Ext Maths · 6 Trigonometry

6.3 Exact trigonometric values

Before you start

SOH CAH TOA in a right-angled triangle (6.2).
Pythagoras' theorem (6.1).
Square roots: $\sqrt{3} \times \sqrt{3} = 3$ and ${(\frac{\sqrt{2}}{2})}^{2} = \frac{2}{4} = \frac{1}{2}$ .

Syllabus objectives covered (2)

Know the exact values of: sin x and cos x for x = 0°, 30°, 45°, 60° and 90°.
Know the exact values of: tan x for x = 0°, 30°, 45° and 60°.

Part 1 of 6

What 'exact' means, and why you must know these

An exact value is written as a whole number, a fraction or a surd, such as $\frac{1}{2}$ or $\frac{\sqrt{3}}{2}$ . A decimal such as $0.866$ is only an approximation.
These values are not on the List of formulas. On the non-calculator paper they are the only way to do trigonometry at all, so you need to know them by heart.
You don't need to memorise a table blindly. Two small triangles produce every value, and you can sketch them in ten seconds.

Part 2 of 6

The 45° triangle: half a square

Take a square with sides $1$ and cut it along a diagonal. Each half is a right-angled triangle with two sides of $1$ and two angles of $45 °$ .
Half a square gives every 45° value; half an equilateral triangle gives every 30° and 60° value.
Find the hypotenuse with Pythagoras: $1^{2} + 1^{2} = 2$ , so the hypotenuse is $\sqrt{2}$ .
Now read the ratios from the $45 °$ angle. Opposite is $1$ , adjacent is $1$ , hypotenuse is $\sqrt{2}$ .
$\sin 45 ° = \frac{1}{\sqrt{2}} \cos 45 ° = \frac{1}{\sqrt{2}} \tan 45 ° = \frac{1}{1} = 1$
$\frac{1}{\sqrt{2}}$ is the same number as $\frac{\sqrt{2}}{2}$ . Multiply top and bottom by $\sqrt{2}$ to see it. Both forms are correct, and $\frac{\sqrt{2}}{2}$ is the tidier one.
Notice $\tan 45 ° = 1$ because the triangle is isosceles: opposite and adjacent are equal.

Part 3 of 6

The 30° and 60° triangle: half an equilateral triangle

Take an equilateral triangle with sides $2$ . All its angles are $60 °$ . Drop a line from the top vertex straight down to the middle of the base.
That cuts it into two right-angled triangles. Each has a base of $1$ (half of $2$ ), a hypotenuse of $2$ , a $60 °$ angle at the bottom and a $30 °$ angle at the top (half of $60 °$ ).
The height by Pythagoras: $2^{2} - 1^{2} = 3$ , so the height is $\sqrt{3}$ .
From the $30 °$ angle at the top: opposite is $1$ , adjacent is $\sqrt{3}$ , hypotenuse is $2$ .
$\sin 30 ° = \frac{1}{2} \cos 30 ° = \frac{\sqrt{3}}{2} \tan 30 ° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
From the $60 °$ angle at the bottom, opposite and adjacent swap: opposite is $\sqrt{3}$ , adjacent is $1$ .
$\sin 60 ° = \frac{\sqrt{3}}{2} \cos 60 ° = \frac{1}{2} \tan 60 ° = \sqrt{3}$
See the swap: $\sin 30 ° = \cos 60 °$ and $\cos 30 ° = \sin 60 °$ . It is the same side of the same triangle, looked at from the other angle.

Part 4 of 6

The values at 0° and 90°

Picture a right-angled triangle with a hypotenuse of $1$ and let the angle shrink towards $0 °$ . The triangle flattens: the opposite side shrinks to $0$ and the adjacent side stretches to $1$ .
So at $0 °$ : $\sin 0 ° = \frac{0}{1} = 0$ , $\cos 0 ° = \frac{1}{1} = 1$ and $\tan 0 ° = \frac{0}{1} = 0$ .
Now let the angle grow towards $90 °$ . The triangle stands up: the opposite side grows to $1$ and the adjacent side shrinks to $0$ .
So at $90 °$ : $\sin 90 ° = 1$ and $\cos 90 ° = 0$ .
There is no value for $\tan 90 °$ , because it would mean dividing by an adjacent side of $0$ . That is why the objective lists tan only up to $60 °$ . You will see the gap as an asymptote on the tan graph in 6.4.

Part 5 of 6

The whole table, and a pattern to rebuild it

Here are all the values you need.
$\begin{matrix} x & 0 ° & 30 ° & 45 ° & 60 ° & 90 ° \\ \sin x & 0 & \frac{1}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{3}}{2} & 1 \\ \cos x & 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2} & 0 \\ \tan x & 0 & \frac{\sqrt{3}}{3} & 1 & \sqrt{3} & none \end{matrix}$
A pattern for the sine row: write it as $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$ . Simplified, that is $0, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, 1$ .
The cosine row is the sine row backwards. Read the sine row from right to left and you have cosine.
Tan is sine divided by cosine, because $\frac{O}{H} \div \frac{A}{H} = \frac{O}{A}$ . For example, $\tan 60 ° = \frac{\sqrt{3}}{2} \div \frac{1}{2} = \sqrt{3}$ .
Sense check: sine goes up from $0$ to $1$ as the angle grows, cosine goes down from $1$ to $0$ , and tan keeps growing.
To learn them, cover the table and rebuild it from the two triangles, then check. Do it daily until it takes under a minute.

Part 6 of 6

Using exact values without a calculator

Finding a side. Use SOH CAH TOA as usual, but replace sin, cos or tan with its exact value. For example, $10 \cos 30 ° = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3}$ .
Leave the answer in surd form unless told otherwise. $5 \sqrt{3}$ is exact; $8.66$ is not, and it can't be worked out without a calculator anyway.
Finding an angle. If a ratio simplifies to one of the table values, read the angle backwards. For example, if $\tan θ = \frac{\sqrt{3}}{3}$ and $θ$ is acute, then $θ = 30 °$ .
The ratio may not look like a table value at first. $\frac{5}{5 \sqrt{3}}$ cancels to $\frac{1}{\sqrt{3}}$ , which is $\frac{\sqrt{3}}{3}$ . Simplify before you look it up.
Evaluating expressions. Substitute each value in brackets, then simplify. Remember that $\sqrt{3} \times \sqrt{3} = 3$ and that $\sin^{2} x$ means ${(\sin x)}^{2}$ .

Worked examples

Example 1. Without a calculator: a right-angled triangle has a hypotenuse of $12$ cm and an angle of $60 °$ . Work out the exact lengths of the other two sides.

Show the working

From the $60 °$ angle, the side opposite uses sine and the side adjacent uses cosine, both with hypotenuse $12$ .
Opposite side:
$12 \sin 60 ° = 12 \times \frac{\sqrt{3}}{2} = 6 \sqrt{3} cm$
Adjacent side:
$12 \cos 60 ° = 12 \times \frac{1}{2} = 6 cm$
Check with Pythagoras: ${(6 \sqrt{3})}^{2} + 6^{2} = 36 \times 3 + 36 = 108 + 36 = 144 = 12^{2}$ . The sides fit.

Answer: Opposite $6 \sqrt{3}$ cm, adjacent $6$ cm.

Example 2. Without a calculator: a right-angled triangle has shorter sides of $5$ cm and $5 \sqrt{3}$ cm. Work out the angle opposite the $5$ cm side.

Show the working

From the angle you want, $5$ is opposite and $5 \sqrt{3}$ is adjacent. Opposite and adjacent means tangent.
$\tan θ = \frac{5}{5 \sqrt{3}}$
Cancel the $5$ s.
$\tan θ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
That is the exact value of $\tan 30 °$ , so $θ = 30 °$ .
Check: the angle is opposite the shorter of the two sides, so it should be the smaller acute angle. $30 °$ is smaller than the other angle, $60 °$ .

Answer: $30 °$ .

Example 3. Without a calculator, work out the exact value of $\cos 30 ° \times \tan 60 ° - \sin 30 °$ .

Show the working

Substitute each exact value.
$\frac{\sqrt{3}}{2} \times \sqrt{3} - \frac{1}{2}$
Multiply first. $\sqrt{3} \times \sqrt{3} = 3$ .
$\frac{3}{2} - \frac{1}{2}$
Subtract.
$\frac{2}{2} = 1$

Answer: $1$ .

Where people go wrong

Mistake: Swapping $\sin 30 °$ and $\cos 30 °$ , writing $\sin 30 ° = \frac{\sqrt{3}}{2}$ .

Instead: $\sin 30 ° = \frac{1}{2}$ and $\cos 30 ° = \frac{\sqrt{3}}{2}$ .

Opposite the small $30 °$ angle is the short side, $1$ . Sketch the half-equilateral triangle if in doubt. Sine also increases with the angle, so $\sin 30 °$ has to be the smaller value.

Mistake: Thinking $\frac{1}{\sqrt{3}}$ and $\frac{\sqrt{3}}{3}$ are different answers.

Instead: They are equal. Multiply top and bottom of $\frac{1}{\sqrt{3}}$ by $\sqrt{3}$ .

Mark schemes accept equivalent forms, but you have to recognise both when finding an angle from a ratio.

Mistake: Giving a decimal such as $10.4$ when an exact answer is asked for.

Instead: Keep the surd, for example $6 \sqrt{3}$ .

A decimal is an approximation. It doesn't answer a question that asks for an exact value, and on the non-calculator paper you can't produce it anyway.

Mistake: Writing a value for $\tan 90 °$ .

Instead: $\tan 90 °$ has no value.

It would need dividing by an adjacent side of $0$ .

Check yourself

Write down the exact values of $\sin 30 °$ , $\sin 45 °$ and $\sin 60 °$ .
Answer
$\frac{1}{2}$ , $\frac{\sqrt{2}}{2}$ , $\frac{\sqrt{3}}{2}$ .
Write down the exact values of $\cos 0 °$ , $\cos 60 °$ and $\cos 90 °$ .
Answer
$1$ , $\frac{1}{2}$ , $0$ .
Write down $\tan 30 °$ in two equivalent forms, and $\tan 60 °$ .
Answer
$\tan 30 ° = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ ; $\tan 60 ° = \sqrt{3}$ .
Explain why $\tan 45 ° = 1$ .
Answer
The $45 °$ triangle is half a square, so the opposite and adjacent sides are equal and their ratio is $1$ .
$x$ is acute and $\sin x = \frac{\sqrt{3}}{2}$ . Write down $x$ .
Answer
$x = 60 °$ .
Explain why $\sin 30 ° = \cos 60 °$ .
Answer
In the half-equilateral triangle, the side opposite the $30 °$ angle is the same side that is adjacent to the $60 °$ angle, and both ratios divide it by the same hypotenuse.
Work out the exact value of ${(\cos 30 °)}^{2} + {(\sin 30 °)}^{2}$ .
Answer
$\frac{3}{4} + \frac{1}{4} = 1$ .
Use the exact values to show that $\tan 60 ° = \frac{\sin 60 °}{\cos 60 °}$ .
Answer
$\frac{\sqrt{3}}{2} \div \frac{1}{2} = \frac{\sqrt{3}}{2} \times 2 = \sqrt{3}$ , which is $\tan 60 °$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. Do not use a calculator in this question.

(a) Write down the exact value of $\cos 30 °$ . 1 mark No calculator

Mark scheme

B1 √3/2 oe

(b) Write down the exact value of $\tan 45 °$ . 1 mark No calculator

Mark scheme

B1 1 cao

(c) Work out the exact value of $4 \sin 30 ° + \tan 60 ° \times \tan 30 °$ . 2 marks No calculator

Mark scheme

M1 4 × 1/2 and √3 × 1/√3 (or √3 × √3/3) seen
A1 3 cao

(d) $y$ is an acute angle and $\tan y = \sqrt{3}$ . Write down the value of $y$ . 1 mark No calculator

Mark scheme

B1 60 cao

Question 2. Do not use a calculator in this question. In triangle $P Q R$ , angle $P Q R = 90 °$ , $P R = 10$ cm and angle $Q P R = 30 °$ .

(a) Work out the exact length of $Q R$ . 2 marks No calculator

Mark scheme

M1 10 sin 30° oe
A1 5 cao (cm)

(b) Work out the exact length of $P Q$ . Give your answer in surd form. 2 marks No calculator

Mark scheme

M1 10 cos 30° oe, or √(10² − 5²) using their (a)
A1 5√3 oe (cm)

(c) Work out the exact area of triangle $P Q R$ . 2 marks No calculator

Mark scheme

M1 ½ × their (a) × their (b)
A1 25√3/2 oe (cm²); FT their (a) and (b)

Question 3. Do not use a calculator in this question. An equilateral triangle has sides of $8$ cm.

(a) Show that the height of the triangle is $4 \sqrt{3}$ cm. 3 marks No calculator

Mark scheme

M1 uses a right-angled triangle with hypotenuse 8 and an angle of 60° (or base 4)
M1 8 sin 60° = 8 × √3/2 oe, or √(8² − 4²) oe
A1 reaches 4√3 with no errors seen; nfww

(b) Work out the exact area of the triangle. 2 marks No calculator

Mark scheme

M1 ½ × 8 × 4√3 oe
A1 16√3 oe (cm²)

How this is examined

Exact side lengths in a triangle with a 30°, 45° or 60° angle: A non-calculator Work out with 'give your answer as an exact value' or 'in surd form', often in two or three parts on one triangle. The method mark is for the ratio with the exact value substituted, such as $10 \times \frac{\sqrt{3}}{2}$ . The accuracy mark is for the simplified surd, $5 \sqrt{3}$ .
Evaluating an expression built from exact values: A 2-mark Work out on the non-calculator paper, such as a sum or product of two or three trig values. Every value correctly substituted is the method mark. A fully simplified final number, using $\sqrt{3} \times \sqrt{3} = 3$ , is the accuracy mark.
Recognising an angle from an exact ratio: A 1- or 2-mark Write down, or the last step of a triangle question, where a ratio such as $\frac{1}{\sqrt{3}}$ has to be turned back into an angle. Simplifying the ratio to a table value, then stating the angle, for example $\tan θ = \frac{\sqrt{3}}{3}$ so $θ = 30 °$ .
A 'Show that' using exact trigonometry: A non-calculator Show (that) where the answer is printed, for example a height of $4 \sqrt{3}$ cm. Every step written, from the triangle through the exact value to the printed answer. Jumping straight to the given result earns little.

Words to know

Exact value: A value written without rounding, as a whole number, a fraction or a surd, such as $\frac{\sqrt{3}}{2}$ .
Surd: A square root that is not a whole number, such as $\sqrt{2}$ or $\sqrt{3}$ , left in root form.
Equilateral triangle: A triangle with all three sides equal and all three angles $60 °$ .

All Ext Maths lessons Next: 6.4 Trigonometric functions