Trigonometry and Functions — Topic Review — Solutions
This review covers all four lessons in the Trigonometry and Functions topic: Pythagorean and Reciprocal Identities, Compound and Double Angle Formulas, Trigonometric Equations with Identities, and Inverse Trigonometric Functions. Questions are exam-style and increase in difficulty. Click each question to reveal the full worked solution.
Mixed Review Questions
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Fluency
Q1 — Pythagorean Identity
Simplify each expression using the Pythagorean identity sin²θ + cos²θ = 1:
(a) 1 − sin²θ (b) sin²θ − 1 (c) 1 − 2cos²θ + cos4θ
(a) 1 − sin²θ = cos²θ
(b) sin²θ − 1 = −(1 − sin²θ) = −cos²θ
(c) 1 − 2cos²θ + cos4θ = (1 − cos²θ)² = sin4θ
Or: let c = cos²θ, so 1 − 2c + c² = (1 − c)² = (sin²θ)² = sin4θ
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Fluency
Q2 — Reciprocal Identities
If tanθ = 3/4 and θ is in the first quadrant, find the exact values of: (a) secθ (b) cosecθ (c) cotθ
Since tanθ = 3/4, use a 3-4-5 right triangle: opposite = 3, adjacent = 4, hypotenuse = 5.
sinθ = 3/5, cosθ = 4/5, tanθ = 3/4 (all positive in Q1).
(a) secθ = 1/cosθ = 5/4
(b) cosecθ = 1/sinθ = 5/3
(c) cotθ = 1/tanθ = 4/3
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Fluency
Q3 — Exact Value Using Compound Angle Formula
Find the exact value of sin 75°, using the compound angle formula for sin(A + B).
Write 75° = 45° + 30°.
sin(A + B) = sin A cos B + cos A sin B
sin 75° = sin 45° cos 30° + cos 45° sin 30°
= (1/√2)(√3/2) + (1/√2)(1/2)
= √3/(2√2) + 1/(2√2)
= (√3 + 1)/(2√2)
Rationalise: × √2/√2 → (√6 + √2)/4
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Fluency
Q4 — Exact Value Using Compound Angle Formula
Find the exact value of cos 15°.
Write 15° = 45° − 30°.
cos(A − B) = cos A cos B + sin A sin B
cos 15° = cos 45° cos 30° + sin 45° sin 30°
= (1/√2)(√3/2) + (1/√2)(1/2)
= √3/(2√2) + 1/(2√2)
= (√3 + 1)/(2√2) = (√6 + √2)/4
Note: cos 15° = sin 75° since they are complementary angles.
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Fluency
Q5 — Double Angle Formula
Given sinθ = 5/13 and θ is in the first quadrant, find the exact values of: (a) sin(2θ) (b) cos(2θ)
sinθ = 5/13, so cosθ = 12/13 (from a 5-12-13 triangle, cos positive in Q1).
(a) sin(2θ) = 2 sinθ cosθ = 2 × (5/13) × (12/13) = 120/169
(b) cos(2θ) = cos²θ − sin²θ = (12/13)² − (5/13)² = (144 − 25)/169 = 119/169
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Understanding
Q6 — Proving a Trig Identity
Prove that (sinθ + cosθ)² = 1 + sin(2θ).
Start with the left-hand side:
(sinθ + cosθ)² = sin²θ + 2 sinθ cosθ + cos²θ
= (sin²θ + cos²θ) + 2 sinθ cosθ
= 1 + 2 sinθ cosθ
= 1 + sin(2θ) [using the double angle formula sin(2θ) = 2 sinθ cosθ]
= RHS ✓
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Understanding
Q7 — Solving a Trig Equation with Identities
Solve 2sin²θ − sinθ − 1 = 0 for θ ∈ [0°, 360°].
This is quadratic in sinθ. Let u = sinθ:
2u² − u − 1 = 0
(2u + 1)(u − 1) = 0
u = −1/2 or u = 1
sinθ = 1: θ = 90°
sinθ = −1/2: Reference angle = 30°. sin is negative in Q3 and Q4:
θ = 180° + 30° = 210° and θ = 360° − 30° = 330°
Solutions: θ = 90°, 210°, 330°
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Understanding
Q8 — Using a Double Angle Identity to Solve an Equation
Solve sin(2θ) = sinθ for θ ∈ [0, 2π].
Use sin(2θ) = 2 sinθ cosθ:
2 sinθ cosθ = sinθ
2 sinθ cosθ − sinθ = 0
sinθ(2 cosθ − 1) = 0
sinθ = 0: θ = 0, π, 2π
2cosθ − 1 = 0 → cosθ = 1/2: θ = π/3, 5π/3
Solutions: θ = 0, π/3, π, 5π/3, 2π
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Understanding
Q9 — Inverse Trig Values
Find the exact value of: (a) arcsin(1/2) (b) arctan(√3) (c) arccos(−1/2) (d) arctan(−1)
Use exact values, remembering the principal value ranges: arcsin and arctan output values in [−π/2, π/2]; arccos outputs values in [0, π].
(a) arcsin(1/2): which angle in [−π/2, π/2] has sine = 1/2? Answer: π/6 (30°)
(b) arctan(√3): which angle in [−π/2, π/2] has tangent = √3? Answer: π/3 (60°)
(c) arccos(−1/2): which angle in [0, π] has cosine = −1/2? Answer: 2π/3 (120°)
(d) arctan(−1): which angle in [−π/2, π/2] has tangent = −1? Answer: −π/4 (−45°)
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Understanding
Q10 — Solving an Inverse Trig Equation
Solve: (a) arccos(x) = π/3 (b) arctan(x) = π/4 (c) arcsin(2x − 1) = π/6
(a) arccos(x) = π/3 → x = cos(π/3) = 1/2
(b) arctan(x) = π/4 → x = tan(π/4) = 1
(c) arcsin(2x − 1) = π/6 → 2x − 1 = sin(π/6) = 1/2
2x = 3/2 → x = 3/4
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Understanding
Q11 — Proving with Reciprocal Identities
Prove that tan²θ + 1 = sec²θ.
Start from the Pythagorean identity: sin²θ + cos²θ = 1
Divide every term by cos²θ (valid when cosθ ≠ 0):
sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ
tan²θ + 1 = sec²θ ✓
Similarly, dividing by sin²θ gives: 1 + cot²θ = cosec²θ.
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Problem Solving
Q12 — Complex Equation Using Multiple Identities
Solve cos(2θ) + cosθ = 0 for θ ∈ [0, 2π].
Use cos(2θ) = 2cos²θ − 1:
(2cos²θ − 1) + cosθ = 0
2cos²θ + cosθ − 1 = 0
(2cosθ − 1)(cosθ + 1) = 0
cosθ = 1/2: θ = π/3, 5π/3
cosθ = −1: θ = π
Solutions: θ = π/3, π, 5π/3
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Problem Solving
Q13 — Reciprocal Identity Proof
Prove that (cosecθ − cotθ)(cosecθ + cotθ) = 1.
The left-hand side is a difference of squares: (A − B)(A + B) = A² − B².
= cosec²θ − cot²θ
= (1/sin²θ) − (cos²θ/sin²θ)
= (1 − cos²θ)/sin²θ
= sin²θ/sin²θ
= 1 ✓
This is the reciprocal-identity analogue of the Pythagorean identity: cosec²θ − cot²θ = 1.
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Problem Solving
Q14 — Compound Angle — Finding sin(A+B) Given Quadrant Conditions
Given sin A = 3/5 with A in Q1, and cos B = −5/13 with B in Q2, find the exact value of sin(A + B).
From sin A = 3/5 (Q1): cos A = 4/5 (positive in Q1).
From cos B = −5/13 (Q2): sin B = 12/13 (positive in Q2).
sin(A + B) = sin A cos B + cos A sin B
= (3/5)(−5/13) + (4/5)(12/13)
= −15/65 + 48/65
= 33/65
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Problem Solving
Q15 — Domain and Range of Inverse Trig
For f(x) = arcsin(x): (a) State the domain and range. (b) Sketch the graph. (c) Solve arcsin(x) = arccos(x) and explain geometrically.
(a) Domain: [−1, 1]. Range: [−π/2, π/2].
(b) The graph passes through (−1, −π/2), (0, 0), (1, π/2). It is a strictly increasing S-shaped curve, concave down for x > 0 and concave up for x < 0.
(c) arcsin(x) = arccos(x): let both equal θ. Then sinθ = x and cosθ = x.
sinθ = cosθ → tanθ = 1 → θ = π/4
So x = sin(π/4) = 1/√2 = √2/2.
Geometrically: arcsin(x) and arccos(x) are both equal at the point where they intersect, and since arcsin(x) + arccos(x) = π/2 for all x ∈ [−1, 1], the intersection occurs at θ = π/4, half of π/2.