Angle x is a third quadrant angle such that cos x = -(2/3).
What is the exact value of cos (x/2) ?
Solve in simplest radical form.
use
cos (2A) = 2cos^2 A − 1
or
cos (x) = 2cos^2 (x/2) − 1
-2/3 = 2cos^2 (x/2) - 1
2cos^2 (x/2) = 1 - 2/3 = 2/3
cos^2 (x/2) = 1/3
cos (x/2) = ± 1/√3
but if x is in quadrant III, then x/2 is in quad II and
cos (x/2) = -1/√3
They are asking for "simplest radical" form, so let's rationalize it
-1/√3 * √3/√3 = -√3 / 3
To find the exact value of cos(x/2), we'll need to use the half-angle identity for cosine, which states that cos(x/2) = ±√[(1 + cos(x))/2].
Given that cos(x) = -(2/3), we can substitute this value into the half-angle identity:
cos(x/2) = ±√[(1 + cos(x))/2]
cos(x/2) = ±√[(1 + (-(2/3)))/2]
cos(x/2) = ±√[(1 - (2/3))/2]
cos(x/2) = ±√[(1/3)/2]
cos(x/2) = ±√(1/6)
Since x is a third quadrant angle, cosine is negative. Therefore, the exact value of cos(x/2) is minus the square root of 1/6:
cos(x/2) = -√(1/6)
Therefore, the exact value of cos(x/2) in simplest radical form is -√(1/6).
To find the exact value of cos(x/2), we'll use the half-angle formula for cosine. The formula states that cos(x/2) = ±sqrt((1 + cos x) / 2).
Since x is a third quadrant angle, the value of cos x should be negative. Given cos x = -(2/3), and we can insert this value into the formula:
cos(x/2) = ±sqrt((1 + cos x) / 2)
= ±sqrt((1 - 2/3) / 2)
= ±sqrt((1/3) / 2)
= ±sqrt(1/6)
= ±(1/√6)
Hence, the exact value of cos(x/2) is ±(1/√6), which is the simplest radical form.
using your half-angle formula,
cos(x/2) = √((1+cosx)/2) = √((1-2/3)/2) = 1/√6
But, since π < x < 3π/2
π/2 < x/2 < 3π/4 so x/2 is in QII and thus
cos x/2 = -1/√6
Note that cosx = 1/√3 is one of our "standard" cosines, and cos(2x) is not 2/3. The mistake was at 1 - 2/3 = 2/3