Shouldn't the answer to this problem be A, not D. Since we are working with the second quadrant, would it be negative... or did I just do it completely wrong? Thank you so much for your help!

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1 Answer
Nghi N.
Jun 23, 2018

#sqrt5/5#

Explanation:

#cos t = -3/5#, and sin t > 0 --> t lies in Quadrant 2.
Use trig identity
#2cos^2 a = 1 + cos 2a#
In this case:
#2cos (t/2) = 1 + cos t = 1 - 3/5 = 2/5#
#cos^2 (t/2) = 2/10#
#cos (t/2) = +- sqrt2/sqrt10 = +- 1/sqrt5 = +- sqrt5/5#
Since t lies in Quadrant 2, so, (t/2) lies in Quadrant 1, therefor, cos (t/2) is positive.
#cos (t/2) = sqrt5/5# --> Answer D.

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