GCSE trigonometry: sin, cos, tan, exact values and the special triangles

Trigonometry is one of the first topics of Year 10 on the Higher tier, and one of the pillars of the GCSE Higher papers. Unlike Year 9, where a calculator and rounded decimals were enough, the Higher tier expects exact results on the non-calculator paper: √3/2 instead of 0.866, and √2/2 instead of 0.707. This guide goes back over the basics of sin, cos and tan in a right-angled triangle, presents the identities you need to know by heart, explains when to use the 30-60-90 and 45-45-90 triangles, and shows how to combine Pythagoras with trigonometry in real exam questions.

Trigonometry is one of the first topics of Year 10 on the GCSE Higher tier, and one of the pillars of the Higher papers. True, you already met it in Year 9 — but in Year 10 the story changes: not only are new identities and combinations with Pythagoras and compound shapes added, but a completely different level of precision is required. The specification expects exact results with surds on the non-calculator paper — √3/2 and not 0.866 — and a student who writes the decimal approximation will lose marks even if the calculation is correct.

This guide sums up everything you need to know about the first geometry chapter of Year 10: definitions, exact values, fundamental identities, the special triangles, and the typical combinations that appear again and again in recent papers.

Why is Year 10 trigonometry harder than Year 9?

In Year 9 the aim was one thing: to understand what sin, cos and tan are, and to solve a simple right-angled triangle with a calculator. In Year 10 the task widens in three directions:

• **Precision instead of approximation** — the Higher tier values exact answers. The value of sin(60°) is √3/2, full stop. Any rounding to a decimal on Paper 1 counts as imprecision and loses marks.
• **Combining with compound shapes** — no more 'a single triangle', but a trapezium in which you need to drop a perpendicular and extract a 30-60-90 triangle, a rhombus where you first prove that the diagonals are perpendicular, or a kite that needs splitting into two right-angled triangles.
• **Algebraic identities** — sin²α + cos²α = 1, tan α = sin α / cos α, sin(90° − α) = cos α. These identities appear in every other question and must be known by heart.

The combination of the three makes this chapter the first filter of the Higher tier — students who skip the exact basics struggle with coordinate geometry and with graphs of trigonometric functions later in the course.

Revision: sin, cos and tan in a right-angled triangle

Let's go back over the foundation for a moment. In a right-angled triangle with an acute angle α, we define:

• **sin α = opposite / hypotenuse** — the ratio of the side opposite the angle α to the hypotenuse.
• **cos α = adjacent / hypotenuse** — the ratio of the side next to the angle (but not the hypotenuse) to the hypotenuse.
• **tan α = opposite / adjacent** — the ratio of the opposite side to the adjacent side.

The standard mnemonic is **SOH-CAH-TOA**: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent. If this base is not solid, it is worth going back to the Year 9 introduction to trigonometry before carrying on.

Two directions of use: finding a **side** when an angle and another side are given (multiply or divide), and finding an **angle** when two sides are given (sin⁻¹, cos⁻¹, tan⁻¹ on the calculator). In Year 10 you are expected to choose the right function within three seconds at most.

Exact values (not 0.866 but √3/2)

This is the critical difference between Year 9 and Year 10. In Year 9 it was fine to use a calculator and write sin(60°) ≈ 0.866. In Year 10, on the Higher tier, you must write the exact value. Why? Because mathematics is a domain of absolute truth, and 0.866 is only an approximation — the digits go on for ever. GCSE Paper 1 is non-calculator, and exact trig values are examined on it directly.

The table every Higher-tier student must know by heart — including in the exams where no calculator is used:

Two tricks that help you remember:

• **The 'growing root' trick**: sin of 0°, 30°, 45°, 60°, 90° is √0/2, √1/2, √2/2, √3/2, √4/2 — that is 0, 1/2, √2/2, √3/2, 1. The same sequence for cos, only in reverse.
• **Symmetry about 45°**: sin(30°) = cos(60°), sin(60°) = cos(30°). That is no accident — they are complementary angles, and sin(α) = cos(90° − α) always.

The iron rule in the exam: **if the question did not explicitly ask for a decimal approximation, do not write a decimal**. An answer like 'the area is 25√3/4 cm²' is always preferable to 'about 10.83 cm²'.

Fundamental identities to know by heart

Four identities are the central tool of every trigonometric solution in Year 10. Anyone who does not know them is left with an equation and no way to simplify it.

1. **The Pythagorean identity: sin²α + cos²α = 1.** Source: Pythagoras' theorem. If sin α = 3/5 is given, you immediately get cos²α = 1 − 9/25 = 16/25, so cos α = 4/5 (for an acute angle).
2. **The tangent ratio: tan α = sin α / cos α.** Lets you move freely between the three functions. If you have sin and cos, you have tan too without going back to the triangle.
3. **Complementary angles: sin(90° − α) = cos α, cos(90° − α) = sin α.** Lets you 'swap' between the two acute angles of the same right-angled triangle.
4. **Tangent and the complementary angle: tan(90° − α) = 1/tan α.** Less common, but appears in questions that mix the two shorter sides.

**An example of using the Pythagorean identity:** tan α = 2 is given for an acute angle. Find sin α and cos α.

Solution: from tan α = sin α / cos α = 2 it follows that sin α = 2cos α. Substitute into the Pythagorean identity: (2cos α)² + cos²α = 1 → 4cos²α + cos²α = 1 → 5cos²α = 1 → cos α = 1/√5 = √5/5. So sin α = 2√5/5. Note — an exact answer, with no approximations.

Pythagoras + trigonometry — the exam combinations

The classic GCSE Higher question does not look like 'calculate sin(30°)'. It looks like this: 'In rectangle ABCD, AB = 8 cm and angle BAC = 30° (where AC is a diagonal). Calculate BC, and the area of triangle ABC.'

In a question like this you go through two stages: first use trigonometry to find a side (tan 30° = BC/AB → BC = 8·tan 30° = 8√3/3), then use Pythagoras or trigonometry again to find the diagonal AC, and finish by calculating the area. This combination is the DNA of the geometry section of the Higher papers.

**Recommended working method:**

1. **Redraw the shape** roughly to scale on your working page, and mark all the information — sides and angles — with circles.
2. **Identify the relevant right-angled triangle** inside the compound shape. In a trapezium it is often a perpendicular dropped from one parallel side to the other; in a rhombus it is the triangle formed by half a diagonal.
3. **Choose between trigonometry and Pythagoras** according to what is given. An angle given? Trigonometry. Three sides and no angle? Pythagoras to calculate the third side.
4. **Write an exact result at every stage** — do not round in the middle, because the error accumulates and leads to a wrong final answer.

An important tip: sometimes the same question can be solved in two ways — once through sin and once through cos. Both get full marks, as long as the steps are justified. Do not look for 'the right way' — look for the way that is clear to you.

The 30-60-90 and 45-45-90 triangles — when to use them?

These two special triangles are an enormous shortcut. Instead of calculating from scratch each time through the definition of sin and cos, you can use the fixed side ratios.

The 30-60-90 triangle

Side ratios: **1 : √3 : 2**. If the short side (opposite 30°) is 1, then the long side (opposite 60°) is √3, and the hypotenuse is 2.

**How to remember:** the short side is always **half the hypotenuse**. The long side is the short side times √3.

**Where it appears:** every time there is an equilateral triangle in which you drop a perpendicular (you get two 30-60-90 triangles), in a rhombus with an angle of 60° or 120°, and in divisions of a regular hexagon.

The 45-45-90 triangle

Side ratios: **1 : 1 : √2**. The two shorter sides are equal (hence 'isosceles right-angled'), and the hypotenuse is a shorter side times √2.

**Where it appears:** in the diagonal of a square (the diagonal splits the square into two 45-45-90 triangles), in half of a rhombus with a 90° angle, and in any problem involving 'an angle of 45°' with the two shorter sides equal.

**Example:** in a square of side 4 cm, the length of the diagonal is 4√2 cm. Many students trip up here because they try to calculate sines — but the ratio is simply side × √2.

Five tips for solving quickly

1. **Learn the table of exact values** until you can write all five rows with your eyes closed. It saves 30 seconds per question.
2. **Identify all the right-angled triangles in the shape first** — mark them in a different pen on your working page. Most questions become simple the moment you isolate the right triangle.
3. **Do not round in the middle of a solution** — keep surds and fractions until the final answer. Early rounding damages accuracy and also hides neat calculations that cancel (for example √3 · √3 = 3).
4. **If you are stuck — try another identity.** sin²α + cos²α = 1 solves half the questions where only sin (or only cos) is given. tan α = sin α / cos α solves the other half.
5. **Check that the answer makes sense** — sin is always between 0 and 1 for an acute angle. If you got sin α = 1.4, there is an error. If you got an angle of 95° in a right-angled triangle — an error.

Summary and the next practice

Trigonometry in Year 10 on the Higher tier is not new material — it is a deepening of what you learnt in Year 9, with a requirement for mathematical precision and command of the identities. Anyone who invests a fortnight at the start of the year in full command of the table of exact values, the four basic identities and the two special triangles — gets a solid base for the rest of the year, including for the coordinate geometry and the trigonometric graphs that follow.

The next step is practice — at least 30 exam questions from recent series, including combinations of trigonometry with a rectangle, rhombus, trapezium and isosceles triangle. The more you solve, the more automatic the recognition of 'which special triangle is hiding here' becomes.

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