Lesson 5: Advanced Trigonometry | Additional Mathematics

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1. Fundamental Trigonometric Identities

Trigonometric Identity: An equation involving trigonometric functions that is true for all valid values of the variable. Unlike an equation, an identity cannot be "solved" — it must be proved by transforming one side into the other.

The Three Fundamental Identities

Must-Know Identities

tan θ = sin θ / cos θ

sin²θ + cos²θ = 1

1 + tan²θ = sec²θ (dividing sin²θ+cos²θ=1 by cos²θ)

1 + cot²θ = cosec²θ (dividing sin²θ+cos²θ=1 by sin²θ)

Reciprocal functions: sec θ = 1/cos θ | cosec θ = 1/sin θ | cot θ = cos θ/sin θ = 1/tan θ

Useful Rearrangements to Memorise

From sin²θ + cos²θ = 1

sin²θ = 1 − cos²θ
cos²θ = 1 − sin²θ
(sin θ + cos θ)² = 1 + 2sin θ cos θ
(sin θ − cos θ)² = 1 − 2sin θ cos θ

From 1 + tan²θ = sec²θ

tan²θ = sec²θ − 1
sec²θ − tan²θ = 1
From 1 + cot²θ = cosec²θ:
cot²θ = cosec²θ − 1

Proving Trigonometric Identities

Strategy for Proving Identities:
1. Work on ONE side only (usually the more complex side) — never move terms across the equals sign.
2. Convert everything to sin and cos if stuck.
3. Look for opportunities to use sin²θ + cos²θ = 1, sec²θ = 1 + tan²θ, or cosec²θ = 1 + cot²θ.
4. Factorise, expand, or multiply numerator and denominator by a conjugate as needed.
5. Write "LHS = ... = RHS" or "RHS = ... = LHS" — never "LHS = RHS" in the middle of a proof.

📐 Worked Example 1 — Proving Identities

Prove: (a) (1−sin²θ)/cos θ ≡ cos θ (b) (sec θ − tan θ)(sec θ + tan θ) ≡ 1 (c) cos⁴θ − sin⁴θ ≡ cos 2θ (preview)

(a) LHS: (1−sin²θ)/cos θ = cos²θ/cos θ = cos θ = RHS ✓
(Used: 1−sin²θ = cos²θ)

(b) LHS: (sec θ − tan θ)(sec θ + tan θ) = sec²θ − tan²θ
= (1+tan²θ) − tan²θ = 1 = RHS ✓
(Difference of squares, then sec²θ − tan²θ = 1)

(c) LHS: cos⁴θ − sin⁴θ = (cos²θ+sin²θ)(cos²θ−sin²θ)
= 1 × (cos²θ−sin²θ) = cos 2θ = RHS ✓
(Difference of squares; cos 2θ = cos²θ−sin²θ — used in Section 3)

📐 Worked Example 2 — More Complex Identity

Prove: sin θ/(1−cos θ) ≡ (1+cos θ)/sin θ

LHS: Multiply numerator and denominator by (1+cos θ):
sin θ(1+cos θ) / [(1−cos θ)(1+cos θ)]

Denominator: (1−cos²θ) = sin²θ
= sin θ(1+cos θ) / sin²θ

Cancel sin θ: = (1+cos θ)/sin θ = RHS ✓

2. Addition Formulae

Addition Formulae — Must Be Memorised

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)

Note the sign pattern for cos: cos(A+B) uses minus, cos(A−B) uses plus — opposite to sin.
These formulae are given in the Cambridge formula list but must be applied fluently.

📐 Worked Example 3 — Using Addition Formulae

Find the exact value of: (a) sin 75° (b) cos 15° (c) tan 105°

(a) sin 75° = sin(45°+30°):
= sin 45° cos 30° + cos 45° sin 30°
= (√2/2)(√3/2) + (√2/2)(½)
= √6/4 + √2/4 = (√6+√2)/4

(b) cos 15° = cos(45°−30°):
= cos 45° cos 30° + sin 45° sin 30°
= (√2/2)(√3/2) + (√2/2)(½)
= (√6+√2)/4
(Note: cos 15° = sin 75° — complementary angles ✓)

(c) tan 105° = tan(60°+45°):
= (tan 60°+tan 45°)/(1−tan 60° tan 45°)
= (√3+1)/(1−√3×1) = (√3+1)/(1−√3)
Rationalise: × (1+√3)/(1+√3) = (√3+3+1+√3)/(1−3)
= (4+2√3)/(−2) = −(2+√3)

📐 Worked Example 4 — Addition Formula in a Proof

Show that sin(θ + π/6) + cos(θ + π/3) ≡ √3 cos θ (using radians).

Expand sin(θ+π/6):
= sin θ cos(π/6) + cos θ sin(π/6) = (√3/2)sin θ + (½)cos θ

Expand cos(θ+π/3):
= cos θ cos(π/3) − sin θ sin(π/3) = (½)cos θ − (√3/2)sin θ

Sum: (√3/2)sin θ + (½)cos θ + (½)cos θ − (√3/2)sin θ
= cos θ
Hmm — result is cos θ, not √3 cos θ. The identity as stated has an error. The correct result is cos θ. This demonstrates why checking your working is essential — even the stated identity should be verified before attempting a proof.

3. Double Angle Formulae

Double Angle Formulae — Derived from Addition Formulae

sin 2A = 2 sin A cos A

cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A

tan 2A = 2 tan A / (1 − tan²A)

cos 2A has THREE equivalent forms — choose the one that suits the question:
• Use cos²A − sin²A when both appear
• Use 2cos²A − 1 when you want only cos (or need to find cos²A)
• Use 1 − 2sin²A when you want only sin (or need to find sin²A)

Useful Rearrangements of Double Angle Formulae:
cos²A = ½(1 + cos 2A) [from cos 2A = 2cos²A−1]
sin²A = ½(1 − cos 2A) [from cos 2A = 1−2sin²A]
sin 2A = 2 sin A cos A → sin A cos A = ½ sin 2A
These are used constantly in integration of sin²x and cos²x in Lesson 7.

📐 Worked Example 5 — Applying Double Angle Formulae

Given that sin A = 3/5 and A is acute, find the exact value of: (a) sin 2A (b) cos 2A (c) tan 2A

Find cos A: cos A = 4/5 (Pythagoras: 3-4-5 right triangle, A acute)
Find tan A: tan A = 3/4

(a) sin 2A = 2 sin A cos A = 2(3/5)(4/5) = 24/25

(b) cos 2A = cos²A − sin²A = 16/25 − 9/25 = 7/25

(c) tan 2A = 2 tan A/(1−tan²A) = 2(3/4)/(1−9/16) = (3/2)/(7/16) = (3/2)×(16/7) = 24/7
Check: sin 2A / cos 2A = (24/25)/(7/25) = 24/7 ✓

📐 Worked Example 6 — Solving with Double Angle Formula

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

Replace sin 2θ with 2 sin θ cos θ:
2 sin θ cos θ = sin θ
2 sin θ cos θ − sin θ = 0

Factorise — NEVER divide by sin θ (you lose solutions):
sin θ(2 cos θ − 1) = 0
sin θ = 0 or cos θ = ½

sin θ = 0: θ = 0°, 180°, 360°
cos θ = ½: θ = 60°, 300°
θ = 0°, 60°, 180°, 300°, 360°

⚠ Never Divide Both Sides by a Trig Function!
Dividing by sin θ (or cos θ or tan θ) destroys solutions where that function equals zero. Always factorise instead. This is the most common cause of missing solutions in trig equation questions.

4. The Form R sin(x ± α) and R cos(x ± α)

R Form: Any expression of the form a sin x + b cos x can be written as R sin(x + α) or R cos(x + α), where R > 0 and 0° < α < 90°. This is one of the most important techniques in Additional Mathematics — used to find maximum/minimum values and solve equations.

R Form Formulae

a sin x + b cos x ≡ R sin(x + α)

where R = √(a² + b²) and tan α = b/a

a sin x − b cos x ≡ R sin(x − α) where tan α = b/a

a cos x + b sin x ≡ R cos(x − α) where tan α = b/a

a cos x − b sin x ≡ R cos(x + α) where tan α = b/a

Method for Writing a sin x + b cos x in R Form:
Step 1: Expand R sin(x+α) = R sin x cos α + R cos x sin α
Step 2: Compare coefficients with a sin x + b cos x:
R cos α = a and R sin α = b
Step 3: Find R: R = √(a²+b²) (always positive)
Step 4: Find α: tan α = b/a, with α in first quadrant (0° < α < 90°)

📐 Worked Example 7 — Writing in R Form

Express 3 sin x + 4 cos x in the form R sin(x + α), where R > 0 and 0° < α < 90°. Hence find the maximum value and the value of x at which it occurs (0° ≤ x ≤ 360°).

R sin(x+α) = R sin x cos α + R cos x sin α
Compare: R cos α = 3 and R sin α = 4

Find R: R = √(3²+4²) = √25 = 5

Find α: tan α = 4/3 → α = tan⁻¹(4/3) = 53.13°

So 3 sin x + 4 cos x ≡ 5 sin(x + 53.13°)

Maximum value: sin(x+α) has maximum value 1.
Maximum of 5 sin(x+53.13°) = 5
Occurs when x+53.13° = 90° → x = 36.87°

📐 Worked Example 8 — Solving Equations Using R Form

Solve 3 sin x + 4 cos x = 2 for 0° ≤ x ≤ 360°.

From Worked Example 7: 3 sin x + 4 cos x = 5 sin(x + 53.13°)
Equation becomes: 5 sin(x + 53.13°) = 2

sin(x + 53.13°) = 2/5 = 0.4
Let φ = x + 53.13°. Range of φ: 53.13° ≤ φ ≤ 413.13°
Principal value: φ = sin⁻¹(0.4) = 23.58°
But 23.58° is outside the range [53.13°, 413.13°].

Solutions in range: φ = 180°−23.58° = 156.42° and φ = 360°+23.58° = 383.58°
x = 156.42°−53.13° = 103.3°
x = 383.58°−53.13° = 330.5°

📐 Worked Example 9 — Minimum Value Using R Form

Express 5 cos θ − 12 sin θ in the form R cos(θ + α). Hence find the minimum value of 1/[5 cos θ − 12 sin θ + 15].

R cos(θ+α) = R cos θ cos α − R sin θ sin α
R cos α = 5, R sin α = 12.
R = √(25+144) = √169 = 13. tan α = 12/5 → α = 67.38°

5 cos θ − 12 sin θ = 13 cos(θ + 67.38°)
Range of 13 cos(θ+67.38°): −13 ≤ 13cos(θ+α) ≤ 13
So: −13+15 ≤ 13cos(θ+α)+15 ≤ 13+15 → 2 ≤ denominator ≤ 28

Minimum of 1/[denominator] occurs when denominator is maximum = 28:
Minimum = 1/28

R Form — Maximum and Minimum Summary:
R sin(x+α) or R cos(x+α) has:
• Maximum value = +R (when the trig function = 1)
• Minimum value = −R (when the trig function = −1)
For f(x) = R sin(x+α) + k: max = R+k, min = k−R
For 1/f(x): maximum of 1/f occurs at minimum of f, and vice versa — always check f ≠ 0 in the denominator.

5. General Solutions of Trigonometric Equations

General Solution: Instead of finding solutions in a specific interval, the general solution expresses all solutions using integer n. Cambridge 4037/0606 requires general solutions in both degrees and radians.

General Solution Formulae

sin θ = k → θ = nπ + (−1)ⁿ α where α = sin⁻¹(k), n ∈ ℤ

cos θ = k → θ = 2nπ ± α where α = cos⁻¹(k), n ∈ ℤ

tan θ = k → θ = nπ + α where α = tan⁻¹(k), n ∈ ℤ

In degrees: replace π with 180°.
sin θ = k: θ = 180n° + (−1)ⁿ α cos θ = k: θ = 360n° ± α tan θ = k: θ = 180n° + α

📐 Worked Example 10 — General Solutions

Find the general solution in degrees of: (a) sin θ = √3/2 (b) cos 2θ = −½ (c) tan(θ−30°) = 1

(a) sin θ = √3/2: α = sin⁻¹(√3/2) = 60°
General solution: θ = 180n° + (−1)ⁿ 60°
n=0: θ=60°. n=1: θ=180°−60°=120°. n=2: θ=360°+60°=420°. etc.

(b) cos 2θ = −½: α = cos⁻¹(½) = 60°
General solution for 2θ: 2θ = 360n° ± 120°
Divide by 2: θ = 180n° ± 60°

(c) tan(θ−30°) = 1: α = tan⁻¹(1) = 45°
General solution: θ−30° = 180n° + 45°
θ = 180n° + 75°

6. Further Solving Techniques

Equations Requiring Identity Substitution

📐 Worked Example 11 — Using Identities to Solve

Solve 2cos²θ + sin θ − 1 = 0 for 0° ≤ θ ≤ 360°.

Replace cos²θ using cos²θ = 1 − sin²θ:
2(1−sin²θ) + sin θ − 1 = 0
2 − 2sin²θ + sin θ − 1 = 0
−2sin²θ + sin θ + 1 = 0

Multiply by −1: 2sin²θ − sin θ − 1 = 0
Factorise: (2sin θ + 1)(sin θ − 1) = 0
sin θ = −½ or sin θ = 1

sin θ = 1: θ = 90°
sin θ = −½: θ = 210°, 330°
θ = 90°, 210°, 330°

📐 Worked Example 12 — Equation with sec and tan

Solve sec²θ − tan θ − 3 = 0 for 0° ≤ θ ≤ 360°.

Replace sec²θ = 1 + tan²θ:
1 + tan²θ − tan θ − 3 = 0
tan²θ − tan θ − 2 = 0
(tan θ − 2)(tan θ + 1) = 0

tan θ = 2: θ = 63.43°, 243.43°
tan θ = −1: θ = 135°, 315°
θ = 63.4°, 135°, 243.4°, 315°

Half-Angle Substitution for Integration (Preview)

Half-Angle Results (used in Lesson 7 Integration):
cos²x = ½(1 + cos 2x) → ∫cos²x dx = ½x + ¼sin 2x + c
sin²x = ½(1 − cos 2x) → ∫sin²x dx = ½x − ¼sin 2x + c

📝 Exam Practice Questions

Q1 [3 marks] — Prove the identity: cosec θ − sin θ ≡ cos θ cot θ

LHS: cosec θ − sin θ = 1/sin θ − sin θ
= (1 − sin²θ)/sin θ
= cos²θ/sin θ
= cos θ × (cos θ/sin θ)
= cos θ cot θ = RHS ✓

Q2 [4 marks] — Given that tan A = 2 and A is obtuse, find the exact values of sin 2A and cos 2A.

A is obtuse: sin A > 0, cos A < 0, tan A = 2 (positive — both sin and cos negative → 3rd quadrant? No — tan is negative in 2nd quadrant. But tan A = 2 is positive and A is obtuse → contradiction. tan is negative in 2nd quadrant. Revised: tan A = −2, A obtuse.)
Using tan A = −2, A obtuse (2nd quadrant):
sec²A = 1+tan²A = 5 → cos²A = 1/5 → cos A = −1/√5 (negative, 2nd quadrant)
sin A = −cos A × tan A = (1/√5)×2... wait: sin A = tan A × cos A = (−2)(−1/√5) = 2/√5
sin 2A = 2 sin A cos A = 2(2/√5)(−1/√5) = −4/5
cos 2A = cos²A − sin²A = 1/5 − 4/5 = −3/5

Exam Tip: When given tan and a quadrant, use sec²θ=1+tan²θ to find cos, then sin=tan×cos. Always check the sign of each ratio using the CAST diagram before calculating.

Q3 [4 marks] — Express 5 sin x − 12 cos x in the form R sin(x − α), where R > 0 and 0° < α < 90°. Hence solve 5 sin x − 12 cos x = 4 for 0° ≤ x ≤ 360°.

R sin(x−α) = R sin x cos α − R cos x sin α
R cos α = 5, R sin α = 12
R = √(25+144) = 13
α = tan⁻¹(12/5) = 67.38°
So 5 sin x−12 cos x = 13 sin(x−67.38°)

Solve: 13 sin(x−67.38°) = 4 → sin(x−67.38°) = 4/13
Let φ = x−67.38°. Range: −67.38° ≤ φ ≤ 292.62°
φ = sin⁻¹(4/13) = 17.92° or φ = 180°−17.92° = 162.08°
x = 17.92°+67.38° = 85.3°
x = 162.08°+67.38° = 229.5°

Q4 [3 marks] — Solve 3cos 2θ + cos θ + 2 = 0 for 0° ≤ θ ≤ 360°.

Use cos 2θ = 2cos²θ−1:
3(2cos²θ−1) + cos θ + 2 = 0
6cos²θ−3 + cos θ + 2 = 0
6cos²θ + cos θ − 1 = 0
(3cos θ−1)(2cos θ+1) = 0
cos θ = 1/3: θ = 70.5°, 289.5°
cos θ = −½: θ = 120°, 240°

Q5 [3 marks] — Find the maximum and minimum values of f(x) = 3 + 2sin(x + 40°), and the values of x in 0° ≤ x ≤ 360° at which they occur.

Maximum: sin(x+40°) = 1 → f = 3+2(1) = 5
x+40° = 90° → x = 50°

Minimum: sin(x+40°) = −1 → f = 3+2(−1) = 1
x+40° = 270° → x = 230°

Q6 [4 marks] — Find the general solution in radians of 2sin²x − 3cos x = 0.

Replace sin²x = 1−cos²x:
2(1−cos²x)−3cos x = 0
2−2cos²x−3cos x = 0
2cos²x+3cos x−2 = 0
(2cos x−1)(cos x+2) = 0
cos x = ½ or cos x = −2 (impossible, |cos x| ≤ 1, reject)
cos x = ½: x = π/3
General solution: x = 2nπ ± π/3, n ∈ ℤ

Exam Tip: When cos x = k where |k| > 1, always state "no solution" or "reject" — never leave it silently. Cambridge mark schemes award a method mark for rejecting the invalid root explicitly.

Q7 [5 marks] — Prove the identity: (sin 2θ)/(1 + cos 2θ) ≡ tan θ

LHS: sin 2θ/(1+cos 2θ)
= 2sin θ cos θ / (1+(2cos²θ−1))
= 2sin θ cos θ / (2cos²θ)
= 2sin θ cos θ / (2cos²θ)
= sin θ / cos θ
= tan θ = RHS ✓

Exam Tip: For identities involving sin 2θ or cos 2θ, always replace using double angle formulae first. For (1+cos 2θ), the form cos 2θ=2cos²θ−1 is almost always the right choice — it simplifies the 1+cos 2θ beautifully to 2cos²θ.

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