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. 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.
2. Find the value of tan 105° using compound angle formulae.
Answer: A
tan 105° = tan (60° + 45°) = tan 60° + tan 45°1 - tan 60° tan 45° = √3 + 11 - √3 = -2 - √3.
3. What is the period of y = 2 tan (3x)?
Answer: A
Period for tangent = 180° / b. Here b = 3, so Period = 180° / 3 = 60°.
4. 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.
5. What is the period of the function y = 3 cos (2x) + 1?
Answer: A
Period for cosine = 360° / b. Here b = 2, so Period = 360° / 2 = 180°.
6. Express 2 sin 15° cos 15° as a single trigonometric ratio and find its exact value.
Answer: A
Using double angle formula 2 sin A cos A = sin 2A: 2 sin 15° cos 15° = sin (2 × 15°) = sin 30° = 12.
7. Simplify sin A cos B + cos A sin B.
Answer: A
This is the addition expansion formula for sine: sin (A + B) = sin A cos B + cos A sin B.
8. 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.
9. 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°.
10. If cos θ = -12 and θ lies in Quadrant III, find the exact value of θ.
Answer: A
Reference angle α = cos⁻¹(12) = 60°. In Quadrant III, θ = 180° + 60° = 240°.