arccosTheta=-1/2
which means it is between 90 and 270
cos(PI/3)=.5
so, the angle has to be PI-+PI/3
so the angle is 4PI/3, or 2PI/3
For the following value of cos 0, determine the rad value of 0 for pi < 0 < 2pi
(greater than or equal to pi and 2pi)
-1/2
I don't understand I found out that the inner circle val is pi over 3. But I can't work out how to find 0
2 answers
cosine is negative in quadrants 2 and 3
so quadrants 1 and 4 are no good
quadrant 2 is from theta = pi/2 to theta = pi
so quadrant 2 is out
quadrant three is from theta = pi to theta = 3/2 pi
so there it is in quadrant 3
well what is the refernce angle, the angle below the -x axis (theta = pi) that has cos = -1/2
x = -1 when hypotenuse = 2
that is 60 degrees or pi/3 radians
so our angle is pi + pi/3 = 4 pi/3
check that with calculator
cos 240 degrees which is 4 pi/3 rad = -.5
so quadrants 1 and 4 are no good
quadrant 2 is from theta = pi/2 to theta = pi
so quadrant 2 is out
quadrant three is from theta = pi to theta = 3/2 pi
so there it is in quadrant 3
well what is the refernce angle, the angle below the -x axis (theta = pi) that has cos = -1/2
x = -1 when hypotenuse = 2
that is 60 degrees or pi/3 radians
so our angle is pi + pi/3 = 4 pi/3
check that with calculator
cos 240 degrees which is 4 pi/3 rad = -.5