How do you find the exact values of sin(theta/2), cos(theta) where cos theta=4/5 0<=theta<=pi/2?

1 Answer
Sep 23, 2016

Answer:

#sin (t/2) = sqrt10/10#
#cos (t/2) = (3sqrt10)/10#

Explanation:

#cos t = 4/5#. Find #sin (t/2) and cos (t/2)#.
Use the 2 trig identities:
#2cos ^2 (t/2) = 1 + cos t#
#2sin^2 (t/2) = 1 - cos t #
We have:
#2cos^2 (t/2) = 1 + cos t = 1 + 4/5 = 9/5#
#cos^2 (t/2) = 9/10#
#cos (t/2) = +- 3/sqrt10 = +- (3sqrt10)/10#
Since t is in Quadrant I, #cos (t/2)# is positive --> #cos t/2 = (3sqrt10)/10#
#2sin^2 (t/2) = 1 - cos t = 1 - 4/5 = 1/5#
#sin^2 (t/2) = 1/10#
#sin (t/2) = +- 1/sqrt10 = +- sqrt10/10#
Since t is in quadrant I, #sin (t/2)# is positive -->
#sin (t/2) = sqrt10/10#