How do you find the exact values of #sin(u/2), cos(u/2), tan(u/2)# using the half angle formulas given #sinu=5/13, pi/2<u<pi#?

1 Answer
Nghi N
Jul 7, 2017

#sin u = 5/13#. First, find cos u.
#cos^2 u = 1 - sin^2 u = 1 - 25/169 = 144/169# --> #cos u = +- 12/13#.
Since u is in Quadrant 2, then, cos u < 0.
#cos u = - 12/13#.
To find #cos (u/2)#, apply trig identity:
#2cos^2 (u/2) = 1 + cos u = 1 - 12/13 = 1/13#
#cos^2 (u/2) = 1/26# --> #cos (u/2) = +- 1/sqrt26#.
Sin u is in Quadrant 2, then #u/2# is in Quadrant 1, and #cos (u/2)# is positive:
#cos (u/2) = 1/(sqrt26) = sqrt26/26#
To find #sin (u/2)#, apply the trig identity:
#sin u = 2sin (u/2).cos (u/2)#
#sin (u/2) = (sin u)/(2cos (u/2)) = (5/13)(sqrt26/2) = (5sqrt26)/26#
#tan (u/2) = (sin (u/2))/(cos (u/2)) = ((5sqrt26)/26)(sqrt26) = 5#

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