Asked by Shirley
If sin x=-.6 and x is in Quadrant 4, find cos x/2
I have no idea how to do it!!!PLEASE HELP!!!!!!!!!!
I have no idea how to do it!!!PLEASE HELP!!!!!!!!!!
Answers
Answered by
Reiny
if sinx = -.6 = -6/10 = -3/5 = y/r
so y=-3, r = 5
then x^2 + y^2 = r^2
---> x = ±4
and in quadrant IV, cosx = 4/5
Using the identity cos 2A = 2cos^2 A - 1
we could say
cos x = 2cos^2 (x/2) - 1
4/5 = 2cos^2 (x/2) - 1
cos (x/2) = ±3/√10
but if x is in quadrant IV, then x/2 is in quadrant II and the cosine is negative, so
cos x/2 = -3/√10
so y=-3, r = 5
then x^2 + y^2 = r^2
---> x = ±4
and in quadrant IV, cosx = 4/5
Using the identity cos 2A = 2cos^2 A - 1
we could say
cos x = 2cos^2 (x/2) - 1
4/5 = 2cos^2 (x/2) - 1
cos (x/2) = ±3/√10
but if x is in quadrant IV, then x/2 is in quadrant II and the cosine is negative, so
cos x/2 = -3/√10
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