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
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
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
To determine the value of θ (0 for this case) such that cos θ = -1/2 for π < θ < 2π, you can use the inverse cosine function, also known as arccos or cos^(-1). Here's how you can find the value:
1. Start by finding the principal angle in the first quadrant where cos θ = -1/2. This is done by taking the inverse cosine of -1/2:
θ = cos^(-1)(-1/2)
2. Use a calculator that has the inverse cosine function to find the principal angle. In this case, you'll have:
θ = cos^(-1)(-1/2) ≈ 120°
3. Since the given range is π < θ < 2π, convert the principal angle from degrees to radians. To convert from degrees to radians, multiply the degree measure by π/180:
θ = 120° * (π/180) = 2π/3
Therefore, the radian value of θ for π < θ < 2π with cos θ = -1/2 is 2π/3.