Exact Trigonometric Values New from 2025
The exact values of sin, cos and tan for 0°, 30°, 45°, 60° and 90°.
0580 Extended · E6.2Why exact values?
When you type \(\sin(60°)\) into a calculator, it shows 0.8660254... This is a decimal approximation. The exact value is \(\frac{\sqrt{3}}{2}\).
On the non-calculator Paper 2, you cannot use a calculator, so you need to know these exact values from memory. If a question says "give your answer in exact form" or "without a calculator", this is what it's asking for.
The two special triangles
All the exact values come from just two triangles. If you can sketch these from memory, you can work out any value you need — even under exam pressure.
An isosceles right-angled triangle with two sides of length 1. By Pythagoras, the hypotenuse is \(\sqrt{2}\).
This gives the exact values for 45°.
Take an equilateral triangle with sides of length 2 and cut it in half. The result is a right-angled triangle with sides 1, \(\sqrt{3}\), and 2.
This gives the exact values for both 30° and 60°.
If you forget the table in the exam, sketch these two triangles and use SOHCAHTOA to read off the values. The 30-60-90 triangle has sides 1, \(\sqrt{3}\), 2 (shortest to longest). The 45-45-90 triangle has sides 1, 1, \(\sqrt{2}\).
The values you must know
These are the values required by the syllabus. You need to memorise all of them.
| 0° | 30° | 45° | 60° | 90° | |
|---|---|---|---|---|---|
| sin | 0 | \(\dfrac{1}{2}\) | \(\dfrac{\sqrt{2}}{2}\) | \(\dfrac{\sqrt{3}}{2}\) | 1 |
| cos | 1 | \(\dfrac{\sqrt{3}}{2}\) | \(\dfrac{\sqrt{2}}{2}\) | \(\dfrac{1}{2}\) | 0 |
| tan | 0 | \(\dfrac{1}{\sqrt{3}}\) | 1 | \(\sqrt{3}\) | undef. |
Notice that the sin row reads 0, \(\frac{1}{2}\), \(\frac{\sqrt{2}}{2}\), \(\frac{\sqrt{3}}{2}\), 1 — and the cos row is exactly the same values in reverse order. This symmetry can help you remember.
You may see \(\sin 45° = \frac{1}{\sqrt{2}}\) instead of \(\frac{\sqrt{2}}{2}\), and \(\tan 30° = \frac{\sqrt{3}}{3}\) instead of \(\frac{1}{\sqrt{3}}\). These are equivalent — the second version has a rationalised denominator. Both are accepted in the exam.
Using exact values
A right-angled triangle has hypotenuse 10 cm and an angle of 30°. Without a calculator, find the exact length of the side opposite the 30° angle.
Using SOHCAHTOA: opposite = hypotenuse × sin(30°).
\[ \begin{aligned} \text{opposite} &= 10 \times \sin 30° \\[6pt] &= 10 \times \frac{1}{2} \\[6pt] &= 5 \text{ cm} \end{aligned} \]A right-angled triangle has hypotenuse 8 cm and an angle of 60°. Find the exact length of the adjacent side.
A right-angled triangle has a side of 6 cm adjacent to a 45° angle. Find the exact length of the hypotenuse.
\(\cos 45° = \frac{\text{adjacent}}{\text{hypotenuse}}\), so hypotenuse = \(\frac{\text{adjacent}}{\cos 45°}\).
\[ \begin{aligned} \text{hypotenuse} &= \frac{6}{\frac{\sqrt{2}}{2}} \\[6pt] &= \frac{6 \times 2}{\sqrt{2}} \\[6pt] &= \frac{12}{\sqrt{2}} \\[6pt] &= \frac{12\sqrt{2}}{2} = 6\sqrt{2} \text{ cm} \end{aligned} \]Two sides of a triangle are 5 cm and 8 cm, with an included angle of 30°. Find the exact area.
Using the area formula \(\frac{1}{2}ab\sin C\):
\[ \begin{aligned} \text{Area} &= \frac{1}{2} \times 5 \times 8 \times \sin 30° \\[6pt] &= \frac{1}{2} \times 5 \times 8 \times \frac{1}{2} \\[6pt] &= 10 \text{ cm}^2 \end{aligned} \]Exam-style question
Do not use a calculator in this question. Show all your working.
In triangle \(ABC\), angle \(B = 90°\), angle \(A = 60°\) and \(AB = 12\) cm.
(a) Find the exact length of \(BC\). [2]
(b) Find the exact length of \(AC\). [2]
(c) Find the exact area of triangle \(ABC\). [3]
Part (a)
\(BC\) is opposite the 60° angle, and \(AB\) is adjacent. Using tan:
\[ \begin{aligned} \tan 60° &= \frac{BC}{AB} \\[6pt] BC &= 12 \times \tan 60° \\[6pt] &= 12\sqrt{3} \text{ cm} \end{aligned} \]Part (b)
\(AC\) is the hypotenuse. Using cos:
\[ \begin{aligned} \cos 60° &= \frac{AB}{AC} \\[6pt] AC &= \frac{12}{\cos 60°} \\[6pt] &= \frac{12}{\frac{1}{2}} = 24 \text{ cm} \end{aligned} \]Part (c)
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AB \times BC\):
\[ \begin{aligned} \text{Area} &= \frac{1}{2} \times 12 \times 12\sqrt{3} \\[6pt] &= 72\sqrt{3} \text{ cm}^2 \end{aligned} \]This is a typical non-calculator question that combines exact trig values with surd arithmetic. Notice the answers are left in exact form (\(12\sqrt{3}\), not 20.78...). If you wrote a decimal, you would lose marks. Practise leaving answers as surds until it feels natural.
Common mistakes
\(\sin 30° = \frac{1}{2}\) but \(\cos 30° = \frac{\sqrt{3}}{2}\). Getting these the wrong way round is the most common error. Remember: the sin row goes 0, \(\frac{1}{2}\), \(\frac{\sqrt{2}}{2}\), \(\frac{\sqrt{3}}{2}\), 1 — and cos is the same values in reverse.
On Paper 2, writing 0.866 instead of \(\frac{\sqrt{3}}{2}\) will lose you the mark, even if it's numerically correct. "Exact form" means surds and fractions, not decimals.
The syllabus requires tan values for 0°, 30°, 45°, and 60° only. \(\tan 90°\) is undefined (division by zero). If a question involves 90°, it will only ask for sin or cos.
If your calculation gives \(\frac{12}{\sqrt{2}}\), rationalise it to \(6\sqrt{2}\). Leaving an unsimplified surd in the denominator may cost a mark.
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