Mathematics — Complete Formula Sheet

$p$: A statement (truth value T or F) · $\sim p$: Negation of p, read 'not p'

Negation reverses the truth value. In words, add 'not' or 'It is false that…' to the statement.

$p,q$: Component statements

Words such as 'and', 'but', 'yet', 'still' all give a conjunction. One false component makes the whole statement false.

Mathematical 'or' is inclusive: p ∨ q is true when at least one of p, q is true, including when both are.

p is the antecedent, q the consequent. Also read as 'p only if q', 'q if p', 'p is sufficient for q', 'q is necessary for p'.

Read 'p if and only if q' or 'p is necessary and sufficient for q'.

$n$: Number of simple statements in the pattern

Two statements need 4 rows, three statements need 8 rows. Use the order TT, TF, FT, FF for two statements.

Tautology: last column all T. Contradiction: last column all F. Contingency: a mixture of T and F.

The most useful equivalence of the chapter. Use it to remove → before applying De Morgan's laws or finding a dual.

All three are formed from the conditional p → q. Write each one in words when the statement is given in words.

A conditional and its contrapositive always have the same truth table. So do the converse and the inverse.

Use this to prove a biconditional or to find its negation.

Replace ∧ by ∨, ∨ by ∧, t by c and c by t (in duality the textbook writes t for a tautology and c for a contradiction). Negation signs and the statements stay as they are. Remove → and ↔ first.

Negate each component AND switch the connective. 'Not (A and B)' means 'not A or not B'.

The negation of 'If p then q' is 'p and not q' — it is NOT another if-then sentence.

Follows from p ↔ q ≡ (p → q) ∧ (q → p), then De Morgan and the negation of each conditional.

$\forall$: Universal quantifier: 'for all', 'for every' · $\exists$: Existential quantifier: 'there exists', 'for some' · $p(x)$: Open sentence in the variable x

'For every' becomes 'there exists' and vice versa; the open sentence p(x) is negated. The set (N, Z, R…) does not change.

Each connective distributes over the other. Read right to left, they let you take a common statement outside a bracket.

T is any tautology and F any contradiction. These laws finish most simplifications after a distributive step.

Row × column dot product. Defined only when columns of A = rows of B.

Matrix multiplication is NOT commutative except in special cases.

Reversal rule for transpose of product.

Cofactor expansion along first row. Use any row/column with zeros for simplicity.

Row swap changes sign; adding multiple of one row to another leaves |A| unchanged.

Transpose of the cofactor matrix. Basis of inverse formula.

Fundamental identity linking adjoint and determinant. Use to verify computed adjoint.

A is invertible iff |A| ≠ 0. Then AA⁻¹ = A⁻¹A = I.

Rewrite the system as AX = B, invert A, multiply by B.

Alternative to A⁻¹B method. Compact for 2×2 and 3×3 systems.

State all three cases in theory questions on consistency.

Preserve solution set. Used in row reduction / Gauss elimination.

Solutions of a trigonometric equation lying in [0, 2π) are its principal solutions. For sin θ = 1/2 they are π/6 and 5π/6.

$n$: Any integer (n ∈ Z)

Here n ∈ Z. Factorise the equation first and set each factor equal to zero.

$\alpha$: A known angle with the same sine (radians)

Choose α in [−π/2, π/2]. For n even the term is +α, for n odd it is −α.

Choose α in [0, π]. Do not mix this up with the sine result.

Choose α in (−π/2, π/2). The period of tan is π, so no ± and no (−1)ⁿ.

All three squared equations have the same general solution. Useful when the equation contains sin²θ = 1/4 or tan²θ = 3.

$r$: Radius vector: distance OP of the point from the pole (origin), r > 0 — the pole itself has no polar co-ordinates · $\theta$: Vectorial angle from the polar axis (positive x-axis), 0 ≤ θ < 2π

Use the signs of x and y to decide the quadrant of θ — tan θ alone does not fix it.

$a,b,c$: Sides opposite angles A, B, C · $R$: Circumradius of the triangle

Writing a = k sin A, b = k sin B, c = k sin C (k = 2R) converts any side expression into angles.

Similarly cos B = (c² + a² − b²)/(2ca) and cos C = (a² + b² − c²)/(2ab). A negative cosine means an obtuse angle.

Each side equals the sum of the projections of the other two sides on it.

$s$: Semi-perimeter, s = (a + b + c)/2

Positive square roots are taken because A/2 is acute. Write the formulas for B/2 and C/2 by cycling a → b → c.

$\Delta$: Area of triangle ABC (square units)

Use ½bc sin A when two sides and the included angle are known, Heron's form when all three sides are known.

Cyclic forms: tan((C − A)/2) = ((c − a)/(c + a)) cot(B/2) and tan((A − B)/2) = ((a − b)/(a + b)) cot(C/2).

Also cot⁻¹x ∈ (0, π), sec⁻¹x ∈ [0, π] − {π/2}, cosec⁻¹x ∈ [−π/2, π/2] − {0}. Domain of sin⁻¹ and cos⁻¹ is [−1, 1].

cos⁻¹ and cot⁻¹ follow the 'π minus' rule because their range is not symmetric about 0: cot⁻¹(−x) = π − cot⁻¹x.

Valid for |x| ≤ 1 (first), all real x (second) and |x| ≥ 1 (third).

Check the condition before using it. If x, y > 0 and xy > 1, the sum is π + tan⁻¹((x + y)/(1 − xy)).

Outside the principal range, sin⁻¹(sin θ) is NOT θ: sin⁻¹(sin 5π/6) = π/6.

Multiply the two equations written in the form u = 0, v = 0. A point lies on the pair if it lies on at least one of the lines.

$m_1,m_2$: Slopes of the two lines

Lines y = m₁x and y = m₂x. Expanding the product gives a homogeneous equation of degree two.

$a,b$: Coefficients of x² and y² · $h$: Half the coefficient of xy

Every term has degree two, so (0, 0) always satisfies it. If h² − ab < 0, only the origin satisfies the equation — there are no real lines.

Its two roots are the slopes m₁ and m₂ of the lines (when b ≠ 0). If b = 0, one line is x = 0.

Sum = −(coefficient of xy)/(coefficient of y²); product = (coefficient of x²)/(coefficient of y²).

Coincident means the equation is a perfect square, e.g. x² − 6xy + 9y² = (x − 3y)².

$\theta$: Acute angle between the two lines

θ is the acute angle. If a + b = 0 the formula breaks down because θ = 90°.

Coefficient of x² + coefficient of y² = 0, e.g. 3x² + 8xy − 3y² = 0. The value of h does not matter.

For ax² + 2hxy + by² = 0: then tan θ = 0, so the two lines coincide. For the general second-degree equation, h² = ab means the two lines are parallel.

Swap the coefficients of x² and y² and change the sign of the xy term. Check: the new slopes are −1/m₁ and −1/m₂.

$g,f$: Half the coefficients of x and y · $c$: Constant term

Read the coefficients in this form: half of the xy, x and y coefficients give h, g and f.

The textbook calls these necessary conditions: use them to find an unknown such as k. To show that an equation does represent a pair of lines, factorise it into two linear factors. The determinant is linear in c, so 'find k' questions with k as the constant are quick.

Solving the two linear equations is usually easier than remembering the formula. Requires ab − h² ≠ 0 (lines not parallel).

The second-degree part alone gives two lines through the origin parallel to the given pair, so the angle formula tan θ = |2√(h² − ab)/(a + b)| also applies to the general equation.

Component-wise addition. Scalar multiplication kâ = ka₁î + ka₂ĵ + ka₃k̂.

Magnitude is nonnegative. Unit vector has magnitude 1.

Scalar quantity. â·b̂ = cos θ; perpendicular ⟺ a·b = 0.

Vector perpendicular to both a and b. |a × b| = |a||b|sin θ. Right-hand rule.

Simple geometric use of cross product.

Volume of parallelepiped. Cyclic permutation preserves value; swap changes sign.

Three vectors are coplanar iff their scalar triple product vanishes.

a = position vector of a point on line; b = direction vector; λ ∈ R.

(l, m, n) direction ratios; (x₀, y₀, z₀) a point on line.

n̂ = unit normal, d = perpendicular distance from origin. Cartesian: lx + my + nz = d.

For planes use normals; for line and plane use sin θ = |n·b|/(|n||b|).

Only for skew lines (non-parallel, non-intersecting). Coplanar ⇒ numerator = 0.

$\vec{a}$: Position vector of a known point A on the line · $\vec{b}$: Direction vector of the line · $\lambda$: Scalar parameter (λ ∈ R)

Each value of the scalar λ gives one point on the line. Any non-zero multiple of b is also a direction vector.

$(x_1,y_1,z_1)$: A point on the line · $a,b,c$: Direction ratios of the line

Symmetric form of the same line. Setting each ratio equal to λ gives the general point (x₁ + aλ, y₁ + bλ, z₁ + cλ).

The direction vector is the vector joining the two points, b − a.

Perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0. Parallel: a₁/a₂ = b₁/b₂ = c₁/c₂. Only the direction vectors matter.

$\vec{\alpha }$: Position vector of the given point P · $\hat{b}$: Unit vector along the line

The second form is Pythagoras: AP² minus the square of the projection of AP on the line.

For lines r = a₁ + λb₁ and r = a₂ + μb₂ that are neither parallel nor intersecting. The numerator is a scalar triple product.

For r = a₁ + λb and r = a₂ + μb. This is the distance of the point A₂ from the first line.

Two non-parallel lines intersect exactly when they are coplanar, i.e. when the shortest distance is zero.

$\hat{n}$: Unit vector normal to the plane · $p$: Distance of the plane from the origin

p ≥ 0 is the perpendicular distance of the plane from the origin and (l, m, n) are the direction cosines of the normal. To reduce ax + by + cz = d (d > 0), divide by √(a² + b² + c²).

Equivalently r · n = a · n. Here (a, b, c) are direction ratios of the normal and (x₁, y₁, z₁) is the given point.

Vector form: (r − a) · [(b − a) × (c − a)] = 0. The cross product (b − a) × (c − a) is a normal to the plane.

Cartesian: (a₁x + b₁y + c₁z − d₁) + λ(a₂x + b₂y + c₂z − d₂) = 0. One more condition (a point, or parallel/perpendicular to something) fixes λ.

The angle between planes is the angle between their normals. Perpendicular planes: n₁ · n₂ = 0; parallel planes: n₁ = kn₂.

$\vec{b}$: Direction vector of the line · $\vec{n}$: Normal vector of the plane

θ is the complement of the angle between the line and the normal, so sin replaces cos.

A line parallel to the plane is perpendicular to its normal; a line perpendicular to the plane is along its normal.

$(x_1,y_1,z_1)$: The given point, with position vector α · $d_0$: Constant term of the plane ax + by + cz + d₀ = 0

First form for the plane r · n = p; second for ax + by + cz + d₀ = 0. Move every term to the left side before substituting.

The line ax + by = c divides the plane into two half-planes; the solution set of the inequation is one of them, including the line.

Valid when a, b, c are all non-zero. If c = 0 the line passes through the origin — use another point such as (1, −a/b).

If the inequality is true at (0, 0), shade the side containing the origin; otherwise shade the other side. If the line passes through the origin, test a point like (1, 0).

Availability of a resource (hours, money, storage) gives '≤'. A minimum requirement (nutrients, demand) gives '≥'.

$x,y$: Decision variables (quantities to be decided) · $Z$: Objective function (profit, cost, …) · $c_1,c_2$: Profit or cost per unit of x and y

Three parts: decision variables x, y; the linear objective function Z; and the linear constraints, including non-negativity.

Quantities produced or bought cannot be negative, so the feasible region always lies in the first quadrant.

The common region of all the half-planes. It is a convex set — bounded (a polygon) or unbounded. Each point of R is a feasible solution.

Cramer's rule for the intersection of two lines. Elimination is equally good — just do not estimate from the graph.

For a bounded feasible region both the maximum and the minimum exist. Evaluate Z at every corner point and compare.

Happens when the objective line is parallel to a boundary edge PQ of the region. Infinitely many optimal solutions.

Otherwise Z has no maximum. For a minimum, take m = smallest corner value and check that c₁x + c₂y < m has no point in R.

If the constraints contradict each other (e.g. x + y ≤ 2 and x + y ≥ 5), there is no region to shade and no optimal solution.

Ab initio definition; used when derivative rules do not apply directly.

Fundamental algebra of derivatives. Do not confuse product and quotient rules.

Differentiate outer function, keep inner unchanged, multiply by inner derivative.

Baseline table; memorise all six trigonometric derivatives and their inverses.

Note the domains: |x| < 1 for sin⁻¹; all real x for tan⁻¹.

Take log both sides, then differentiate implicitly. Essential for x^x-type problems.

Alternative: differentiate both sides w.r.t. x treating y as function of x; then solve for dy/dx.

Cancel dt; use chain rule. For d²y/dx² differentiate w.r.t. x again and use dt/dx.

Do NOT just do (d²y/dt²)/(d²x/dt²) — that gives a wrong answer.

Apply the derivative n times. Used in Taylor series, mean-value theorem.

Useful in log-differentiation to simplify products/quotients before differentiating.

Applies to related-rates: chain rule links two changing quantities.

First-order Taylor expansion. dy = f'(x) dx is the differential of y.

Normal is perpendicular to tangent. Slope product = −1.

Point-slope form. Verify the point (x₀, y₀) lies on the curve first.

Establish sign of f'(x) on the interval. Zero at endpoints doesn't spoil monotonicity.

Sign change + to − ⇒ local maximum. Sign change − to + ⇒ local minimum.

Inconclusive if f''(x₀) = 0; use higher-order test or first-derivative test.

c_i are critical points inside [a,b]. Compare all these values.

State all three conditions explicitly for full marks.

Generalisation of Rolle's; no requirement f(a) = f(b).

Useful for volume of a sphere, area of a circle, cube — small measurement errors.

m₁, m₂ are slopes of the two curves at the point of intersection.

Links definite integration to antiderivatives. F is any antiderivative of f.

Basic table; add +C in indefinite form.

Memorise all six. Sign errors are the most common HSC mistake.

Note the ln a in denominator for general base a.

Change variable so that du = g'(x) dx and integrand becomes tractable.

ILATE order: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. Choose u higher up in the list.

Compute A, B by cover-up rule or comparing coefficients. For repeated roots include (x−a)² term as well.

Powerful trick for symmetric integrands. Also (0,a): ∫f(x) = ∫f(a−x).

Recognise sin, cos, x symmetries. Saves half the work.

Repeated integration by parts. Similar formula for cosⁿ x.

Riemann sum definition. Use Σr = n(n+1)/2, Σr² = n(n+1)(2n+1)/6.

Choose based on the surd. Simplifies to a·cos, a·sec, a·tan respectively.

$a,b$: Lower and upper limits of integration · $n$: Number of equal sub-intervals · $h$: Width of each sub-interval, (b − a)/n

Divide [a, b] into n equal strips of width h, add the strip areas h·f(a + rh), then let n → ∞. Simplify the sum with the standard Σ results before taking the limit.

For an exponential integrand the sum is a G.P.: use a(rⁿ − 1)/(r − 1) and the limit (eʰ − 1)/h → 1 as h → 0.

$F(x)$: Any antiderivative (indefinite integral) of f(x)

Find any antiderivative F, substitute the upper limit, then subtract the value at the lower limit. No constant of integration is needed — it cancels.

The variable of integration is a dummy variable. This is what lets you rename t back to x at the end of every property proof.

Swapping the limits changes the sign. Used in almost every proof after a substitution reverses the limits.

Use it for modulus functions, greatest-integer functions and piecewise definitions — split at the point where the formula of f changes.

Works for any limits a to b. Typical use: an integrand like √x/(√x + √(a + b − x)), where the two integrands add up to 1, so 2I = b − a.

The special case of the a + b − x property with lower limit 0. With a = π/2 it swaps sin x and cos x.

Split at a, then substitute x = 2a − t in the second piece. Leads directly to the next property.

Example: sin(π − x) = sin x, so ∫₀^π sin x dx = 2∫₀^{π/2} sin x dx = 2. And cos(π − x) = −cos x, so ∫₀^π cos x dx = 0.

Test f(−x) for the whole integrand. x², cos x, |x| are even; x³, sin x, tan x are odd; odd × even is odd, odd × odd is even.

Change the limits to u = g(a) and u = g(b) as soon as you substitute; then you never need to return to x.

Choose u by the ILATE order, as in indefinite integration. Apply the limits to the first term as well as to the remaining integral.

Holds for any power n (including n = 1/2, i.e. √sin x). The same idea works with tan x and cot x.

Here log means natural logarithm (base e), as in the textbook. Learn the proof — it uses three different properties.

$A$: Area of the region (sq. units) · $y=f(x)$: Equation of the curve · $a,b$: x-coordinates of the bounding vertical lines

Region between y = f(x), the x-axis and the lines x = a, x = b. Uses vertical strips of height y and width dx.

$x=g(y)$: Equation of the curve solved for x · $c,d$: y-coordinates of the bounding horizontal lines

Region between x = g(y), the y-axis and the lines y = c, y = d. Uses horizontal strips; the limits are y-values.

The integral comes out negative when the curve lies below the axis; the area is its absolute value.

Here f ≥ 0 on [a, c] and f ≤ 0 on [c, b]. Find every root c between a and b and split there; otherwise positive and negative parts cancel.

$f(x)$: Upper curve · $g(x)$: Lower curve

Always upper curve minus lower curve. The limits a and b are usually the x-coordinates of the points of intersection.

$\varphi (y)$: Right-hand curve, x = φ(y) · $\psi (y)$: Left-hand curve, x = ψ(y)

Right curve minus left curve, with y-limits. Useful when the curves are naturally given as x in terms of y.

Solve the two equations simultaneously. Check which curve is on top by substituting one x-value between a and b.

Comes from ∫√(a² − x²) dx = (x/2)√(a² − x²) + (a²/2) sin⁻¹(x/a) + c. Needed for every circle and ellipse question.

The circle is symmetric about both axes, so take 4 × the first-quadrant area, where y = √(a² − x²).

$a,b$: Semi-axes along the x- and y-axes (semi-major and semi-minor when a > b)

In the first quadrant y = (b/a)√(a² − x²). For a = b it reduces to the circle πa².

The parabola is symmetric about the x-axis, so double the area above it, where y = 2√(ax).

The curves meet at (0, 0) and (4a/m², 4a/m). The parabola is the upper curve between them. Use it to check your answer, but show the integration in the exam.

The parabolas meet at (0, 0) and (4a, 4a). y² = 4ax is the upper curve between them. More generally, y² = 4ax and x² = 4by enclose 16ab/3.