Posted by Yadira on Monday, April 30, 2012 at 12:22am.
Given cos x=2/3 and 3pi/2<x<2pi, find the exact value of cos2x?

PreCalc/Math  Steve, Monday, April 30, 2012 at 11:46am
cos2x = 2cos^2(x)  1 = 2(4/9)1 = 8/91 = 1/9
Posted by Yadira on Monday, April 30, 2012 at 12:22am.
Given cos x=2/3 and 3pi/2<x<2pi, find the exact value of cos2x?
cos2x = 2cos^2(x)  1 = 2(4/9)1 = 8/91 = 1/9