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°

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.

Solve cot²θ + 3 cosec θ + 3 = 0 for 0° ≤ θ < 360°.

• Aθ = 30°, 90°, 150°
• Bθ = 210°, 330°
• Cθ = 210°, 270°, 330°✓
• Dθ = 30°, 150°, 210°, 330°

Answer: Use cot²θ = cosec²θ − 1: cosec²θ + 3 cosec θ + 2 = 0, so (cosec θ + 1)(cosec θ + 2) = 0. cosec θ = −1 gives sin θ = −1, θ = 270°; cosec θ = −2 gives sin θ = −1/2, θ = 210° or 330°. Listing only 210° and 330° loses the root from the first factor; 30°, 90°, 150° solve the equations with the signs reversed.

Find the exact value of sec²(π/6).

• A3/4
• B4/3✓
• C4
• D2

Answer: cos(π/6) = √3/2, so sec(π/6) = 2/√3 and sec²(π/6) = 4/3. The value 3/4 is cos²(π/6), the reciprocal of the answer; 4 is sec²(π/3).

Given that tan θ = −3/4 and 90° < θ < 180°, find the exact value of sec θ.

• A−5/4✓
• B5/4
• C−4/5
• D−5/3

Answer: 1 + tan²θ = sec²θ gives sec²θ = 1 + 9/16 = 25/16, so sec θ = ±5/4. In the second quadrant cos θ is negative, so sec θ = −5/4. The value 5/4 ignores the quadrant; −4/5 is cos θ, not its reciprocal.

Use a compound-angle formula to find the exact value of cos 75°.

• A(√6 − √2)/4✓
• B(√6 + √2)/4
• C(√3 − 1)/4
• D(√2 − √6)/4

Answer: cos 75° = cos(45° + 30°) = cos 45° cos 30° − sin 45° sin 30° = (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4. The value (√6 + √2)/4 uses + in the cosine formula, which gives cos 15°; (√2 − √6)/4 is negative, but cos 75° > 0.

Angle A is acute with sin A = 3/5. Angle B is obtuse with cos B = −5/13. Find the exact value of sin(A + B).

• A63/65
• B−63/65
• C−56/65
• D33/65✓

Answer: cos A = 4/5 and, since B is obtuse, sin B = 12/13 (positive). sin(A + B) = sin A cos B + cos A sin B = (3/5)(−5/13) + (4/5)(12/13) = (−15 + 48)/65 = 33/65. Taking cos B as +5/13 gives 63/65; −56/65 is cos(A + B).

Solve sin 2θ = sin θ for 0° ≤ θ < 360°.

• Aθ = 60°, 300°
• Bθ = 0°, 180°, 300°
• Cθ = 0°, 60°, 180°, 300°✓
• Dθ = 0°, 30°, 180°, 330°

Answer: Write 2 sin θ cos θ − sin θ = 0, so sin θ(2cos θ − 1) = 0. sin θ = 0 gives 0° and 180°; cos θ = 1/2 gives 60° and 300°. Dividing both sides by sin θ loses the solutions 0° and 180°, leaving only 60° and 300°.

Express 8 cos θ + 15 sin θ in the form R cos(θ − α), where R > 0 and 0° < α < 90°. Hence solve 8 cos θ + 15 sin θ = 10 for 0° ≤ θ < 360°, giving answers to 1 decimal place.

• Aθ = 8.0° or θ = 115.9°✓
• Bθ = 82.0° or θ = 334.1°
• Cθ = 61.9° or θ = 115.9°
• Dθ = 126.2° or θ = 357.7°

Answer: R = √(8² + 15²) = 17 and tan α = 15/8, so α = 61.93°. Then cos(θ − 61.93°) = 10/17, so θ − 61.93° = ±53.97°, giving θ = 115.9° or θ = 7.96° ≈ 8.0°. Taking tan α = 8/15 (α = 28.1°) gives 82.0° and 334.1°, which do not satisfy the equation.

Express 7 cos θ + 24 sin θ in the form R cos(θ − α), where R > 0 and 0° < α < 90°, giving α to 1 decimal place.

• AR = 25, α = 73.7°✓
• BR = 25, α = 16.3°
• CR = 31, α = 73.7°
• DR = 625, α = 73.7°

Answer: Expanding, R cos α = 7 and R sin α = 24. So R = √(7² + 24²) = 25 and tan α = 24/7, giving α = 73.7°. The angle 16.3° comes from tan α = 7/24, which belongs to the form R sin(θ + α); 31 adds the coefficients instead of using Pythagoras.

Solve √3 sin θ − cos θ = 1 for 0 ≤ θ < 2π.

• Aθ = π/6 or θ = 5π/6
• Bθ = π/3 or θ = π✓
• Cθ = π/3 or θ = 2π/3
• Dθ = 2π/3 or θ = π

Answer: √3 sin θ − cos θ = 2 sin(θ − π/6), since R = √(3 + 1) = 2 and tan α = 1/√3. So sin(θ − π/6) = 1/2, giving θ − π/6 = π/6 or 5π/6, i.e. θ = π/3 or π. Check: at π, 0 − (−1) = 1. The pair π/6, 5π/6 solves for θ − π/6 but forgets to add π/6 back.

Using small-angle approximations, where θ is in radians, find an approximation for sin 3θ + cos θ.

• A1 + 3θ − θ²/2✓
• B1 + 3θ + θ²/2
• C1 + θ − θ²/2
• D1 + 3θ − θ²

Answer: For small θ, sin 3θ ≈ 3θ and cos θ ≈ 1 − θ²/2. Adding gives 1 + 3θ − θ²/2. The approximation 1 + θ − θ²/2 uses sin 3θ ≈ θ, forgetting that the angle is 3θ; the cosine term is subtracted, not added.

For small values of θ, in radians, find the value that (1 − cos 4θ)/(θ tan θ) approaches.

• A4
• B16
• C8✓
• D2

Answer: cos 4θ ≈ 1 − (4θ)²/2 = 1 − 8θ², so the numerator ≈ 8θ². tan θ ≈ θ, so the denominator ≈ θ². The ratio is 8. Using 1 − 4θ²/2 forgets to square the 4 and gives 2; 16 squares the 4 but drops the halving.