10 questions · Form 5 Additional Mathematics Bab 6: Trigonometric Functions
Given tan A = 43 where A is acute, find sec A.
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. 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.
3. Which identity is equivalent to cos 2A?
Answer: A
Double angle formula for cosine: cos 2A = cos² A - sin² A = 2 cos² A - 1 = 1 - 2 sin² A.
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. 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.
6. Which of the following is equivalent to csc θ?
Answer: A
By definition, cosecant (csc θ) is the reciprocal of sine, so csc θ = 1 / sin θ.
7. 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°.
8. Solve cos θ = 0 for 0° ≤ θ ≤ 360°.
Answer: A
On the unit circle, cos θ = 0 at θ = 90° and θ = 270°.
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. 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°.