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.

A circular arc has radius 8 cm and subtends an angle of 1.5 radians at the centre. Find the length of the arc.

• A12 cm✓
• B6 cm
• C48 cm
• D24 cm

Answer: Arc length s = rθ = 8 × 1.5 = 12 cm. The distractor 48 cm comes from using ½r²θ (the sector area formula) instead of rθ.

A sector of a circle has radius 10 cm and angle 1.2 radians. Find the area of the sector.

• A12 cm²
• B60 cm²✓
• C6 cm²
• D120 cm²

Answer: Sector area = ½r²θ = ½ × 10² × 1.2 = ½ × 100 × 1.2 = 60 cm². The distractor 12 cm² is the arc length rθ.

State the exact value of sin(π/6).

• A√(3)/2
• B1/2✓
• C√(3)/3
• D2

Answer: From the standard exact-value triangles, sin(π/6) = 1/2 and cos(π/6) = √(3)/2. Mixing sin and cos at these angles is the most frequent exact-value error, because they swap between π/6 and π/3.

State the exact value of sin(π/4).

• A1
• B√(2)/2✓
• C√(2)
• D√(2)/4

Answer: From the standard exact-value triangles, sin(π/4) = √(2)/2 and cos(π/4) = √(2)/2. Mixing sin and cos at these angles is the most frequent exact-value error, because they swap between π/6 and π/3.

State the exact value of sin(π/3).

• A2√(3)/3
• B1/2
• C√(3)
• D√(3)/2✓

Answer: From the standard exact-value triangles, sin(π/3) = √(3)/2 and cos(π/3) = 1/2. Mixing sin and cos at these angles is the most frequent exact-value error, because they swap between π/6 and π/3.

How many solutions does sin(2x) = 0 have in the interval 0 ≤ x < 2π?

• A2
• B4✓
• C5
• D8

Answer: sin(2x) = 0 when 2x = 0, π, 2π, … Over 0 ≤ x < 2π the argument 2x runs from 0 to 4π, giving 4 solutions. Multiplying the angle by 2 compresses the graph, so it crosses zero 2 times as often as sin x does.

How many solutions does sin(3x) = 0 have in the interval 0 ≤ x < 2π?

• A7
• B3
• C6✓
• D12

Answer: sin(3x) = 0 when 3x = 0, π, 2π, … Over 0 ≤ x < 2π the argument 3x runs from 0 to 6π, giving 6 solutions. Multiplying the angle by 3 compresses the graph, so it crosses zero 3 times as often as sin x does.

Using the identity sin²θ + cos²θ = 1, simplify (1 − cos²θ)/sinθ for sinθ ≠ 0.

• Asinθ✓
• Bcosθ
• Ctanθ
• D1/sinθ

Answer: From the identity, 1 − cos²θ = sin²θ. So the expression is sin²θ/sinθ = sinθ. Cancelling one power of sinθ is the whole step; writing tanθ suggests dividing by cosθ, which is not what is here.

Solve tanθ = 1 for 0° ≤ θ < 360°.

• Aθ = 45° only
• Bθ = 45° and 225°✓
• Cθ = 45° and 135°
• Dθ = 45° and 315°

Answer: tan is positive in the first and third quadrants and has period 180°, so from the principal value 45° the second solution is 45° + 180° = 225°. Using 180° − 45° gives 135°, which is the rule for sine, not tangent — each function has its own symmetry.

Which identity correctly expresses tanθ?

• Atanθ = sinθ / cosθ✓
• Btanθ = cosθ / sinθ
• Ctanθ = sinθ × cosθ
• Dtanθ = 1 − sinθ

Answer: By definition tanθ = sinθ/cosθ, which is why tan is undefined wherever cosθ = 0, at 90° and 270°. The reciprocal cosθ/sinθ is cotθ.

Express 3sinθ + 4cosθ in the form Rsin(θ + α), giving R.

• AR = 5✓
• BR = 7
• CR = 12
• DR = 25

Answer: R = √(3² + 4²) = √25 = 5. The coefficients form a right triangle and R is its hypotenuse, so it is found by Pythagoras — adding them gives 7, which is the usual wrong move.

What is the period of y = sin x in degrees?

• A90°, corresponding to one quarter of a complete revolution
• B180°, because the curve is symmetrical about its own midpoint
• C360°✓
• D720°, since the curve completes two full oscillations

Answer: The sine curve repeats every full revolution, so its period is 360° (or 2π radians). tan x is the one with a period of 180°, which is why it is so often confused with this.