O-Level A-Math Formula Sheet: PDF and Worked Examples

Additional Mathematics Formulae PDF

Use this PDF as your clean reference copy when practising O-Level A-Math questions. This one-page reference contains algebra and trigonometry formulas; it is not a complete set of revision notes.

• Format: PDF
• Use: O-Level Additional Mathematics revision
• Best for: Printing, annotation, and timed practice

Download AMF PDF

Examination reference versus revision guide: The formula content in this hosted PDF matches the mathematical-formulae page in SEAB’s 2026 syllabus (page 9), checked on 18 September 2026. The worked examples below are original TGM revision material, not SEAB questions or extra formulas supplied in the exam. Check your own examination year’s syllabus if you are taking a different examination.

The AMF formula sheet includes key results from algebra and trigonometry. The formula page contains:

• Quadratic equation formula

• Binomial expansion
• Basic trigonometric identities
• Compound angle formulas
• Double angle formulas
• Sine rule, cosine rule, and triangle area formula

For example, the sheet gives the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

It also gives the binomial expansion. If this topic is still shaky, revise the full Binomial Theorem study guide alongside the formula sheet. The most useful part to understand during practice is the general term:

$${n \choose r}a^{n-r}b^r$$

and common trigonometric identities such as:

$$\sin^2 A + \cos^2 A = 1$$

The sheet is useful because it reduces memory load. But it does not remove the need for understanding. A student still has to know which topic is being tested and which formula fits the question.

Use this map to find the right part of the Additional Mathematics Formulae sheet quickly during timed practice.

The laws of indices guide is a useful companion for algebraic simplification. For trigonometry, revise the R-formula and trigonometric functions guide; the R-formula method still requires coefficient comparison even when supporting identities are provided.

A tempting assumption is: "If the formula is given, I do not need to learn it properly." The missing skill is deciding when it applies.

In A Math, marks are often lost before the formula is even used. Students may choose the wrong identity, substitute the wrong value, miss a condition, or write unclear working.

So the goal is not just to "know where the formula is". The goal is to know why the formula applies.

The algebra section looks short, but it appears across many chapters. For algebra foundations behind these formulas, use the Indices, Surds, and Logarithms study guides as supporting practice.

Quadratic formula

For $ax^2 + bx + c = 0$, where $a\ne0$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Students should not use this blindly. If the quadratic factorises neatly, factorisation is usually faster. If the question is about the nature of roots, tangent conditions, or intersections, the discriminant $b^2 - 4ac$ may be the main idea.

Binomial expansion

The formula sheet gives the expansion, but students still need to handle the general term:

$${n \choose r}a^{n-r}b^r$$

This is where many mistakes happen. Students mix up the term number with the value of $r$, forget negative signs, or substitute the wrong expression for $a$ and $b$.

Question: Find the minimum value of $y=2x^2-12x+11$ and the value of $x$ where it occurs. Then solve ${2x^2-12x+11=0}$ exactly.

Decision: Completing the square exposes the minimum and can then give the roots. The quadratic formula finds roots directly, but roots alone do not answer the minimum-value question.

Working: Factor out ${2}$ from the quadratic and linear terms first:

$$y=2(x^2-6x)+11$$

$$y=2\bigl((x-3)^2-9\bigr)+11$$

$$y=2(x-3)^2-7.$$

Since $(x-3)^2\ge0$ for real $x$, the minimum is $-7$ at $x=3$.

For the roots, set $y=0$:

$$2(x-3)^2-7=0$$

$$(x-3)^2=\frac72$$

$$x=3\pm\sqrt{\frac72}=3\pm\frac{\sqrt{14}}2.$$

As a check, the quadratic formula with $a=2$, $b=-12$, $c=11$ gives:

$$x=\frac{12\pm\sqrt{144-88}}4=3\pm\frac{\sqrt{14}}2.$$

Plausible error: Writing $y=2(x-3)^2+11-9$ and claiming the minimum is ${2}$. The $-9$ is inside brackets multiplied by ${2}$, so the adjustment is $-18$, not $-9$.

Correction: Expand your completed-square form: ${2(x-3)^2-7=2x^2-12x+11}$. The two roots lie equally far from $x=3$, which also checks the axis of symmetry.

Method reminder: Factorise when the factors are easy to spot; complete the square for a turning point; use the discriminant for the number or nature of real roots. Choose from the question’s target, not merely from the formula available.

Challenge the incorrect expression: At $x=3$, the original expression gives ${2(9)-12(3)+11=-7}$. The incorrect expression gives ${2(0)^2+11-9=2}$. They give different outputs for the same input, so they cannot be equivalent. One counterexample disproves an identity; one matching value would not prove it.

Another way to see the minimum: Start with the graph $y=x^2$. The form $y=2(x-3)^2-7$ shifts its vertex three units right and seven units down, and doubles its vertical scale. Its lowest point is $(3,-7)$. The algebra and the graph describe the same feature.

Question: Find the coefficient of $x^3$ in $(2-x)^7$, and identify the term’s position when the expansion is written in ascending powers of $x$.

Decision: Only one coefficient is needed, so use the general term instead of expanding all eight terms. Here $a=2$, $b=-x$ and $n=7$.

Working: Counting from $r=0$,

$$T_{r+1}=\binom7r2^{7-r}(-x)^r.$$

The power of $x$ is $r$, so choose $r=3$:

$$T_4=\binom732^4(-x)^3$$

$$T_4=35(16)(-x^3)=-560x^3.$$

The coefficient is $-560$; the term is $-560x^3$; its position is fourth, because the constant term comes first.

Plausible error: Taking $r=2$ because “$x^3$ means the third term”. That actually gives

$$T_3=\binom722^5(-x)^2=672x^2,$$

which has the wrong power.

Correction: Match the power before identifying the position. Here $r=3$ and the position is $r+1=4$. Keep the minus sign inside $(-x)^3$: an odd power leaves it negative.

If both parts of a binomial contain $x$, do not assume the resulting power is $r$. Combine their powers first, then solve for $r$. The Binomial Theorem study guide extends this to other term and coefficient questions.

Build the term rather than memorise its position: Think of $(2-x)^7$ as seven factors $(2-x)$ multiplied together. To produce $x^3$, choose $-x$ from exactly three factors and ${2}$ from the other four. Each choice contributes $(-x)^3\,2^4=-16x^3$. There are $\binom73=35$ ways to choose those three factors, so the total is $-560x^3$.

For one possible choice, the seven contributions look like this:

$$(-x)(-x)(-x)(2)(2)(2)(2)=-16x^3.$$

Challenge yourself: If you choose $-x$ from only two factors, can the product contain $x^3$? No: it contains two factors of $x$, so its power is $x^2$. At $x=1$, the three selected negative factors also give a negative product. This checks the sign of that contribution, not the whole expansion.

The trigonometry part of the sheet can feel crowded because many identities are printed together. Students should not wait until the exam to scan the whole page. For topic revision beyond the formula list, pair this section with the Trigonometric Functions study guide and the Sketching of Trigonometric Curves study guide.

Important identities include:

$$\sin^2 A + \cos^2 A = 1$$

$$\sec^2 A = 1 + \tan^2 A$$

$$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$$

$$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$$

For double angles, students should be comfortable moving between forms:

$$\begin{aligned}\cos 2A &= \cos^2 A - \sin^2 A \\ &= 2\cos^2 A - 1 \\ &= 1 - 2\sin^2 A\end{aligned}$$

The key question is not "Which formula did I memorise?" The better question is "What expression do I need to create?" That shift helps students choose identities more calmly.

Question: Solve $\cos2x=\sin x$ for ${0^\circ\le x\le360^\circ}$.

Start with a graph: An equation asks where two expressions have the same value. Plot the left side as $y=\cos2x$ and the right side as $y=\sin x$, with $x$ in degrees. At any solution, the curves have the same height at the same horizontal position.

On a narrow screen, scroll the graph sideways to read its labels. The blue solid and orange dashed curves meet at ${30^\circ}$ and ${150^\circ}$, and touch at ${270^\circ}$. A touching point is still a solution: the curves do not have to cross. Their shared heights are one half, one half and minus one respectively.

Use this picture to understand what the answers mean. The graph suggests where to look; the algebra below establishes the exact angles and checks that no solutions are missed.

Decision: The right side contains $\sin x$, so choose $\cos2x=1-2\sin^2x$. This produces a quadratic in one trigonometric function. The equivalent form ${2\cos^2x-1}$ is valid, but leaves both sine and cosine in this equation.

Working: Substitute and rearrange:

$$1-2\sin^2x=\sin x$$

$$2\sin^2x+\sin x-1=0$$

$$(2\sin x-1)(\sin x+1)=0.$$

Hence $\sin x=\frac12$ or $\sin x=-1$.

Within the stated interval, $\sin x=\frac12$ gives $x=30^\circ,150^\circ$, while $\sin x=-1$ gives $x=270^\circ$. Therefore

$$x=30^\circ,\ 150^\circ,\ 270^\circ.$$

Plausible error: Using $\cos2x=1-2\sin x$. The square belongs to the sine value; $\sin^2x=(\sin x)^2$ and is not $\sin x$ or $\sin(x^2)$.

Correction: Copy the identity with its square before substituting. Keep both factors when solving, then use the interval to find all angles. A calculator’s inverse-sine answer alone would miss ${150^\circ}$ and would need adjustment from $-90^\circ$ to ${270^\circ}$.

Check in the original equation: $\cos60^\circ=\sin30^\circ=\frac12$, $\cos300^\circ=\sin150^\circ=\frac12$, and $\cos540^\circ=\sin270^\circ=-1$.

This is an equation to solve, so it holds only at the answers in the interval. An identity holds wherever both sides are defined. For more background, use the trigonometric functions guide.

Challenge the missing square with a value: Is $\cos2x=1-2\sin x$ really an identity? Try $x=30^\circ$. The left side is $\cos60^\circ=\frac12$, but the proposed right side is ${1-2(\frac12)=0}$. They disagree. With the square restored, ${1-2(\frac12)^2=\frac12}$, as required. Testing $x=0^\circ$ alone would not expose the error because both versions give ${1}$ there.

When your working fails, ask what each expression means: a squared sine value, a choice from repeated factors, or the height of a graph. Fixing that meaning is more useful than simply memorising the corrected line.

If the question asks you to prove an identity rather than solve an equation, read how to start a trigonometric proof for ways to choose and justify your first step.

The AMF formula sheet also includes formulas for a triangle $ABC$:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

$$a^2 = b^2 + c^2 - 2bc\cos A$$

$$\Delta = \frac{1}{2}bc\sin A$$

A quick decision rule helps:

• Use sine rule when there is a matching side-angle pair.

• Use cosine rule when there are two sides and the included angle, or all three sides.
• Use area formula when there are two sides and the included angle.

If the angle is not between the two sides, pause. Many careless errors come from substituting a non-included angle into the area formula.

The AMF sheet does not replace topic mastery. It does not show every algebraic manipulation, proof step, or exam pattern.

Students still need to know how to:

• Factorise and simplify expressions accurately

• Complete the square when graph interpretation is required
• Solve inequalities and state ranges correctly
• Handle logarithmic and exponential equations
• Differentiate and integrate with correct notation
• Prove trigonometric identities step by step
• Check whether an answer satisfies the domain or context

When the gap is calculus rather than formula recall, revise the Differentiation, Techniques of Integration, and Applications of Integration guides before returning to exam questions.

This is why memorising math formulas alone is not enough. A Math rewards students who understand how formulas connect.

Use this practice loop:

1. Attempt with the formula sheet open. Focus on recognising the topic and selecting the correct formula.
2. Mark the question and label the mistake. Was it formula choice, substitution, algebra, sign error, or presentation?
3. Redo the same question without looking at the solution. This checks whether the correction actually stuck.
4. Attempt a similar question without opening the formula sheet immediately. Only check the sheet after deciding the method.
5. Write one sentence beside the formula. For example: "Use cosine rule when I know two sides and the included angle."

When a mistake points to a topic gap instead of formula recall, go back to the matching A-Math study guide and practise that skill directly. This turns the sheet from a crutch into a training tool. If A Math feels overwhelming beyond formulas, pair this with our guide on how to do well in Additional Mathematics.

A Math formula habits matter later. In A-Level H2 Math, students use the MF27 formula list, but the principle is the same: the booklet gives results, not strategy.

These three habits are useful preparation:

• Identify the structure of the question before choosing a formula

• Understand the conditions behind each formula
• Present working clearly enough to earn method marks

This is why A Math is a useful foundation for students who intend to take H2 Math. The formulas matter, but the thinking habits matter more.