Trigonometry

Find the exact value of sin 60°.

• A√3
• B√2/2
• C√3/2✓
• D1/2

Answer: From the standard 30-60-90 triangle, sin 60° = √3/2.

Given that cos θ = 3/5 and θ is acute, find the exact value of sin θ.

• A4/5✓
• B3/4
• C4/3
• D5/4

Answer: Using sin²θ + cos²θ = 1, sin²θ = 1 − 9/25 = 16/25, so sin θ = 4/5 since θ is acute.

Solve tan θ = 1 for θ in the interval 0° ≤ θ < 360°. Which set gives all solutions?

• A45°, 225°✓
• B135°, 315°
• C45°, 135°
• D45° only

Answer: tan θ = 1 has solutions where θ = 45°, and since tan has period 180°, the next is 45° + 180° = 225°.

Simplify the expression (1 − cos²θ)/sin θ for sin θ ≠ 0.

• A1
• Bcos θ
• Csin θ✓
• Dtan θ

Answer: Since 1 − cos²θ = sin²θ, the expression becomes sin²θ/sin θ = sin θ.

Find the exact value of tan 30°.

• A√3
• B√3/3✓
• C1/2
• D√3/2

Answer: From the 30-60-90 triangle, tan 30° = 1/√3, which rationalises to √3/3. The distractor √3 is tan 60°.

Given that sin θ = 5/13 and θ is acute, find the exact value of tan θ.

• A5/12✓
• B12/5
• C5/13
• D13/12

Answer: Since θ is acute, cos θ = √(1 − 25/169) = 12/13. Then tan θ = sin θ / cos θ = (5/13)/(12/13) = 5/12. The distractor 12/5 is cot θ.

Solve sin θ = −1/2 for 0° ≤ θ < 360°. Which set gives all solutions?

• A30°, 150°
• B30°, 330°
• C210°, 330°✓
• D150°, 210°

Answer: The reference angle is 30°. Sine is negative in the third and fourth quadrants, giving θ = 180° + 30° = 210° and θ = 360° − 30° = 330°. The set 30°, 150° solves sin θ = +1/2.

Solve cos θ = √3/2 for 0 ≤ θ < 2π. Which set gives all solutions (in radians)?

• Aπ/6, 5π/6
• Bπ/3, 5π/3
• Cπ/6 only
• Dπ/6, 11π/6✓

Answer: cos θ = √3/2 has reference angle π/6, and cosine is positive in the first and fourth quadrants, giving θ = π/6 and θ = 2π − π/6 = 11π/6. The pair π/6, 5π/6 wrongly uses the second quadrant (where cosine is negative).

Given that cos θ = 1/4, use the identity cos 2θ = 2cos²θ − 1 to find cos 2θ.

• A1/2
• B−7/8✓
• C7/8
• D−1/2

Answer: cos 2θ = 2(1/4)² − 1 = 2(1/16) − 1 = 1/8 − 1 = −7/8. The distractor 7/8 comes from a sign slip (1 − 1/8).

Given that sin θ = 3/5 and θ is acute, find the exact value of sin 2θ.

• A6/5
• B12/25
• C−7/25
• D24/25✓

Answer: With cos θ = 4/5, sin 2θ = 2 sin θ cos θ = 2 × (3/5) × (4/5) = 24/25. The distractor 6/5 wrongly uses 2 sin θ; −7/25 is cos 2θ.

The expression 3 sin θ + 4 cos θ can be written in the form R sin(θ + α) with R > 0 and α acute. Find the value of R.

• A5✓
• B7
• C√7
• D25

Answer: R = √(3² + 4²) = √(9 + 16) = √25 = 5. The distractor 7 comes from adding 3 + 4 instead of squaring.

For small θ (in radians), use the approximations cos θ ≈ 1 − θ²/2 to find an approximation for cos 3θ.

• A1 − 3θ²/2
• B1 − 3θ²
• C1 − 9θ²/2✓
• D3 − 9θ²/2

Answer: Replace θ with 3θ: cos 3θ ≈ 1 − (3θ)²/2 = 1 − 9θ²/2. The distractor 1 − 3θ²/2 forgets to square the 3.