Given cos x=2/3 and 3pi/2<x<2pi, find the exact value of cos2x?
2 answers
cos2x = 2cos^2(x) - 1 = 2(4/9)-1 = 8/9-1 = -1/9
sinx=-2/3 3pi/2 greater than or equal to x less than or equal to 2pi. determine cos2x
cos2x=cos^2x-sin^2x
cosx=square root of 5/2, sinx=-2/3
x=square root of 5/2^2-(-2/3)^2=1/9
cos2x=cos^2x-sin^2x
cosx=square root of 5/2, sinx=-2/3
x=square root of 5/2^2-(-2/3)^2=1/9