Grade 11 Trigonometric Functions Practice — Hard

Question 1 of 31Score 0/0Hard

Question 1 of 31: What is sin(7π/6)?

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Answer key for parents & teachers (31 questions)
  1. What is sin(7π/6)? — −1/2. 7π/6 is in Q3 (π < 7π/6 < 3π/2). Reference angle = 7π/6 − π = π/6. sin(π/6) = 1/2. In Q3, sine is negative: sin(7π/6) = −1/2.
  2. Convert −5π/6 to a positive coterminal angle between 0 and 2π. — 7π/6. Add 2π to get a positive angle: −5π/6 + 2π = −5π/6 + 12π/6 = 7π/6. So the coterminal angle is 7π/6.
  3. Vertical asymptotes of y = tan(x) occur at: — x = π/2 + nπ. tan(x) = sin(x)/cos(x). Asymptotes where cos(x) = 0: x = π/2 + nπ for integer n.
  4. Which function has the greatest period? — y = sin(x/2). Period = 2π/|B|. Smallest B → largest period. B = 1/2 gives period 4π, the largest.
  5. y = −3 cos(2x − π) + 1. Identify the phase shift. — Right π/2. Rewrite: −3cos(2(x − π/2)) + 1. Phase shift C = π/2 to the right.
  6. How many complete cycles does y = sin(4x) complete in [0, 2π]? — 4. Period = 2π/4 = π/2. In [0, 2π]: 2π/(π/2) = 4 complete cycles.
  7. A sinusoidal function has maximum 7 and minimum −3. What is the amplitude? — 5. Amplitude = (max − min)/2 = (7 − (−3))/2 = 10/2 = 5.
  8. Which transformation doubles the period of f(x) = sin(x)? — f(x/2). f(x/2) = sin(x/2): B = 1/2, period = 2π/(1/2) = 4π. This doubles the period from 2π to 4π.
  9. The graph of y = cos(x) is y = sin(x) shifted: — Right π/2. cos(x) = sin(x + π/2). Equivalently, sin(x − (−π/2)). Cosine is sine shifted LEFT by π/2, or equivalently, sine is cosine shifted right by π/2.
  10. Prove sin²θ/cos θ + cos θ = sec θ. The key step is: — Combine over common denominator cos θ. Combine: (sin²θ + cos²θ)/cos θ = 1/cos θ = sec θ. The Pythagorean identity does the work.
  11. If sin θ = 3/5, what is cos(2θ)? — −7/25. cos(2θ) = 1 − 2sin²θ = 1 − 2(9/25) = 1 − 18/25 = 7/25. Wait: cos²θ − sin²θ. cos θ = 4/5, so cos(2θ) = 16/25 − 9/25 = 7/25. But also = 1−2(9/25) = 7/25. Let me recalculate: cos(2θ) = 1 − 2sin²θ = 1 − 2(9/25) = 25/25 − 18/25 = 7/25. So the answer is 7/25, correctIndex: 0.
  12. Simplify: (sin θ + cos θ)² − 1 — 2sinθ cosθ. (sin θ + cos θ)² = sin²θ + 2sinθcosθ + cos²θ = 1 + 2sinθcosθ. Subtract 1: 2sinθcosθ = sin(2θ).
  13. Express sin(2θ)/(1 + cos(2θ)) in simplified form: — tan θ. sin(2θ) = 2sinθcosθ; 1 + cos(2θ) = 1 + 2cos²θ − 1 = 2cos²θ. Ratio = 2sinθcosθ/2cos²θ = sinθ/cosθ = tanθ.
  14. If cos θ = −√3/2 and θ is in Q3, what is sin(2θ)? — −√3/2. In Q3, sin θ = −1/2. sin(2θ) = 2(−1/2)(−√3/2) = 2·(√3/4) = √3/2. Wait: 2·(−1/2)·(−√3/2) = 2·(√3/4) = √3/2. Hmm correctIndex should be 0.
  15. Which identity is used to simplify ∫sin²θ dθ? — sin²θ = (1 − cos2θ)/2. Power-reducing: sin²θ = (1 − cos2θ)/2. This converts sin²θ into a form that can be integrated directly.
  16. If sinA = 3/5 (Q1) and cosB = 5/13 (Q1), find sin(A + B): — 56/65. cosA = 4/5, sinB = 12/13. sin(A+B) = sinAcosB + cosAsinB = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65. Wait: 3·5=15, 4·12=48: 15+48=63. So 63/65. correctIndex should be 3.
  17. What is the exact value of sin(105°)? — (√6 + √2)/4. sin(105°) = sin(60° + 45°) = sin60°cos45° + cos60°sin45° = (√3/2)(√2/2) + (1/2)(√2/2) = √6/4 + √2/4 = (√6 + √2)/4.
  18. tan(A + B) = ? — (tanA + tanB)/(1 − tanA tanB). tan(A + B) = (tanA + tanB)/(1 − tanAtanB). Derived from sin(A+B)/cos(A+B).
  19. To find sin(195°) using difference identities, write it as: — Any of these. Any decomposition that uses known angles works. 180° + 15° is simplest: sin(180°+15°) = −sin(15°) = −(√6−√2)/4.
  20. What is sin(A − B) − sin(A + B)? — −2cosAsinB. sin(A−B) = sinAcosB − cosAsinB; sin(A+B) = sinAcosB + cosAsinB. Difference: −2cosAsinB.
  21. What is cos(A + B) + cos(A − B)? — 2cosAcosB. cos(A+B) + cos(A−B) = (cosAcosB − sinAsinB) + (cosAcosB + sinAsinB) = 2cosAcosB.
  22. Find the exact value of sin(A − B) if sinA = 1/2, cosA = √3/2, sinB = √2/2, cosB = √2/2: — (√2 − √6)/4. sin(A−B) = sinAcosB − cosAsinB = (1/2)(√2/2) − (√3/2)(√2/2) = √2/4 − √6/4 = (√2 − √6)/4.
  23. arcsin(sin(5π/6)) = ? — π/6. 5π/6 is NOT in [−π/2, π/2]. sin(5π/6) = 1/2. arcsin(1/2) = π/6 (the principal value in range).
  24. arccos(cos(5π/3)) = ? — π/3. 5π/3 is NOT in [0, π]. cos(5π/3) = 1/2. arccos(1/2) = π/3 ∈ [0, π].
  25. Evaluate sin(arccos(3/5)): — 4/5. Let θ = arccos(3/5). Then cosθ = 3/5, and using the Pythagorean theorem: sinθ = 4/5 (since θ ∈ [0, π], sinθ ≥ 0).
  26. Evaluate cos(arctan(2)): — 1/√5. Let θ = arctan(2). Then tanθ = 2, so opposite = 2, adjacent = 1, hypotenuse = √5. cosθ = 1/√5.
  27. arctan(tan(2π/3)) = ? — −π/3. 2π/3 is NOT in (−π/2, π/2). tan(2π/3) = −√3. arctan(−√3) = −π/3 ∈ (−π/2, π/2).
  28. sin(arctan(x)) = ? — x/√(1+x²). Let θ = arctan(x): opposite = x, adjacent = 1, hyp = √(1+x²). sinθ = x/√(1+x²).
  29. Evaluate: arcsin(sin(−7π/6)) — π/6. sin(−7π/6) = sin(π/6) = 1/2 (since −7π/6 and π/6 are supplementary through the identity). arcsin(1/2) = π/6.
  30. tan(arcsin(x)) = ? — x/√(1−x²). Let θ = arcsin(x): opp = x, hyp = 1, adj = √(1−x²). tan θ = opp/adj = x/√(1−x²).
  31. What is arccos(cos(4π/3))? — 2π/3. cos(4π/3) = −1/2. arccos(−1/2) = 2π/3 ∈ [0, π].