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Trigonometry

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.

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  1. 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

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    Reiny
  2. 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

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