I do not understand for my life how to solve this problem.
Evaluate the expression without the aid of a calculator.
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 - Reiny, Monday, May 18, 2009 at 9:11pm
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
theta is pi + pi/6 = 7pi/6
then arccos[(-3^1/2)/2) = 5pi/6 or 7pi/6