1
UNIT SIX
UNIVERSITY OF THE PEOPLE
1201-01: COLLEGE ALGEBRA
Task 1
, 2
(i) The point A has coordinates (-√3/2, 1/2) on the unit circle. For a point (x, y) on the unit circle, the six
trigonometric functions are defined as:
sin θ = y, cos θ = x, tan θ = y/x, csc θ = 1/y, sec θ = 1/x, cot θ = x/y.
Substituting the given values:
sin θ = 1/2, cos θ = -√3/2, tan θ = (1/2) / (-√3/2) = -1/√3 = -√3/3,
csc θ = 2, sec θ = -2/√3 = -2√3/3, cot θ = -√3.
(ii) The point has a negative x-coordinate and a positive y-coordinate, so it lies in Quadrant II. The reason is that in
Quadrant II, x < 0 and y > 0.
(iii) The reference angle is the acute angle between the terminal side and the x-axis. It can be found using tan⁻¹(|y|/|
x|) = tan⁻¹((1/2)/(√3/2)) = tan⁻¹(1/√3) = π/6 (30°). Since the point is in Quadrant II, the actual angle is π - π/6 =
5π/6 (150°). Thus, the angle formed by point A is 5π/6 and the reference angle is π/6.
UNIT SIX
UNIVERSITY OF THE PEOPLE
1201-01: COLLEGE ALGEBRA
Task 1
, 2
(i) The point A has coordinates (-√3/2, 1/2) on the unit circle. For a point (x, y) on the unit circle, the six
trigonometric functions are defined as:
sin θ = y, cos θ = x, tan θ = y/x, csc θ = 1/y, sec θ = 1/x, cot θ = x/y.
Substituting the given values:
sin θ = 1/2, cos θ = -√3/2, tan θ = (1/2) / (-√3/2) = -1/√3 = -√3/3,
csc θ = 2, sec θ = -2/√3 = -2√3/3, cot θ = -√3.
(ii) The point has a negative x-coordinate and a positive y-coordinate, so it lies in Quadrant II. The reason is that in
Quadrant II, x < 0 and y > 0.
(iii) The reference angle is the acute angle between the terminal side and the x-axis. It can be found using tan⁻¹(|y|/|
x|) = tan⁻¹((1/2)/(√3/2)) = tan⁻¹(1/√3) = π/6 (30°). Since the point is in Quadrant II, the actual angle is π - π/6 =
5π/6 (150°). Thus, the angle formed by point A is 5π/6 and the reference angle is π/6.
