MAT1510 Assignment 04 (751721) Due 16 July 2025.

,MAT1510 – Assignment 04
UNIQUE No.: 751721
Due Date: Wednesday, 16 July 2025


Question 1 (36 Marks)


1.1 Find the terminal point P(x, y) on the unit circle determined by the given value of t

(a) t = π/6
x = cos(π/6) = √3/2 ≈ 0.866
y = sin(π/6) = 1/2 = 0.5
P(x, y) = (√3/2, 1/2) ≈ (0.866, 0.5)



(b) t = 3π/4
x = cos(3π/4) = -√2/2 ≈ -0.707
y = sin(3π/4) = √2/2 ≈ 0.707
P(x, y) = (-√2/2, √2/2) ≈ (-0.707, 0.707)



(c) t = 7π/6
x = cos(7π/6) = -√3/2 ≈ -0.866
y = sin(7π/6) = -1/2 = -0.5
P(x, y) = (-√3/2, -1/2) ≈ (-0.866, -0.5)



(d) t = 11π/6
x = cos(11π/6) = √3/2 ≈ 0.866
y = sin(11π/6) = -1/2 = -0.5
P(x, y) = (√3/2, -1/2) ≈ (0.866, -0.5)

,1.2 Suppose P(x, y) is the terminal point determined by t. Find the terminal point for:

(a) t + π
cos(t + π) = -cos(t), sin(t + π) = -sin(t)
→ Terminal point: (-x, -y)



(b) 2π − t
cos(2π - t) = cos(t), sin(2π - t) = -sin(t)
→ Terminal point: (x, -y)



(c) -t
cos(-t) = cos(t), sin(-t) = -sin(t)
→ Terminal point: (x, -y)



(d) π − t
cos(π - t) = -cos(t), sin(π - t) = sin(t)
→ Terminal point: (-x, y)

, 1.3 Find the reference number t̄ for each value of t:

(a) t = 5π/6
Quadrant II → t̄ = π - t = π - 5π/6 = π/6



(b) t = 7π/4
Quadrant IV → t̄ = 2π - t = 2π - 7π/4 = π/4



(c) t = 3π/2
Lies on negative y-axis → t̄ = π/2