Lesson 4: Algebra, Logarithms & Trigonometry (P2) | A Level Mathematics

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1. Polynomials and Rational Functions P2

Polynomial Division

Polynomial long division divides a polynomial f(x) by a divisor d(x), giving a quotient q(x) and remainder r(x): f(x) = d(x)·q(x) + r(x). The degree of r(x) is always less than the degree of d(x).

📐 Worked Example 1 — Polynomial Long Division

Divide 2x³ + 3x² − 5x + 1 by (x + 2).

Set up long division: 2x³ ÷ x = 2x². Write 2x²(x+2) = 2x³+4x² and subtract:
(2x³+3x²−5x+1) − (2x³+4x²) = −x²−5x+1

−x² ÷ x = −x. Write −x(x+2) = −x²−2x and subtract:
(−x²−5x+1) − (−x²−2x) = −3x+1

−3x ÷ x = −3. Write −3(x+2) = −3x−6 and subtract:
(−3x+1) − (−3x−6) = 7
2x³+3x²−5x+1 = (x+2)(2x²−x−3) + 7
Quotient = 2x²−x−3, Remainder = 7. Check: f(−2) = −16+12+10+1 = 7 ✓

Partial Fractions

Partial fractions decompose a proper rational function into a sum of simpler fractions. Three forms — always check that the fraction is proper (degree of numerator < degree of denominator) before decomposing; if improper, divide first.

Partial Fraction Forms

Distinct linear: (px+q)/[(ax+b)(cx+d)] = A/(ax+b) + B/(cx+d)

Repeated factor: (px+q)/[(ax+b)²(cx+d)] = A/(ax+b) + B/(ax+b)² + C/(cx+d)

Irreducible quadratic: (px²+qx+r)/[(ax²+b)(cx+d)] = (Ax+B)/(ax²+b) + C/(cx+d)

Improper: if degree(num) ≥ degree(denom), divide first → quotient + proper remainder

📐 Worked Example 2 — Partial Fractions

Express (5x−1)/[(2x+1)(x−2)] in partial fractions.

(5x−1)/[(2x+1)(x−2)] = A/(2x+1) + B/(x−2)
Multiply both sides by (2x+1)(x−2):
5x−1 = A(x−2) + B(2x+1)

x=2: 10−1 = B(5) → B = 9/5
x=−½: −5/2−1 = A(−5/2) → −7/2 = −5A/2 → A = 7/5

(5x−1)/[(2x+1)(x−2)] = 7/[5(2x+1)] + 9/[5(x−2)]

📐 Worked Example 3 — Improper Fraction

Express (2x²+3x−1)/[(x+1)(x−1)] in partial fractions.

Degree of numerator (2) = degree of denominator (2) → improper. Divide first.
(2x²+3x−1)/(x²−1) = 2 + (3x+1)/(x²−1) (since 2x²+3x−1 = 2(x²−1) + 3x+1)

Decompose (3x+1)/[(x+1)(x−1)] = A/(x+1) + B/(x−1)
3x+1=A(x−1)+B(x+1): x=1: B=2; x=−1: A=1

= 2 + 1/(x+1) + 2/(x−1)

2. The Modulus Function P2

|f(x)| gives the absolute value — all negative outputs are reflected to positive. For equations and inequalities involving modulus, consider both cases.

Solving Modulus Equations and Inequalities:
|f(x)| = g(x) → f(x) = g(x) OR f(x) = −g(x) (check both; reject if g(x) < 0)
|f(x)| < g(x) → −g(x) < f(x) < g(x) (connected interval, valid when g(x) > 0)
|f(x)| > g(x) → f(x) > g(x) OR f(x) < −g(x) (two separate regions)
Squaring method: |f(x)| < |g(x)| ⟺ [f(x)]² < [g(x)]² — avoids sign issues.

📐 Worked Example 4 — Modulus Equations

Solve: (a) |2x−3| = x+1 (b) |3x+1| > |x−2|

(a) Case 1: 2x−3 = x+1 → x = 4. Check: |5|=5, x+1=5 ✓
Case 2: 2x−3 = −(x+1) → 3x=2 → x=2/3. Check: |−5/3|=5/3, x+1=5/3 ✓
Both valid. Solutions: x = 4 and x = 2/3.

(b) Square both sides: (3x+1)² > (x−2)²
9x²+6x+1 > x²−4x+4 → 8x²+10x−3 > 0 → (4x−1)(2x+3) > 0
Roots: x=1/4 and x=−3/2
x < −3/2 or x > 1/4

3. Logarithms — Laws and Equations P2

Laws of Logarithms (any valid base)

log(xy) = log x + log y log(x/y) = log x − log y

log(xⁿ) = n log x log_a b = ln b / ln a (change of base)

log_a aˣ = x a^(log_a x) = x (inverse relationship)

ln eˣ = x e^(ln x) = x

Solving Exponential and Logarithmic Equations

📐 Worked Example 5 — Exponential Equations

Solve: (a) 3²ˣ⁻¹ = 5ˣ⁺² (b) e²ˣ − 5eˣ + 4 = 0 (c) 2ˣ + 2⁻ˣ = 3

(a) Take ln both sides: (2x−1)ln3 = (x+2)ln5
2x ln3 − ln3 = x ln5 + 2ln5
x(2ln3−ln5) = 2ln5+ln3
x = (2ln5+ln3)/(2ln3−ln5) = ln(75)/ln(9/5) ≈ 5.07

(b) Let u = eˣ: u²−5u+4=0 → (u−1)(u−4)=0
u=1: eˣ=1 → x=0 u=4: eˣ=4 → x=ln4

(c) Multiply by 2ˣ: (2ˣ)² − 3(2ˣ) + 1 = 0. Let u=2ˣ:
u = (3±√5)/2. Both positive → valid.
x = log₂((3+√5)/2) and x = log₂((3−√5)/2)
= ln((3+√5)/2)/ln2 ≈ 1.19 and ≈ −1.19

Logarithmic Modelling

📐 Worked Example 6 — Linearisation

A quantity P is modelled by P = Aeᵇᵗ. When t = 2, P = 50 and when t = 5, P = 200. Find A and b.

ln P = bt + ln A. Two equations:
ln 50 = 2b + ln A ...(1) ln 200 = 5b + ln A ...(2)

(2)−(1): 3b = ln200 − ln50 = ln4 → b = ln4/3 ≈ 0.462

ln A = ln50 − 2b = ln50 − 2ln4/3 = ln50 − ln4^(2/3)
A = 50/4^(2/3) = 50/∛16 ≈ 12.5

4. Addition Formulae and Double Angle P2

Addition Formulae

sin(A±B) = sinA cosB ± cosA sinB

cos(A±B) = cosA cosB ∓ sinA sinB

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

Double Angle Formulae

sin 2A = 2 sinA cosA

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

tan 2A = 2tanA/(1 − tan²A)

Half-angle: cos²A = ½(1+cos2A) | sin²A = ½(1−cos2A)

📐 Worked Example 7 — Exact Values Using Addition Formulae

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

(a) sin 75° = sin(45°+30°) = sin45°cos30° + cos45°sin30°
= (√2/2)(√3/2) + (√2/2)(½) = √6/4 + √2/4 = (√6+√2)/4

(b) cos105° = cos(60°+45°) = cos60°cos45° − sin60°sin45°
= (½)(√2/2) − (√3/2)(√2/2) = (√2−√6)/4 = −(√6−√2)/4

(c) tan15° = tan(45°−30°) = (1−1/√3)/(1+1/√3) = (√3−1)/(√3+1)
Rationalise: (√3−1)²/(3−1) = (4−2√3)/2 = 2−√3

R sin(x + α) Form

Harmonic Form — R sin(x ± α) and R cos(x ± α)

a sinx + b cosx ≡ R sin(x+α) where R=√(a²+b²), tanα=b/a

a sinx − b cosx ≡ R sin(x−α) where R=√(a²+b²), tanα=b/a

a cosx + b sinx ≡ R cos(x−α) where R=√(a²+b²), tanα=b/a

Maximum value = R, Minimum value = −R

📐 Worked Example 8 — R Form and Solving Equations

Express 5sinx + 12cosx in the form R sin(x+α). Hence solve 5sinx + 12cosx = 6 for 0 ≤ x ≤ 2π.

R = √(25+144) = √169 = 13. tanα = 12/5 → α = arctan(12/5) = 1.176 rad

5sinx + 12cosx = 13 sin(x+1.176)

13 sin(x+1.176) = 6 → sin(x+1.176) = 6/13
Let φ = x+1.176. Range of φ: [1.176, 2π+1.176] = [1.176, 7.459]
Principal value: φ = arcsin(6/13) = 0.4759 (not in range)
φ = π−0.4759 = 2.666 and φ = 2π+0.4759 = 6.759

x = φ − 1.176:
x = 2.666−1.176 = 1.49 rad and x = 6.759−1.176 = 5.58 rad

📐 Worked Example 9 — Maximum of a Trig Expression

Find the maximum value of 1/(3 + 4sinx − 3cosx) and the value of x in [0, 2π] at which it occurs.

Express 4sinx−3cosx = R sin(x−α): R=√(16+9)=5, tanα=3/4, α=0.6435 rad
4sinx−3cosx = 5sin(x−0.6435)

Expression = 1/(3+5sin(x−0.6435)).
Denominator range: 3+5×[−1,1] = [−2, 8]. Since denom can be negative, expression is unbounded.
For the maximum of the expression where denominator > 0: denominator minimum (positive) → maximum of expression.
Minimum positive denominator: denominator → 0⁺ (expression → ∞). For the minimum positive finite value: denom=3−5=−2 (negative, excluded). The denominator is minimum positive at the boundary.

For maximum value of the expression with bounded denominator: the maximum occurs when the denominator is minimum but still well-defined. If the question intends minimum value of the expression (most common): denom maximum = 8 → minimum = 1/8.
Denom max=8 when sin(x−0.6435)=1 → x−0.6435=π/2 → x = π/2+0.6435 = 2.21 rad.
Minimum of expression = 1/8.

5. Further Trigonometric Identities P2

📐 Worked Example 10 — Solving Using Double Angle

Solve sin2x + cosx = 0 for 0 ≤ x ≤ 2π.

Replace sin2x = 2sinx cosx:
2sinx cosx + cosx = 0 → cosx(2sinx+1) = 0

cosx = 0: x = π/2, 3π/2
sinx = −½: x = 7π/6, 11π/6
x = π/2, 7π/6, 3π/2, 11π/6

📐 Worked Example 11 — Proving an Identity

Prove: cos4θ ≡ 8cos⁴θ − 8cos²θ + 1

cos4θ = cos2(2θ) = 2cos²(2θ) − 1

cos2θ = 2cos²θ−1, so cos²(2θ) = (2cos²θ−1)² = 4cos⁴θ−4cos²θ+1

cos4θ = 2(4cos⁴θ−4cos²θ+1)−1 = 8cos⁴θ−8cos²θ+1 ≡ RHS ✓

The t-substitution (Half-Angle)

Weierstrass substitution t = tan(x/2):
sinx = 2t/(1+t²) cosx = (1−t²)/(1+t²) tanx = 2t/(1−t²)
Used to solve equations of the form a cosx + b sinx = c when R-form is inconvenient, or to integrate rational functions of sin and cos.

📐 Worked Example 12 — t-Substitution

Solve 3cosx + sinx = 2 for 0 < x < 2π using the t-substitution t = tan(x/2).

Substitute: 3(1−t²)/(1+t²) + 2t/(1+t²) = 2
3−3t²+2t = 2(1+t²) = 2+2t²
5t²−2t−1 = 0

t = (2±√(4+20))/10 = (2±√24)/10 = (1±√6)/5
t₁ = (1+√6)/5 ≈ 0.690 → x/2 = arctan(0.690) → x ≈ 1.09 rad
t₂ = (1−√6)/5 ≈ −0.290 → x/2 = arctan(−0.290) → x ≈ 2π−0.567 ≈ 5.72 rad

6. Numerical Methods — Location of Roots P2

Locating roots: If f(a) and f(b) have opposite signs and f is continuous on [a,b], then by the Intermediate Value Theorem there is at least one root in (a,b). This is shown by computing f(a) and f(b) and showing they have opposite signs.

Iterative Methods

Fixed-point iteration: x_{n+1} = g(xₙ)

Convergence condition: |g'(x)| < 1 near the root

Newton-Raphson: x_{n+1} = xₙ − f(xₙ)/f'(xₙ)

Newton-Raphson converges quickly (quadratically) near a simple root

📐 Worked Example 13 — Fixed-Point Iteration

Show that x³ + 3x − 5 = 0 has a root in (1, 2). Use the iteration xₙ₊₁ = (5−xₙ³)/3, starting from x₁=1, to find the root correct to 3 decimal places.

f(1)=1+3−5=−1 < 0; f(2)=8+6−5=9 > 0. Sign change → root in (1,2) ✓

x₁=1.000
x₂=(5−1)/3=1.333
x₃=(5−2.370)/3=0.877... → wait, 1.333³=2.370, (5−2.370)/3=0.877 (diverging?)
Check convergence: g(x)=(5−x³)/3, g'(x)=−x²; |g'(1.2)|=1.44/3=0.48 <1 ✓ (converges near root)
x₁=1.2: x₂=(5−1.728)/3=1.091; x₃=(5−1.299)/3=1.234; x₄=1.099; ... converges to x ≈ 1.154

Verify: f(1.154)≈1.537+3.462−5=−0.001≈0 ✓
Root ≈ 1.154

📐 Worked Example 14 — Newton-Raphson

Use Newton-Raphson to solve eˣ = 4 − x starting from x₀ = 1.

f(x) = eˣ + x − 4, f'(x) = eˣ + 1
x₁ = x₀ − f(x₀)/f'(x₀) = 1 − (e+1−4)/(e+1) = 1 − (2.718−3)/(3.718)
= 1 − (−0.282)/3.718 = 1 + 0.0759 = 1.0759

x₂ = 1.0759 − f(1.0759)/f'(1.0759)
f(1.0759) = e^1.0759 + 1.0759 − 4 = 2.9324+1.0759−4 = 0.0083
x₂ ≈ 1.0759 − 0.0083/3.9324 ≈ 1.0738

One more iteration gives x ≈ 1.0737 (converged).
Root ≈ 1.074 (3 d.p.)

📝 Exam Practice Questions

Q1 [4 marks] — Express (3x²+x+2)/[(x+1)(x²+1)] in partial fractions.

(3x²+x+2)/[(x+1)(x²+1)] = A/(x+1) + (Bx+C)/(x²+1)
3x²+x+2 = A(x²+1) + (Bx+C)(x+1)
x=−1: 3−1+2=2A → A=2
Expand: 3x²+x+2 = 2x²+2+Bx²+Bx+Cx+C
Coeff x²: 3=2+B → B=1. Coeff x: 1=B+C → C=0. Const: 2=2+C ✓
2/(x+1) + x/(x²+1)

Q2 [3 marks] — Solve |5x−2| = |2x+7|.

Square: (5x−2)²=(2x+7)² → 25x²−20x+4=4x²+28x+49
21x²−48x−45=0 → 7x²−16x−15=0 → (7x+5)(x−3)=0
x = 3 or x = −5/7

Q3 [4 marks] — Given that sinA = 3/5 and A is obtuse, find the exact values of sin2A, cos2A, and tan(A/2).

A obtuse: cosA = −4/5 (negative in 2nd quadrant)
sin2A = 2sinA cosA = 2(3/5)(−4/5) = −24/25
cos2A = 1−2sin²A = 1−18/25 = 7/25
tan(A/2): Since A obtuse, A/2 ∈ (π/4, π/2), so tan(A/2) > 0.
tan(A/2) = sinA/(1+cosA) = (3/5)/(1−4/5) = (3/5)/(1/5) = 3

Exam Tip: The half-angle formula tan(A/2) = sinA/(1+cosA) is very useful. Alternatively tan(A/2) = (1−cosA)/sinA. Both are derived from the double angle formulae for sin and cos.

Q4 [4 marks] — Express 3sinθ − 4cosθ in the form R sin(θ−α). Hence find the maximum and minimum values of 10/(7 + 3sinθ − 4cosθ), and the values of θ ∈ [0, 2π] at which they occur.

R=√(9+16)=5, tanα=4/3, α=0.927 rad
3sinθ−4cosθ = 5sin(θ−0.927)

Range of 5sin(θ−0.927): [−5, 5]. Denominator: [7−5, 7+5]=[2, 12]
Maximum of fraction: denominator minimum=2 → max=10/2=5
When 5sin(θ−0.927)=−5 → θ−0.927=−π/2 → θ=0.927−π/2≈−0.644 (outside range)
θ=0.927−π/2+2π≈5.64 rad
Minimum of fraction: denominator maximum=12 → min=10/12=5/6
When 5sin(θ−0.927)=5 → θ−0.927=π/2 → θ=π/2+0.927≈2.50 rad

Q5 [4 marks] — Show that x³ − 2x − 2 = 0 has a root α between 1 and 2. Use the Newton-Raphson method with x₀ = 1.5 to find α correct to 4 significant figures.

f(1)=1−2−2=−3<0, f(2)=8−4−2=2>0. Sign change → root in (1,2) ✓
f(x)=x³−2x−2, f'(x)=3x²−2
x₁=1.5−(3.375−3−2)/(6.75−2)=1.5−(−1.625)/4.75=1.5+0.342=1.842
x₂=1.842−(6.243−3.684−2)/(10.174−2)=1.842−0.559/8.174=1.842−0.068=1.774
x₃=1.774−(5.583−3.548−2)/(9.447−2)=1.774−0.035/7.447=1.774−0.005=1.769
x₄≈1.7693 (converged)
α ≈ 1.769

Q6 [3 marks] — Prove the identity: (sin2x)/(1 + cos2x) ≡ tanx

LHS = 2sinx cosx / (1 + 2cos²x − 1) = 2sinx cosx / 2cos²x
= sinx/cosx = tanx = RHS ✓

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