How do you find the exact values of #sin(u/2), cos(u/2), tan(u/2)# using the half angle formulas given #cosu=3/5, 0<u<pi/2#?

1 Answer
Jan 9, 2017

#sin (u/2) = sqrt5/3#
#cos (u/2) = (2sqrt5)/5#
#tan (u/2) = 1/2#

Explanation:

Use trig identities:
#2cos^2 (u/2) = 1 + cos 2u#
#2sin^2 (u/2) = 1 - cos 2u#
In this case:
#2cos^2 (u/2) = 1 + 3/5 = 8/5#
#cos^2 (u/2) = 8/10 = 4/5#
#cos (u/2) = +- 2/sqrt5 = +- (2sqrt5)/5#
Since u is in Quadrant I, then #cos (u/2)# is positive.
#2sin^2 (u/2) = 1 - 3/5 = 2/5#
#sin^2 (u/2) = 2/10 = 1/5#
#sin (u/2) = +- 1/sqrt5 = +- sqrt5/5#
Since u is in Quadrant I, then #sin (u/2)# is positive.
#tan (u/2) = sin/(cos) = (sqrt5/5)(5/(2sqrt5)) = 1/2#