Extended. On the non-calculator paper you still need sin, cos and tan of 30°, 45° and 60°. You do not have to memorise a table: two small triangles hold every value.
Half a square gives sides \(1, 1, \sqrt{2}\) and the \(45^\circ\) values. Half an equilateral triangle of side 2 gives sides \(1, \sqrt{3}, 2\) and the \(30^\circ\) and \(60^\circ\) values. Read off SOH CAH TOA from these triangles.
Fig. 1: the two special triangles, drawn to scale. Left: half a unit square (1, 1, √2). Right: half an equilateral triangle of side 2 (1, √3, 2)
Section 01
Where the values come from.
In the lab, slide the angle to \(30^\circ\), \(45^\circ\) and \(60^\circ\) and compare each decimal with its exact form: \(0.7071\ldots\) is \(\dfrac{1}{\sqrt{2}}\) and \(0.8660\ldots\) is \(\dfrac{\sqrt{3}}{2}\). Then build the two triangles of Fig. 1 yourself.
Half a square: cut a square of side 1 along a diagonal. The angles are \(45^\circ\), \(45^\circ\), \(90^\circ\), and Pythagoras gives the hypotenuse \(\sqrt{1^2 + 1^2} = \sqrt{2}\).
Half an equilateral triangle: cut an equilateral triangle of side 2 down its line of symmetry. The base halves to 1, the angles are \(30^\circ\), \(60^\circ\), \(90^\circ\), and the height is \(\sqrt{2^2 - 1^2} = \sqrt{3}\).
\(0^\circ\) and \(90^\circ\): picture a triangle squashed flat. \(\sin 0^\circ = 0\), \(\cos 0^\circ = 1\), \(\tan 0^\circ = 0\); \(\sin 90^\circ = 1\), \(\cos 90^\circ = 0\). (\(\tan 90^\circ\) does not exist.)
Worked example Reading values from a triangle
Use the triangles in Fig. 1 to find the exact values of \(\sin 60^\circ\), \(\cos 60^\circ\) and \(\tan 60^\circ\).
In the right-hand triangle, from the \(60^\circ\) angle: opposite \(= \sqrt{3}\), adjacent \(= 1\), hypotenuse \(= 2\).
Show that \(\sin 45^\circ = \dfrac{\sqrt{2}}{2}\).
From the left-hand triangle: \(\sin 45^\circ = \dfrac{1}{\sqrt{2}}\).
Multiply top and bottom by \(\sqrt{2}\): \(\dfrac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \dfrac{\sqrt{2}}{2}\).
Answer: \(\sin 45^\circ = \cos 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\). Both forms are accepted.
Examiner trap
Mixing up \(\sin 30^\circ\) and \(\sin 60^\circ\). Sine grows with the angle, so the smaller angle has the smaller value: \(\sin 30^\circ = \dfrac{1}{2} < \sin 60^\circ = \dfrac{\sqrt{3}}{2}\).
Section 02
The table of exact values.
Here is everything the syllabus asks for in one place. Notice the patterns: sine reads \(\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}\), and cosine is the same list backwards.
\(x\)
\(0^\circ\)
\(30^\circ\)
\(45^\circ\)
\(60^\circ\)
\(90^\circ\)
\(\sin x\)
\(0\)
\(\dfrac{1}{2}\)
\(\dfrac{\sqrt{2}}{2}\)
\(\dfrac{\sqrt{3}}{2}\)
\(1\)
\(\cos x\)
\(1\)
\(\dfrac{\sqrt{3}}{2}\)
\(\dfrac{\sqrt{2}}{2}\)
\(\dfrac{1}{2}\)
\(0\)
\(\tan x\)
\(0\)
\(\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}\)
\(1\)
\(\sqrt{3}\)
not defined
Worked example Evaluating an expression
Without a calculator, find the exact value of \(\tan 60^\circ \times \cos 30^\circ\).
\(\tan 60^\circ = \sqrt{3}\) and \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\).
Without a calculator, find the acute angle \(x\) such that \(2\cos x = \sqrt{3}\).
Divide by 2: \(\cos x = \dfrac{\sqrt{3}}{2}\).
From the table, \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\).
Answer: \(x = 30^\circ\)
Your turn
Without a calculator, find the exact value of \(\sin 30^\circ + \cos 60^\circ + \tan 45^\circ\).
\(\dfrac{1}{2} + \dfrac{1}{2} + 1 = 2\).
Examiner trap
\(\sin^2 60^\circ\) means \((\sin 60^\circ)^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4}\). It is not \(\sin(60^2)\) and not \(\dfrac{\sqrt{3}}{4}\): square the top AND the bottom.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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What is the exact value of \(\sin 30^\circ\)?
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What is the exact value of \(\tan 45^\circ\)?
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What is the exact value of \(\cos 30^\circ\)?
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Which acute angle \(x\) has \(\tan x = \dfrac{1}{\sqrt{3}}\)?
Section 03
Exact sides without a calculator.
When a right-angled triangle has a \(30^\circ\), \(45^\circ\) or \(60^\circ\) angle, SOH CAH TOA gives exact lengths in surd form.
Worked example Exact sides of a 30-60-90 triangle
In triangle \(ABC\), angle \(B = 90^\circ\), angle \(A = 30^\circ\) and \(AC = 10\) cm. Without a calculator, find \(BC\) and \(AB\) exactly.
\(BC\) is opposite \(30^\circ\): \(BC = 10\sin 30^\circ = 10 \times \dfrac{1}{2} = 5\) cm.
\(AB\) is adjacent to \(30^\circ\): \(AB = 10\cos 30^\circ = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}\) cm.
Answer: \(BC = 5\) cm, \(AB = 5\sqrt{3}\) cm
Worked example Unknown on the bottom
A right-angled triangle has an angle of \(60^\circ\) with an adjacent side of 6 cm. Find the hypotenuse and the opposite side exactly.
\(\cos 60^\circ = \dfrac{6}{h}\), so \(h = \dfrac{6}{\cos 60^\circ} = \dfrac{6}{\frac{1}{2}} = 12\) cm.
Opposite \(= 6\tan 60^\circ = 6\sqrt{3}\) cm.
Answer: Hypotenuse 12 cm, opposite \(6\sqrt{3}\) cm
Examiner trap
Dividing by a fraction: \(\dfrac{6}{\frac{1}{2}} = 12\), not 3. Dividing by \(\dfrac{1}{2}\) is the same as multiplying by 2.
Stage 2 · Exam
Exam-style questions
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Without a calculator, find the exact value of \(\sin 60^\circ \times \tan 30^\circ\).
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A right-angled triangle has hypotenuse 8 cm and an angle of \(60^\circ\). Find the exact length of the side opposite the \(60^\circ\) angle.
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Without a calculator, find the exact value of \(\cos^2 30^\circ + \sin^2 45^\circ\).
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A straight ramp 6 m long makes an angle of \(30^\circ\) with the horizontal ground. Find, exactly, the horizontal distance it covers.
Skill unlocked
Exact trigonometric values, sorted.
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