10 questions · Form 5 Additional Mathematics Bab 6: Trigonometric Functions
10 questions · plays in fullscreen
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. Solve cos θ = 0 for 0° ≤ θ ≤ 360°.
Answer: A
On the unit circle, cos θ = 0 at θ = 90° and θ = 270°.
2. The graph of y = a sin (bx) has 3 complete cycles in 360°. What is the value of b?
Answer: A
The number of complete cycles in 360° (or 2π) is directly given by the parameter b. Thus, b = 3.
3. Given tan θ = -1 and 90° ≤ θ ≤ 180°, find the value of θ.
Answer: A
Reference angle α = 45°. Since θ is in Quadrant II (90° ≤ θ ≤ 180°), θ = 180° - 45° = 135°.
4. State the maximum value of y = 5 cos (x) - 3.
Answer: A
Maximum value of cos x is 1. So Max y = 5(1) - 3 = 2.
5. Simplify (1 - cos² θ) / sin θ.
Answer: A
Using Pythagorean identity sin² θ + cos² θ = 1 => 1 - cos² θ = sin² θ. Therefore, sin² θ / sin θ = sin θ.
6. Find the amplitude of the function y = -4 sin (3x) - 2.
Answer: A
Amplitude is the absolute value of coefficient a. Amplitude = |-4| = 4.
7. Given tan A = 43 where A is acute, find sec A.
Answer: A
1 + tan² A = sec² A => 1 + (43)² = 1 + 169 = 259. Since A is acute, sec A = √(259) = 53.
8. Solve 2 sin θ = 1 for 0° ≤ θ ≤ 360°.
Answer: A
sin θ = 12. Reference angle α = 30°. Sine is positive in Quadrants I and II. θ = 30° and (180° - 30°) = 150°.
9. Simplify sin 2x1 + cos 2x.
Answer: A
sin 2x1 + cos 2x = 2 sin x cos x1 + 2 cos² x - 1 = 2 sin x cos x2 cos² x = sin x / cos x = tan x.
10. Find the acute reference angle for θ = 220°.
Answer: A
Since 220° is in Quadrant III, the reference angle α = θ - 180° = 220° - 180° = 40°.