Read the core explanations from Exact Values of Trigonometric Ratios before testing yourself from memory.
1. Why Exact Values Matter
In questions without a calculator, your calculator is gone — but the examiner can still ask you to work with trigonometry. That is where exact values come in. Instead of typing $\sin 30°$ into a calculator and getting $0.5$, you need to know that result and write it down instantly.
There are five angles for sine and cosine ($0°, 30°, 45°, 60°, 90°$) and four angles for tangent ($0°, 30°, 45°, 60°$) that the examiner expects you to have memorised. These values appear in calculation questions, "show that" questions, and simplification tasks — so fluency with them is a genuine mark-winner.
2. The Complete Exact Values Table
Here is every value you need. Notice the patterns — they make memorisation far more manageable than learning fifteen separate facts.
| $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|
| $0°$ | $0$ | $1$ | $0$ |
| $30°$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{3}}$ |
| $45°$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $1$ |
| $60°$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ |
| $90°$ | $1$ | $0$ | undefined |
Two patterns worth locking in: first, $\sin$ and $\cos$ swap as you go from $30°$ to $60°$ — $\sin 30° = \cos 60° = \frac{1}{2}$ and $\sin 60° = \cos 30° = \frac{\sqrt{3}}{2}$. Second, $\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$ because a $45°$-$45°$-$90°$ triangle is isosceles.
$\tan 90°$ is left off the list entirely because it is undefined — dividing by zero. Examiners will not ask for $\tan 90°$, but knowing why it is missing shows real understanding.
3. Where These Values Come From
You do not need to derive these from scratch in the exam, but understanding the geometry behind them makes the values stick. Two special right-angled triangles generate every value in the table.
For the $30°$-$60°$-$90°$ triangle: start with an equilateral triangle of side length $2$ and cut it in half. The half-triangle has a short side of $1$, a hypotenuse of $2$, and — by Pythagoras — a vertical side of $\sqrt{2^2 - 1^2} = \sqrt{3}$. Reading off the ratios gives every $30°$ and $60°$ value directly.
For the $45°$-$45°$-$90°$ triangle: take an isosceles right-angled triangle with legs of length $1$. The hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$. Because both legs are equal, $\sin 45° = \cos 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
4. Applying Exact Values in Calculations
Knowing the table is only half the job. The exam will ask you to substitute exact values into calculations and leave your answer in exact form — often as a surd or fraction. Here is how that looks in practice.
Worked example: Finding a Side Using an Exact Value
A right-angled triangle has a hypotenuse of $10$ cm and one angle of $30°$. Find the length of the side opposite the $30°$ angle. Give your answer as an exact value. [2 marks]
- $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ — We know the hypotenuse and want the opposite side, so sine is the correct ratio. [M1 — correct trig ratio selected]
- $\sin 30° = \dfrac{\text{opposite}}{10}$ — Replace $\theta$ with $30°$ and the hypotenuse with $10$.
- $\dfrac{1}{2} = \dfrac{\text{opposite}}{10}$ — Use the known exact value $\sin 30° = \dfrac{1}{2}$ — no calculator needed.
- $\text{opposite} = 10 \times \dfrac{1}{2} = 5 \text{ cm}$ — [A1 — correct exact answer with units]
Answer: The opposite side is $5$ cm.
Worked example: Exact Value Calculation Involving a Surd
A right-angled triangle has a hypotenuse of $8$ cm and one angle of $60°$. Find the length of the side adjacent to the $60°$ angle. Give your answer as an exact value. [2 marks]
- $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ — We know the hypotenuse and want the adjacent side, so cosine is the correct ratio. [M1]
- $\cos 60° = \dfrac{\text{adjacent}}{8}$ — Replace $\theta$ with $60°$ and the hypotenuse with $8$.
- $\dfrac{1}{2} = \dfrac{\text{adjacent}}{8}$ — Use the known exact value $\cos 60° = \dfrac{1}{2}$.
- $\text{adjacent} = 8 \times \dfrac{1}{2} = 4 \text{ cm}$ — [A1 — correct exact answer with units]
Answer: The adjacent side is $4$ cm.
Worked example: Exact Value Calculation with a Surd Answer
A right-angled triangle has a hypotenuse of $6$ cm and one angle of $45°$. Find the length of the side opposite the $45°$ angle. Give your answer as an exact value. [2 marks]
- $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ — We know the hypotenuse and want the opposite side, so sine is correct. [M1]
- $\sin 45° = \dfrac{\text{opposite}}{6}$ — Replace $\theta$ with $45°$ and the hypotenuse with $6$.
- $\dfrac{\sqrt{2}}{2} = \dfrac{\text{opposite}}{6}$ — Use the known exact value $\sin 45° = \dfrac{\sqrt{2}}{2}$.
- $\text{opposite} = 6 \times \dfrac{\sqrt{2}}{2} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2} \text{ cm}$ — Multiply and simplify the fraction. $\dfrac{6}{2} = 3$, so the answer is $3\sqrt{2}$. [A1]
Answer: The opposite side is $3\sqrt{2}$ cm.
5. Deliberate Practice
These problems are structured to build fluency across all the required angles. Work through each one without a calculator, leaving surds in simplified form.
6. Exam Technique and Common Mistakes
Exact values questions are reliable mark-winners once you know the table — but there are specific traps that cost students marks every year. The guidance below consolidates the most important exam technique for this topic.
Exact trigonometric values are easier to remember when you connect them to the special triangles that generate them. The 45-45-90 triangle gives equal shorter sides, leading to sine and cosine values involving root 2, while the 30-60-90 triangle produces the familiar fractions involving 1, 2 and root 3. In the exam, these values are not approximations; they are exact and should stay in surd form unless the question explicitly asks for a decimal. A good check is symmetry: sine and cosine swap roles between complementary acute angles, which helps confirm whether a value has been matched to the correct angle.