Solutions — Compound and Double Angle Formulas
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Q1 — Expanding Compound Angles
Apply the compound angle formulas directly, substituting in known exact values where applicable.
(a) sin(A + 30°) = sin A cos 30° + cos A sin 30°
cos 30° = √3/2, sin 30° = 1/2
= (√3/2) sin A + (1/2) cos A
(b) cos(45° − B) = cos 45° cos B + sin 45° sin B (note: cos(A−B) uses plus)
cos 45° = sin 45° = 1/√2
= (1/√2) cos B + (1/√2) sin B = (cos B + sin B)/√2
(c) tan(π/4 + x) = (tan(π/4) + tan x) / (1 − tan(π/4) tan x)
tan(π/4) = 1
= (1 + tan x) / (1 − 1 × tan x) = (1 + tan x)/(1 − tan x)
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Q2 — Exact Values Using Compound Angles
(a) cos 75° = cos(45° + 30°) = 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)
Rationalise: × (√2/√2) = (√6 − √2)/4
cos 75° = (√6 − √2)/4
(b) π/12 = π/4 − π/6 = 15°
sin(π/12) = sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30°
= (1/√2)(√3/2) − (1/√2)(1/2) = (√3 − 1)/(2√2) = (√6 − √2)/4
Note: sin 15° = cos 75° since they are complementary — a useful check.
(c) tan 15° = tan(45° − 30°) = (tan 45° − tan 30°) / (1 + tan 45° tan 30°)
= (1 − 1/√3) / (1 + 1/√3) = (√3 − 1)/(√3 + 1)
Rationalise by multiplying by (√3 − 1)/(√3 − 1):
= (√3 − 1)² / ((√3)² − 1²) = (3 − 2√3 + 1)/(3 − 1) = (4 − 2√3)/2 = 2 − √3
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Q3 — Double Angle Values
Given sinθ = 5/13, θ acute. Find cosθ first.
cos²θ = 1 − sin²θ = 1 − 25/169 = 144/169
Since θ is acute: cosθ = 12/13
(a) sin 2θ = 2 sinθ cosθ = 2 × (5/13) × (12/13) = 120/169 sin 2θ = 120/169
(b) cos 2θ = cos²θ − sin²θ = 144/169 − 25/169 = 119/169 cos 2θ = 119/169
(c) tan 2θ = sin 2θ / cos 2θ = (120/169) ÷ (119/169) = 120/119
Check: since θ is acute, 2θ < π, so both sin 2θ and cos 2θ could be positive, which is consistent here.
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Q4 — Three Forms of cos 2A
Given cosθ = 3/5, θ acute → sinθ = 4/5 (from sin²θ = 1 − 9/25 = 16/25, positive in Q1).
Form 1 (cos²θ − sin²θ):
cos 2θ = (3/5)² − (4/5)² = 9/25 − 16/25 = −7/25
Form 2 (2cos²θ − 1):
cos 2θ = 2(3/5)² − 1 = 2(9/25) − 1 = 18/25 − 25/25 = −7/25 ✓
Form 3 (1 − 2sin²θ):
cos 2θ = 1 − 2(4/5)² = 1 − 2(16/25) = 1 − 32/25 = 25/25 − 32/25 = −7/25 ✓
All three forms give cos 2θ = −7/25. Note: 2θ is in Q2 (since θ ≈ 53.1°, 2θ ≈ 106.2°), so cos 2θ negative is expected.
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Q5 — Using sin(A + B) to Simplify
Proof: Apply the compound angle formula with A = x, B = π:
sin(x + π) = sin x cos π + cos x sin π
Substituting exact values: cos π = −1, sin π = 0:
= sin x (−1) + cos x (0)
= −sin x
Therefore sin(x + π) = −sin x ✓
This confirms the familiar fact that shifting a sine graph by π reflects it vertically.
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Q6 — Proving a Double Angle Identity
Proof: Work on the LHS only and transform it into the RHS.
Numerator: sin 2θ = 2 sinθ cosθ
Denominator: 1 + cos 2θ = 1 + (2cos²θ − 1) = 2cos²θ
LHS = (2 sinθ cosθ) / (2cos²θ)
= sinθ / cosθ (cancel factor of 2cosθ, valid when cosθ ≠ 0)
= tanθ = RHS ✓
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Q7 — Exact Value for a Non-Standard Angle
Write 105° = 60° + 45° (both standard angles).
cos 105° = cos(60° + 45°) = cos 60° cos 45° − sin 60° sin 45°
= (1/2)(1/√2) − (√3/2)(1/√2)
= 1/(2√2) − √3/(2√2)
= (1 − √3)/(2√2)
Rationalise by multiplying by √2/√2:
= (1 − √3)√2/4 = (√2 − √6)/4
cos 105° = (√2 − √6)/4
Check: 105° is in Q2, so cosine should be negative. Since √6 > √2, this is indeed negative. ✓
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Q8 — Double Angle with Obtuse Angle
Given cosα = −3/5 and π/2 < α < π (Q2), so sinα > 0.
sin²α = 1 − cos²α = 1 − 9/25 = 16/25 → sinα = 4/5
sin 2α = 2 sinα cosα = 2(4/5)(−3/5) = −24/25
cos 2α = 1 − 2sin²α = 1 − 2(16/25) = 1 − 32/25 = −7/25
Alternatively: cos 2α = cos²α − sin²α = 9/25 − 16/25 = −7/25 ✓
Quadrant check: since π/2 < α < π, we have π < 2α < 2π. With α ≈ 126.9°, 2α ≈ 253.7° which is in Q3. In Q3 both sin and cos are negative — consistent. ✓
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Q9 — Compound Angle Identity Proof
The identity to prove is: cos 2A / (1 + sin 2A) = (cos A − sin A)/(cos A + sin A)
Work on the LHS. Apply double angle formulas:
Numerator: cos 2A = cos²A − sin²A = (cos A + sin A)(cos A − sin A)
Denominator: 1 + sin 2A = 1 + 2 sin A cos A
= sin²A + cos²A + 2 sin A cos A (since sin²A + cos²A = 1)
= (sin A + cos A)² = (cos A + sin A)²
Therefore:
LHS = (cos A + sin A)(cos A − sin A) / (cos A + sin A)²
= (cos A − sin A) / (cos A + sin A) (cancel one factor of (cos A + sin A))
= RHS ✓
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Q10 — Finding Exact Value with Compound Angle Strategy
Given sinθ = 7/25, θ acute. Find cosθ first.
cos²θ = 1 − 49/625 = 576/625 → cosθ = 24/25 (positive, Q1)
Apply sin(A + B) = sin A cos B + cos A sin B with A = θ, B = 45°:
sin(θ + 45°) = sinθ cos 45° + cosθ sin 45°
= (7/25)(1/√2) + (24/25)(1/√2)
= 7/(25√2) + 24/(25√2)
= (7 + 24)/(25√2)
= 31/(25√2)
Rationalise by multiplying by √2/√2:
sin(θ + 45°) = 31√2/50