IGCSE Math RevisionExample 1
Find the exact value of \(\sin 60^\circ \times \tan 30^\circ\).
- \(\sin 60^\circ = \tfrac{\sqrt{3}}{2}\) and \(\tan 30^\circ = \tfrac{1}{\sqrt{3}}\)
- \(\tfrac{\sqrt{3}}{2} \times \tfrac{1}{\sqrt{3}} = \tfrac{1}{2}\)
Answer: \(\tfrac{1}{2}\)
Cambridge IGCSE Maths (0580) · Non-calculator · 6.3
Exact trigonometric values without a calculator
Extended students need the exact values of sin, cos and tan for the special angles. Without a calculator these are the only angles a trigonometry question can use, so they turn up in right-angled triangles, the area of a triangle and simple equations.
- Extended only
- Paper 2 (Extended)
- 8 questions
- Free, no sign-in
The method
- Draw two triangles instead of memorising a list. Half of a square with sides \(1\) has sides \(1\), \(1\) and \(\sqrt{2}\), and angles \(45^\circ\).
- Half of an equilateral triangle with sides \(2\) has sides \(1\), \(\sqrt{3}\) and \(2\), and angles \(30^\circ\) and \(60^\circ\).
- Read off SOH CAH TOA from the triangles: for example \(\sin 30^\circ = \tfrac{1}{2}\) and \(\tan 60^\circ = \sqrt{3}\).
- \(0^\circ\) and \(90^\circ\): \(\sin 0^\circ = 0\), \(\cos 0^\circ = 1\), \(\tan 0^\circ = 0\), \(\sin 90^\circ = 1\), \(\cos 90^\circ = 0\).
- Use the values exactly in the working, and simplify any surds at the end.
| Angle | sin | cos | tan |
|---|---|---|---|
| 0° | 0 | 1 | 0 |
| 30° | 1/2 | √3/2 | √3/3 (= 1/√3) |
| 45° | √2/2 (= 1/√2) | √2/2 (= 1/√2) | 1 |
| 60° | √3/2 | 1/2 | √3 |
| 90° | 1 | 0 | not defined |
Worked examples
Example 2
The longest side of a right-angled triangle is \(14\) cm. One of its angles is \(60^\circ\).
Work out the exact lengths of the two shorter sides.
- Opposite the \(60^\circ\) angle: \(14 \sin 60^\circ = 14 \times \tfrac{\sqrt{3}}{2} = 7\sqrt{3}\) cm
- Next to it: \(14 \cos 60^\circ = 14 \times \tfrac{1}{2} = 7\) cm
Answer: \(7\sqrt{3}\) cm and \(7\) cm
Common slips
- Swapping sin and cos: \(\sin 30^\circ = \cos 60^\circ = \tfrac{1}{2}\), but \(\sin 60^\circ = \tfrac{\sqrt{3}}{2}\).
- Writing a rounded decimal such as \(0.866\) when the question asks for an exact value.
- \(\sin^2 45^\circ\) means \((\sin 45^\circ)^2\), which is \(\tfrac{1}{2}\).
Practice questions
No calculator. Write down every step, then open the worked answer.
Question 1
Write down the exact value of \(\cos 45^\circ\).
Show the worked answer
- From the half-square triangle: adjacent \(1\), hypotenuse \(\sqrt{2}\), so \(\cos 45^\circ = \tfrac{1}{\sqrt{2}}\).
Answer: \(\tfrac{\sqrt{2}}{2}\) (or \(\tfrac{1}{\sqrt{2}}\))
Question 2
Find the exact value of \(2\sin 30^\circ + \cos 60^\circ\).
Show the worked answer
- \(2 \times \tfrac{1}{2} + \tfrac{1}{2} = 1 + \tfrac{1}{2}\)
Answer: \(\tfrac{3}{2}\)
Question 3
Find the exact value of \(\tan 60^\circ \times \sin 60^\circ\).
Show the worked answer
- \(\sqrt{3} \times \tfrac{\sqrt{3}}{2} = \tfrac{3}{2}\)
Answer: \(\tfrac{3}{2}\)
Question 4
In triangle \(ABC\), angle \(B = 90^\circ\), angle \(A = 60^\circ\) and \(AB = 5\) cm.
Find the exact length of \(BC\).
Show the worked answer
- \(BC\) is opposite angle \(A\) and \(AB\) is adjacent to it, so \(\tan 60^\circ = \tfrac{BC}{5}\).
- \(BC = 5 \tan 60^\circ = 5\sqrt{3}\)
Answer: \(5\sqrt{3}\) cm
Question 5
In triangle \(PQR\), angle \(R = 90^\circ\), angle \(P = 45^\circ\) and \(PQ = 10\) cm.
Find the exact length of \(PR\). Give your answer in its simplest form.
Show the worked answer
- \(PQ\) is the hypotenuse and \(PR\) is adjacent to angle \(P\), so \(PR = 10 \cos 45^\circ\).
- \(10 \times \tfrac{\sqrt{2}}{2} = 5\sqrt{2}\)
Answer: \(5\sqrt{2}\) cm
Question 6
Find the exact value of \(\tan^2 30^\circ + \cos^2 60^\circ\). Give your answer as a fraction.
Show the worked answer
- \(\tan^2 30^\circ = \left(\tfrac{1}{\sqrt{3}}\right)^2 = \tfrac{1}{3}\)
- \(\cos^2 60^\circ = \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}\)
- \(\tfrac{1}{3} + \tfrac{1}{4} = \tfrac{4}{12} + \tfrac{3}{12} = \tfrac{7}{12}\)
Answer: \(\tfrac{7}{12}\)
Question 7
Solve \(2\cos x = \sqrt{3}\) for \(0^\circ \le x \le 90^\circ\).
Show the worked answer
- \(\cos x = \tfrac{\sqrt{3}}{2}\)
- From the half-equilateral triangle, \(\cos 30^\circ = \tfrac{\sqrt{3}}{2}\).
Answer: \(x = 30^\circ\)
Question 8
Two sides of a triangle are \(6\) cm and \(8\) cm, and the angle between them is \(60^\circ\).
Find the exact area of the triangle.
Show the worked answer
- Area \(= \tfrac{1}{2}ab\sin C = \tfrac{1}{2} \times 6 \times 8 \times \sin 60^\circ\)
- \(= 24 \times \tfrac{\sqrt{3}}{2} = 12\sqrt{3}\)
Answer: \(12\sqrt{3}\) cm²
Every example and question on this page is original, written for IGCSE Math Revision rather than taken from Cambridge papers, and each answer was re-solved independently before publishing. The syllabus is summarised in our own words: check the official 0580 syllabus.
Keep going
- Exact Trigonometric Values questions for 0580: Core and Extended, each labelled calculator or non-calculator
- Trigonometry practice for 0580, then My mistakes and a Skill Builder on what you missed
- Exact trigonometric values notes and the 0580 formula sheet
- 0580 past papers: time yourself on a specimen Paper 2
- All six non-calculator methods
Questions students ask
Which 0580 papers test exact trigonometric values without a calculator?
Exact trigonometric values is Extended content (syllabus 6.3), so Extended candidates can meet it on Paper 2, the Extended non-calculator paper, as well as on Paper 4 with a calculator. Core candidates do not need it.
Are these Cambridge questions?
No. The method is explained in our own words, and every example and question was written for IGCSE Math Revision and checked independently before publishing. For real papers, use the Cambridge copies linked from our 0580 past papers page.