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 = 8/9-1 = -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
To find the exact value of cos 2x, we will use the double-angle formula for cosine.
The double-angle formula for cosine states that cos 2x = 2 * cos^2(x) - 1.
Given that cos x = 2/3, we can find cos^2(x) by squaring the given value:
cos^2(x) = (2/3)^2 = 4/9
Now, substituting this value into the double-angle formula:
cos 2x = 2 * (4/9) - 1
= 8/9 - 1
= 8/9 - 9/9
= -1/9
So, the exact value of cos 2x is -1/9.
To find the exact value of cos 2x, we can use the double-angle formula for cosine:
cos 2x = 2(cos^2 x) - 1
Given that cos x = 2/3, we can substitute this value into the formula:
cos 2x = 2((2/3)^2) - 1
= 2(4/9) - 1
= 8/9 - 1
= 8/9 - 9/9
= -1/9
Hence, the exact value of cos 2x is -1/9.