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SEC 2027 · K341 · G3 SEC 3 replaces 4049

Sec 3 A Math
Visual Notes

Pictures first. Formulas boxed. Every exam keyword highlighted.
HOW TO STUDY WITH THIS
Read the 💡 box
the whole topic in 1 line
Study the diagram
cover it, redraw it
Learn the yellow
formulas & keywords
Redo the worked example
without looking
THE CHAPTERS
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A1
Quadratic Functions
Complete the square → turning point.
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A2
Equations & Inequalities
b² − 4ac decides it.
√
A3
Surds
Rationalise with the conjugate.
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A4
Polynomials
f(a) = 0 → (x − a) is a factor.
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A4
Partial Fractions
Split by the denominator type.
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A5
Binomial Theorem
General term ⁿCᵣ aⁿ⁻ʳ bʳ.
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A6
Exponentials & Logs
log = power.
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G1
Trig Functions & Graphs
ASTC + amplitude, period.
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G1
Trig Identities & Equations
Prove one side only.
⭕
G2
Coordinate Geometry & Circles
m₁m₂ = −1.
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G2
Linear Law
Make it Y = mX + c.
🖨 PRINTCtrl+P → Save as PDF · Scale 100% · Margins None · Background graphics ON
A Math · Sec 3 · SEC K341 (G3)
⛰️
A1

Quadratic Functions

max/min, always positive/negative, models
BIG IDEAComplete the square: y = a(x − h)² + k → turning point (h, k). a > 0: minimum k. a < 0: maximum k.
xymin (2, −5)(0, 3)
y = 2x² − 8x + 3 = 2(x − 2)² − 5
Completing the square — steps
① take out a from the x terms
② halve the x coefficient, square it, add & subtract
③ tidy up
2x² − 8x + 3 = 2(x² − 4x) + 3
= 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5
ConditionFor y = ax² + bx + c
always positive (above x-axis)a > 0 and b² − 4ac < 0
always negative (below x-axis)a < 0 and b² − 4ac < 0
As a model
ball height h = −5t² + 20t + 1 → h = −5(t − 2)² + 21
max height 21 m at t = 2 s
⚠ TRAPWith a ≠ 1, factor a out before halving. The min/max value is k; it occurs at x = h.
🔎 WHY IT WORKSWhy completing the square finds the min: (x − h)² is a square, so it is never negative — its smallest value is 0, when x = h. So a(x − h)² + k is smallest (a > 0) when the bracket is 0, giving y = k. That is why the turning point is (h, k).
🧠 Worked example — follow the thinking
Find the range of k for which kx² + 4x + k is always positive.
StepWhy I do this
Need a > 0 → k > 0💭 "Always positive" means the graph is a ∪ that never touches the x-axis. A ∪ needs a positive x² coefficient.
Need D < 0: 4² − 4(k)(k) < 0💭 "Never touches the x-axis" = no real roots = discriminant negative.
16 − 4k² < 0 → k² > 4 → k < −2 or k > 2💭 Solve like a quadratic inequality: k² > 4 means k is further than 2 from 0.
Combine with k > 0 → k > 2💭 Both conditions must be true, so take the overlap.
A Math · Sec 3 · SEC K341 (G3)
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A2

Equations & Inequalities

discriminant, line–curve, simultaneous, inequalities
BIG IDEAThe discriminant D = b² − 4ac tells you how many real roots — and whether a line meets a curve.
D > 0 · 2 real roots
D = 0 · 2 equal roots
D < 0 · no real roots
Line & curveSub line into curve, then
meet at 2 pointsD > 0
tangent (touch once)D = 0
do not meetD < 0
real roots / meetsD ≥ 0
Quadratic inequality
x² − x − 6 < 0 → (x − 3)(x + 2) < 0
sketch ∪ → below axis between roots
−2 < x < 3
> 0 would give x < −2 or x > 3
xy
< 0 → between the roots
Simultaneous (one linear)
x + y = 5, x² + y² = 13 → y = 5 − x
x² + (5 − x)² = 13 → x² − 5x + 6 = 0 → (2, 3) and (3, 2)
🔎 WHY IT WORKSWhy b² − 4ac counts the roots: in the formula x = [−b ± √(b² − 4ac)]/2a, the ± gives two answers only if the square root is a real non-zero number. If b² − 4ac > 0 → 2 roots; = 0 → the ± adds nothing → 1 (equal) root; < 0 → you cannot square-root a negative → no real roots. For a line and a curve, substituting gives a quadratic whose roots are the x-values where they meet.
🧠 Worked example — follow the thinking
Find m if y = mx + 1 is a tangent to y = x² + 2x + 2.
StepWhy I do this
x² + 2x + 2 = mx + 1💭 Where they meet, the y-values are equal, so set the two equations equal.
x² + (2 − m)x + 1 = 0💭 Bring everything to one side to get a quadratic in x. Group the x terms: coefficient is (2 − m).
Tangent → D = 0: (2 − m)² − 4(1)(1) = 0💭 A tangent touches at exactly ONE point = one (repeated) root = discriminant 0.
(2 − m)² = 4 → 2 − m = ±2 → m = 0 or 4💭 Square root both sides — remember ± gives 2 cases.
A Math · Sec 3 · SEC K341 (G3)
√
A3

Surds

operations, rationalising, equations
BIG IDEATreat √a like a letter. To rationalise, multiply by the conjugate: (a + √b) ↔ (a − √b).
RuleExample
√a × √b = √(ab)√3 × √12 = √36 = 6
√a ÷ √b = √(a/b)√50 ÷ √2 = 5
m√a + n√a = (m + n)√a3√2 + 5√2 = 8√2
√(a²b) = a√b√72 = 6√2
(√a + √b)(√a − √b) = a − b(√5 + 1)(√5 − 1) = 4
Rationalise
6/√3 = 6√3/3 = 2√3
(2 + √3)/(2 − √3)
× (2 + √3)/(2 + √3) = (7 + 4√3)/1 = 7 + 4√3
Surd equations
square both sides → solve → CHECK every answer
squaring can create fake roots
⚠ TRAP√a + √b ≠ √(a + b). And (2√3)² = 4 × 3 = 12, not 6.
🔎 WHY IT WORKSWhy the conjugate removes the surd: (a + √b)(a − √b) = a² − (√b)² = a² − b — this is the difference of two squares, and squaring √b leaves a whole number. So multiplying top AND bottom by the conjugate clears the surd from the denominator without changing the value (you are multiplying by 1).
🧠 Worked example — follow the thinking
Solve √(x + 5) = x − 1.
StepWhy I do this
x + 5 = (x − 1)² = x² − 2x + 1💭 Square both sides to remove the square root. Square the WHOLE right side, not each term.
x² − 3x − 4 = 0 → (x − 4)(x + 1) = 0💭 Rearrange to = 0 and factorise.
x = 4 or x = −1💭 These are only candidates — squaring can create fake answers because (−2)² = 2² too.
Check x = −1: √4 = 2 but −1 − 1 = −2 ✗ · x = 4: √9 = 3 = 3 ✓💭 √ always means the POSITIVE root, so x = −1 fails. x = 4 only.
A Math · Sec 3 · SEC K341 (G3)
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A4

Polynomials

remainder & factor theorems, cubics, a³ ± b³
BIG IDEARemainder theorem: f(x) ÷ (x − a) leaves remainder f(a). If f(a) = 0, then (x − a) is a factor.
Remainder & factor theorems
divisor (x − a) → remainder f(a)
divisor (ax − b) → remainder f(b/a)
f(a) = 0 ⇔ (x − a) is a factor
Identities (memorise — not given)
a³ + b³ = (a + b)(a² − ab + b²)
a³ − b³ = (a − b)(a² + ab + b²)
8x³ − 27 = (2x − 3)(4x² + 6x + 9)

Solving a cubic

Try x = ±1, ±2, ±3 … (factors of the constant) until f(a) = 0
Divide (long division or compare coefficients) → quadratic
Factorise / use formula on the quadratic
List all roots
Long division layout
        2x² + 3x − 2
x − 3 ) 2x³ − 3x² − 11x + 6
        2x³ − 6x²
              3x² − 11x
              3x² −  9x
                   −2x + 6
                   −2x + 6
                         0
Compare coefficients
(x − 3)(2x² + bx + c) = 2x³ − 3x² − 11x + 6
x²: b − 6 = −3 → b = 3
const: −3c = 6 → c = −2
🔎 WHY IT WORKSWhy the remainder theorem works: division always gives f(x) = (x − a) × Q(x) + R. Put x = a: the bracket becomes 0, so the whole first part vanishes and f(a) = R. If R = 0, then (x − a) divides exactly — it is a factor.
🧠 Worked example — follow the thinking
f(x) = 2x³ − 3x² − 11x + 6. Solve f(x) = 0.
StepWhy I do this
Try f(1) = −6, f(−1) = 18, f(3) = 0 ✓💭 Only try factors of the constant (±1, ±2, ±3, ±6) — a root must divide 6. f(3) = 0 means (x − 3) is a factor.
f(x) = (x − 3)(2x² + bx + c)💭 If (x − 3) is a factor, the rest must be a quadratic. 2x² so the x³ terms match; −3c = 6 so c = −2.
Compare x²: −3 = b − 6 → b = 3💭 Expand mentally: x² terms come from x·bx and −3·2x². Match with −3x².
2x² + 3x − 2 = (2x − 1)(x + 2) → x = 3, ½, −2💭 Factorise the quadratic and set each factor = 0.
A Math · Sec 3 · SEC K341 (G3)
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A4

Partial Fractions

3 denominator types, improper fractions
BIG IDEASplit one fraction into simpler ones. The denominator decides the form. Top degree must be lower than bottom (else divide first).
DenominatorWrite as
(ax + b)(cx + d)A/(ax + b) + B/(cx + d)
(ax + b)(cx + d)²A/(ax + b) + B/(cx + d) + C/(cx + d)²
(ax + b)(x² + c²)A/(ax + b) + (Bx + C)/(x² + c²)
Linear factors (cover-up)
(5x + 1)/((x − 1)(x + 1))
x = 1: A = 6/2 = 3
x = −1: B = −4/−2 = 2
→ 3/(x − 1) + 2/(x + 1)
Repeated factor
1/((x − 1)(x + 1)²)
x = 1: A = ¼ · x = −1: C = −½
compare x²: 0 = A + B → B = −¼
check x = 0: −¼ − ¼ − ½ = −1 ✓
Improper fraction
top degree ≥ bottom → long divide first
(x² + 1)/(x² − 1) = 1 + 2/(x² − 1) = 1 + 1/(x − 1) − 1/(x + 1)
⚠ TRAPA repeated factor needs both B/(x + 1) and C/(x + 1)². A quadratic factor x² + c² needs Bx + C on top.
🔎 WHY IT WORKSWhy cover-up works: partial fractions reverse adding fractions. Multiply both sides by the denominator and you get an identity true for EVERY x — so choose x-values that make brackets 0: each one "kills" all terms except one, leaving one unknown at a time.
🧠 Worked example — follow the thinking
Express (x² + 4x + 1)/((x − 1)(x² + 1)) in partial fractions.
StepWhy I do this
= A/(x − 1) + (Bx + C)/(x² + 1)💭 x² + 1 cannot factorise, so its numerator must be linear (Bx + C) — one degree less than the bottom.
x² + 4x + 1 ≡ A(x² + 1) + (Bx + C)(x − 1)💭 Multiply both sides by the whole denominator.
x = 1: 6 = 2A → A = 3💭 x = 1 makes (x − 1) = 0, killing the Bx + C term.
x²: 1 = A + B → B = −2 · constant: 1 = A − C → C = 2💭 No more "nice" x-values, so compare coefficients of x² and the constants.
→ 3/(x − 1) + (2 − 2x)/(x² + 1)💭 Check x-terms: 4 = −B + C = 2 + 2 ✓
A Math · Sec 3 · SEC K341 (G3)
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A5

Binomial Theorem

expansion, ⁿCᵣ, general term
BIG IDEA(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ. Use the general term Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ to find any one term.
11112113311464115101051
Pascal's triangle — row 4: 1 4 6 4 1
ⁿCᵣ
ⁿCᵣ = n! / (r!(n − r)!)
⁵C₂ = 10 · ⁸C₃ = 56
ⁿC₀ = ⁿCₙ = 1 · ⁿC₁ = n
(given in the formula list)
Full expansion
(2 + x)⁴ = 16 + 32x + 24x² + 8x³ + x⁴
powers of a go down, powers of b go up, they always add to n
Write Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
Collect powers of x
Set the power = what you want (x³, x⁰ …)
Solve for r, then substitute
Term independent of x
(x + 2/x²)⁶: Tᵣ₊₁ = ⁶Cᵣ x⁶⁻ʳ (2)ʳ x⁻²ʳ
6 − 3r = 0 → r = 2
⁶C₂ × 2² = 60
⚠ TRAPKeep the sign and coefficient inside b: (2 − x/2) means b = −x/2, so bʳ = (−½)ʳ xʳ. Not tested: greatest term, properties of coefficients.
🔎 WHY IT WORKSWhy ⁿCᵣ appears: (a + b)ⁿ = (a + b)(a + b)…(a + b). To make the term with bʳ you pick b from r of the n brackets and a from the rest. The number of ways to choose those r brackets is ⁿCᵣ — that is the coefficient.
🧠 Worked example — follow the thinking
Find the coefficient of x³ in (2 − x/2)⁸.
StepWhy I do this
Tᵣ₊₁ = ⁸Cᵣ (2)⁸⁻ʳ (−x/2)ʳ💭 Write the general term with a = 2 and b = −x/2 (keep the minus inside b).
Power of x is r → need r = 3💭 Only the b part contains x, and it is raised to r.
⁸C₃ × 2⁵ × (−½)³💭 Substitute r = 3. (−x/2)³ = (−½)³x³.
56 × 32 × (−⅛) = −224💭 Odd power of a negative stays negative.
A Math · Sec 3 · SEC K341 (G3)
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A6

Exponential & Logarithmic Functions

graphs, laws, change of base, equations, models
BIG IDEAA log is a power: y = aˣ ⇔ x = logₐ y. lg = log₁₀, ln = logₑ.
xyy = aˣy = logₐx
reflections in y = x · (0, 1) and (1, 0)
LawRule
productlog(xy) = log x + log y
quotientlog(x/y) = log x − log y
powerlog xⁿ = n log x
change baselogₐ b = log b / log a
specialslogₐ a = 1 · logₐ 1 = 0
Solve exponentials
take lg/ln of both sides
2ˣ⁺¹ = 5ˣ → (x + 1) lg 2 = x lg 5
x = lg 2 / (lg 5 − lg 2) = 0.756
disguised quadratic: 4ˣ − 3(2ˣ) − 4 = 0 → let u = 2ˣ
Solve logs
combine into one log, then remove it
log₂x + log₂(x − 2) = 3
x(x − 2) = 8 → x = 4 (reject −2)
📈 As a model
N = 500e⁰·²ᵗ → when N = 2000: e⁰·²ᵗ = 4
t = ln 4 ÷ 0.2 = 6.93
N₀ = value at t = 0
⚠ TRAPlog(x + y) ≠ log x + log y. You can only take logs of positive numbers — reject answers that make a log argument ≤ 0.
🔎 WHY IT WORKSWhy the log laws work: a log just asks "what power?" — log₂8 = 3 because 2³ = 8. Multiplying numbers ADDS their powers (2³ × 2⁴ = 2⁷), so log(xy) = log x + log y. Dividing subtracts powers, and a power multiplies them — every log law is an index law in disguise.
🧠 Worked example — follow the thinking
Solve lg(x + 3) − lg(x − 1) = lg 5.
StepWhy I do this
lg[(x + 3)/(x − 1)] = lg 5💭 Quotient law: subtracting logs = log of a division. Now one log on each side.
(x + 3)/(x − 1) = 5💭 If lg A = lg B then A = B (log is one-to-one).
x + 3 = 5x − 5 → x = 2💭 Cross-multiply and solve.
Check: x + 3 = 5 > 0 and x − 1 = 1 > 0 ✓💭 You can only take logs of positive numbers — always check.
A Math · Sec 3 · SEC K341 (G3)
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G1

Trig for Any Angle

unit circle → ASTC → basic angle method
BIG IDEAFor ANY angle, a point on a circle of radius 1 is (cos θ, sin θ). So cos is the x-coordinate and sin is the y-coordinate — that is the whole reason for ASTC.
θ (cos θ, sin θ)cos θ = xsin θ = yr = 1 180° − θ
rotate anticlockwise from the +x-axis
Read the signs off the circle
Q1 (0–90°): x +, y + → All positive
Q2 (90–180°): x −, y + → only Sin positive
Q3 (180–270°): x −, y − → only Tan positive (− ÷ − = +)
Q4 (270–360°): x +, y − → only Cos positive

So ASTC is not magic — it is just the signs of x and y.
🔎 WHY IT WORKSSymmetry: the point for 180° − θ is the mirror image of θ in the y-axis — same height (same sin), opposite x (cos flips sign). That is why sin(180° − θ) = sin θ. Same idea gives 180° + θ (both flip) and 360° − θ (only y flips).

Exact values — derive them, don't just memorise

11√245°
half a square: sin 45° = 1/√2 = √2/2, tan 45° = 1
√31230°60°
half an equilateral △ (side 2): sin 30° = ½, cos 30° = √3/2
🧠 Worked example — follow the thinking
Solve sin x = −0.5 for 0° ≤ x ≤ 360°.
StepWhy I do this
Basic angle α = sin⁻¹(0.5) = 30°💭 Always use the POSITIVE value to get the acute basic angle. The sign only tells me which quadrants.
sin is negative → Q3 and Q4💭 From the circle: y is negative below the x-axis, i.e. Q3 and Q4.
Q3: 180° + 30° = 210°💭 Q3 angles are 180° plus the basic angle.
Q4: 360° − 30° = 330°💭 Q4 angles are 360° minus the basic angle.
x = 210°, 330°💭 Check with calculator: sin 210° = −0.5 ✓
A Math · Sec 3 · SEC K341 (G3)
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G1

Trig Graphs — what a, b, c do

sketching y = a sin bx + c, a cos bx + c, a tan bx
BIG IDEAIn y = a sin bx + c: a stretches up/down, b squashes sideways (more waves), c lifts the whole graph.
a = 3: taller · amplitude 3
b = 2: 2 waves in 360° · period 180°
c = 2: lifted up 2
🔎 WHY IT WORKSWhy period = 360°/b: sin bx repeats when bx goes through 360°, so x only needs to go through 360° ÷ b. For sin(x/b), x must go through 360° × b. Max = c + |a|, min = c − |a| because sin only ranges −1 to 1.
🧠 Worked example — follow the thinking
Sketch y = 2 cos 2x − 1 for 0° ≤ x ≤ 360°.
StepWhy I do this
Amplitude = 2, period = 360° ÷ 2 = 180°, shift c = −1💭 Read a, b, c first — they fix the shape before plotting anything.
Max = −1 + 2 = 1, min = −1 − 2 = −3💭 cos 2x goes from −1 to 1, so ×2 then −1 gives −3 to 1.
Cos starts at its MAX: (0, 1)💭 cos 0 = 1, so a cosine graph starts at the top (sine starts at the middle).
Quarter-period = 45°: points at 0, 45, 90, 135, 180 → max, mid, min, mid, max💭 Every wave has 4 equal quarters: max → middle → min → middle → max.
Repeat for 180° to 360°, join smoothly💭 Period 180° means 2 full waves fit in 0–360°.
x°y
y = 2 cos 2x − 1 · dashed line = centre y = −1
⚠ TRAPb must change the period, not the height. A negative a flips the graph: y = −2 sin x starts going down.
A Math · Sec 3 · SEC K341 (G3)
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G1

Trig Identities — how to use them

turning trig into algebra, proving
BIG IDEAIdentities let you swap one trig function for another so an equation has only ONE function — then it becomes algebra.
🔎 WHY IT WORKSWhere sin² + cos² = 1 comes from: on the unit circle the point (cos θ, sin θ) is 1 unit from O, so by Pythagoras cos²θ + sin²θ = 1. Divide every term by cos²θ → 1 + tan²θ = sec²θ. Divide by sin²θ → cot²θ + 1 = cosec²θ. That is all three given identities from one picture.
If the equation has…Swap usingTo get only…
sin² and cossin² = 1 − cos²cos
cos² and sincos² = 1 − sin²sin
tan² and sectan² = sec² − 1sec
cot² and coseccot² = cosec² − 1cosec
sin and cos (no squares)divide by cos → tantan
🧠 Worked example — follow the thinking
Solve tan²x + sec x = 1 for 0° ≤ x ≤ 360°.
StepWhy I do this
tan²x = sec²x − 1💭 The equation has tan² AND sec — swap tan² so only sec is left.
sec²x − 1 + sec x = 1 → sec²x + sec x − 2 = 0💭 Now it is a quadratic in sec x. Let u = sec x if it helps: u² + u − 2 = 0.
(sec x + 2)(sec x − 1) = 0💭 Factorise like normal algebra.
sec x = 1 → cos x = 1 → x = 0°, 360°💭 sec = 1/cos, so flip it: always convert sec, cosec, cot back to cos, sin, tan to use the calculator.
sec x = −2 → cos x = −½ → basic 60°, Q2 & Q3 → x = 120°, 240°💭 cos negative → Q2, Q3. Q2 = 180 − 60, Q3 = 180 + 60.
x = 0°, 120°, 240°, 360°💭 Collect every answer inside the interval.
🧠 Proof — follow the thinking
Prove (1 − cos 2x)/sin 2x ≡ tan x.
StepWhy I do this
Start with LHS (the messier side)💭 Work on ONE side only. Pick the side with more to simplify.
1 − cos 2x = 1 − (1 − 2sin²x) = 2sin²x💭 There are 3 forms of cos 2x — choose 1 − 2sin²x because it CANCELS the 1.
sin 2x = 2 sin x cos x💭 Double-angle formula from the sheet.
LHS = 2sin²x / (2 sin x cos x) = sin x / cos x = tan x = RHS ∎💭 Cancel 2 sin x top and bottom, then sin/cos = tan.
A Math · Sec 3 · SEC K341 (G3)
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G1

Compound & Double Angle — how to use them

exact values, given ratios, equations
BIG IDEACompound-angle formulas break a "weird" angle into two nice angles. Double-angle is just compound with A = B.
🔎 WHY IT WORKSDouble angle comes free: put B = A in sin(A + B) = sin A cos B + cos A sin B → sin 2A = 2 sin A cos A. Put B = A in cos(A + B) → cos 2A = cos²A − sin²A, then use sin² + cos² = 1 to get the other two forms. You don't need to memorise them separately — they're on the sheet anyway.
Which cos 2A form?Use when
cos²A − sin²Aboth sin and cos appear
2cos²A − 1the rest of the equation is in cos
1 − 2sin²Athe rest of the equation is in sin, or there is a "1 −" to cancel
🧠 Worked example — follow the thinking
Find the exact value of sin 75°.
StepWhy I do this
75° = 45° + 30°💭 Split into two angles whose exact values I know.
sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°💭 Compound formula from the sheet, with A = 45°, B = 30°.
= (√2/2)(√3/2) + (√2/2)(½)💭 Substitute exact values (from the special triangles).
= (√6 + √2)/4💭 Combine over 4. Calculator check: 0.966 ✓
435A51213B
draw a right △ to find the missing ratio
🧠 Use a triangle
sin A = 3/5, cos B = 5/13, A and B acute. Find sin(A + B).
StepWhy I do this
cos A = 4/5, sin B = 12/13💭 Draw the triangles: 3-4-5 and 5-12-13 (Pythagoras).
sin(A + B) = (3/5)(5/13) + (4/5)(12/13)💭 Substitute into the compound formula.
= 15/65 + 48/65 = 63/65💭 If A were obtuse, cos A would be −4/5 (Q2) — always check the quadrant.
🧠 Worked example — follow the thinking
Solve sin 2x = cos x for 0° ≤ x ≤ 360°.
StepWhy I do this
2 sin x cos x = cos x💭 Expand sin 2x so every term is in x, not 2x.
2 sin x cos x − cos x = 0 → cos x(2 sin x − 1) = 0💭 Move everything to one side and FACTORISE — never divide by cos x (you would lose cos x = 0).
cos x = 0 → x = 90°, 270°💭 Read straight off the graph of cos.
sin x = ½ → x = 30°, 150°💭 Basic 30°, sin + → Q1, Q2.
x = 30°, 90°, 150°, 270°💭 4 answers. Dividing by cos x would have lost 90° and 270°.
A Math · Sec 3 · SEC K341 (G3)
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G1

R-Formula — how and why

a sin x + b cos x as one wave
BIG IDEAa sin x + b cos x is secretly ONE sine wave. Rewrite it as R sin(x + α) — then max, min and solving become easy.
a = 3b = 4R = 5α
R = √(a² + b²), tan α = b/a
🔎 WHY IT WORKSWhy it works: expand R sin(x + α) = R sin x cos α + R cos x sin α. Match with 3 sin x + 4 cos x: R cos α = 3 and R sin α = 4. That is a right triangle with sides 3 and 4 → R = 5 (Pythagoras), tan α = 4/3 (opp/adj).
ExpressionWrite asR cos α = …, R sin α = …
a sin x + b cos xR sin(x + α)a, b
a sin x − b cos xR sin(x − α)a, b
a cos x + b sin xR cos(x − α)a, b
a cos x − b sin xR cos(x + α)a, b
🧠 Worked example — follow the thinking
Solve 3 sin x + 4 cos x = 2 for 0° ≤ x ≤ 360°.
StepWhy I do this
3 sin x + 4 cos x = 5 sin(x + 53.13°)💭 R = √(9 + 16) = 5, α = tan⁻¹(4/3) = 53.13°. Keep 2 d.p. for α in working.
5 sin(x + 53.13°) = 2 → sin(x + 53.13°) = 0.4💭 Now it is a single sine — solve for the whole bracket first.
Range: 53.13° ≤ x + 53.13° ≤ 413.13°💭 Add α to BOTH ends of the interval — the bracket has a different range from x.
Basic 23.58° · sin + → Q1, Q2 → 23.58°, 156.42°, 383.58°💭 Keep only the ones inside 53.13°–413.13°: 156.42° and 383.58° (23.58° is too small).
x = 156.42° − 53.13° = 103.3°, x = 383.58° − 53.13° = 330.4°💭 Subtract α to get back to x. Check: 3 sin 103.3° + 4 cos 103.3° = 2 ✓
Max / min for free
max of 5 sin(x + 53.13°) = 5 when x + 53.13° = 90° → x = 36.9°
min = −5 when x + 53.13° = 270° → x = 216.9°
⚠ TRAPR-formula is NOT on the formula sheet. Always state R exactly (√ form if needed) and α to 2 d.p. in working.
A Math · Sec 3 · SEC K341 (G3)
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G2

Coordinate Geometry & Circles

parallel/perpendicular, midpoint, area, circles
BIG IDEAParallel: m₁ = m₂. Perpendicular: m₁m₂ = −1. Circle: centre (a, b), radius r.
Formulas
midpoint ((x₁ + x₂)/2, (y₁ + y₂)/2)
gradient (y₂ − y₁)/(x₂ − x₁)
perpendicular gradient = −1/m
line y − y₁ = m(x − x₁)
Area (shoelace)
list vertices anticlockwise, repeat the first:
½ |(x₁y₂ + x₂y₃ + …) − (y₁x₂ + y₂x₃ + …)|
(0,0), (4,0), (5,3), (1,4) → ½|32 − 3| = 14.5
(a, b)r
(x − a)² + (y − b)² = r²
General form
x² + y² + 2gx + 2fy + c = 0
centre (−g, −f) · radius √(g² + f² − c)
x² + y² − 6x + 4y − 12 = 0
→ centre (3, −2), r = √(9 + 4 + 12) = 5
tangent ⊥ radius at the point of contact
⚠ TRAPCoefficients of x² and y² must be 1 before reading g, f, c — divide through first. Problems with two circles are not tested.
🔎 WHY IT WORKSWhy m₁m₂ = −1: turning a line 90° swaps the "rise" and "run" and flips one sign: gradient a/b becomes −b/a. Multiply them: (a/b)(−b/a) = −1. A circle equation is just Pythagoras: every point (x, y) is distance r from the centre (a, b), so (x − a)² + (y − b)² = r².
🧠 Worked example — follow the thinking
A(1, 2), B(5, −4). Find the perpendicular bisector of AB.
StepWhy I do this
Midpoint = ((1 + 5)/2, (2 − 4)/2) = (3, −1)💭 A bisector passes through the MIDDLE of AB.
m_AB = (−4 − 2)/(5 − 1) = −3/2💭 Gradient = change in y ÷ change in x.
Perpendicular m = 2/3💭 Flip and change sign so that m₁m₂ = −1.
y + 1 = ⅔(x − 3) → y = ⅔x − 3💭 Use y − y₁ = m(x − x₁) with the midpoint.
A Math · Sec 3 · SEC K341 (G3)
📏
G2

Linear Law

converting to Y = mX + c
BIG IDEATurn a curved relationship into a straight line Y = mX + c, read m and c from the graph, then find the constants.
GivenRearrangeYXmc
y = axⁿlg y = n lg x + lg alg ylg xnlg a
y = kbˣlg y = (lg b)x + lg klg yxlg blg k
y = ax² + bxy/x = ax + by/xxab
y = a/x + by = a(1/x) + by1/xab
lg xlg y(0.2, 1.1)(0.8, 2.3)c = 0.7
straight line → read gradient & intercept
Steps
① rearrange to Y = mX + c — Y and X must contain only x, y (no unknown constants)
② make a table of X, Y
③ plot, draw line of best fit
④ m = gradient, c = Y-intercept
⑤ solve for the constants
⚠ TRAPThe Y-axis intercept is only "c" if the X-axis starts at 0. If not, use c = Y − mX with a point on the line.
🔎 WHY IT WORKSWhy taking lg makes a straight line: y = axⁿ is a curve. Take lg of both sides: lg y = lg a + n lg x. If you call Y = lg y and X = lg x, this is Y = nX + lg a — exactly y = mx + c. So the gradient is n and the intercept is lg a.
🧠 Worked example — follow the thinking
y = axⁿ. The graph of lg y against lg x passes through (0.2, 1.1) and (0.8, 2.3). Find a and n.
StepWhy I do this
lg y = n lg x + lg a💭 Take lg of both sides and use the log laws to make it Y = mX + c.
n = gradient = (2.3 − 1.1)/(0.8 − 0.2) = 2💭 The gradient of the straight line IS n.
lg a = 1.1 − 2(0.2) = 0.7💭 Use c = Y − mX with a point (safe even if the axis does not start at 0).
a = 10⁰·⁷ = 5.01💭 Undo lg by raising 10 to that power.
A Math · Sec 3 · SEC K341 (G3)
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