How do you find the exact value of sinx/2, cosx/2, & tanx/2 given that #sinx= 5/13# is in Quadrant 2?

1 Answer
Apr 26, 2015

First find cos x by using trig identity: #cos^2 x + sin^2 x = 1.#
#cos^2 x = 1 - 25/169 = (169 - 25)/169 = 144/169 #
--> cos x = -12/13. (Quadrant II)
Next, use the trig identity: #cos 2a = 2.(cos^2 a) - 1#.
#cos x = 2cos^2 (x/2) - 1# -->
# 2cos^2 (x/2) = 1 - 12/13 = 1/13 #-->
#cos^2 (x/2) = 1/26 #
--> #cos x/2 = -1/5.1# (Quadrant II)
Next, use the trig identity: #cos x = cos^2 (x/2) - sin^2 (x/2) #-->
#sin^2 (x/2) = 1/26 + 12/13 = 25/26# -->
#sin (x/2) = 5/5.1# (quadrant II)
Find #tan (x/2) = sin (x/2)/cos (x/2) = (5/5.1)/(-1/5.1) = -5/1 = -5#