and tangent #(19pi)/12#?

1 Answer
Nghi N
Feb 5, 2018

#sin ((19pi)/12) = - sqrt(2 + sqrt3)/2#
#cos ((19pi)/12) = sqrt(2 - sqrt3)/2#

Explanation:

First, call #(t/2) = (19pi)/12# -->
#t = (38pi)/12 = (2pi)/12 + 3pi = pi/6 + 3pi#
Next, use the half angle formulas:
#sin (t/2) = +- sqrt((1 - cos t)/2)#
#cos (t/2) = +- sqrt((1 + cos t)/2)#
Note that
#cos t = cos ((pi)/6 + 3pi) = - cos (pi/6) = - sqrt3/2#.
Therefor, since #(t/2)# is in Quadrant 4 -->
#sin (t/2) = - sqrt((1 + sqrt3/2)/2) = - sqrt(2 + sqrt3)/2#
By the same process, with #cos (t/2)# positive -->
#cos (t/2) = + sqrt((1 - sqrt3/2)/2) = sqrt(2 - sqrt3)/2#
From there -->
#tan (t/2) = sin/(cos) = - (2 + sqrt3)/(2 - sqrt3)#

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