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trig

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I do not understand for my life how to solve this problem.

Evaluate the expression without the aid of a calculator.

arccos[(-3^1/2)/2)

I'm not sure how my teacher did it. I know she said to find which quadrant that cos can be in and graph the triangle to find the angle.

  • trig -

    first of all
    3^(1/2) = √3

    arccos[(-3^1/2)/2) really says :

    find the angle theta, so that
    cos theta = -√3/2

    are you familiar with the ratio of sides of the 30-60-90 triangle ?
    if so, then you should recognize that cos 30º = √3/2

    but our cosine is negative so the angle must be in quadrants II or III by the CAST rule. (and 30º = pi/6 radians)

    so theta is pi - pi/6 = 5pi/6
    or
    theta is pi + pi/6 = 7pi/6

    then arccos[(-3^1/2)/2) = 5pi/6 or 7pi/6

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