Locate the following on the unit circle and find the coordinates. For each of the numbers t in the above problem, find a number s in [0, 2 pi) such that W(s) = W(t). Part a is done for you. For each of the following values of t, find two other values, one positive and one negative, that have the value. Solution Use the following method: x = rcosQ y = rsinQ here r = sqrt(x^2+y^2) as given a unit circle then r = 1 thus, x = cosQ y = sinQ Question 1 now Q = 24pi x = cos(24pi) = 1 y = sin(24pi) = 0, thus the point is (1,0) Q = 9pi/2 or pi/2 + 4pi x = cos(4pi + pi/2) = cos(pi/2) = 0 y = sin(4pi + pi/2) = sin(pi/2) = 1 thus the point is (0,1) Question 2: W(9pi/2) = W(4pi + pi/2) = W(pi/2) W(-15pi) = W(-pi) = W(pi-2pi) = W(pi) W(-13pi/3) = W(-4pi-pi/3) = W(-pi/3) W(-8pi/6) = W(-2pi-pi/6) = W(-pi/6) Question 3: t = pi/4 the two values would be t = -pi/4 and 3pi/4 t = 2pi/3 the other two values would be t = -pi/3 and 5pi/3.

1. Locate the following on the unit circle and find the coordinates. For each of the numbers t in the
above problem, find a number s in [0, 2 pi) such that W(s) = W(t). Part a is done for you. For
each of the following values of t, find two other values, one positive and one negative, that have
the value.
Solution
Use the following method:
x = rcosQ
y = rsinQ
here r = sqrt(x^2+y^2) as given a unit circle then r = 1
thus,
x = cosQ
y = sinQ
Question 1
now Q = 24pi
x = cos(24pi) = 1
y = sin(24pi) = 0,
thus the point is (1,0)
Q = 9pi/2 or pi/2 + 4pi
x = cos(4pi + pi/2) = cos(pi/2) = 0
y = sin(4pi + pi/2) = sin(pi/2) = 1
thus the point is (0,1)

2. Question 2:
W(9pi/2) = W(4pi + pi/2) = W(pi/2)
W(-15pi) = W(-pi) = W(pi-2pi) = W(pi)
W(-13pi/3) = W(-4pi-pi/3) = W(-pi/3)
W(-8pi/6) = W(-2pi-pi/6) = W(-pi/6)
Question 3:
t = pi/4
the two values would be t = -pi/4 and 3pi/4
t = 2pi/3
the other two values would be t = -pi/3 and 5pi/3