Maths · 7 Trigonometry

7.3 Exact trigonometric values

Before you start

The three trigonometric ratios and how to label a right-angled triangle (7.2).
Pythagoras' theorem (7.1).
Simplifying surds, e.g. $\sqrt{12} = 2 \sqrt{3}$ , and rationalising $\frac{1}{\sqrt{2}}$ .

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 4

Why these values matter

The word in the objective is know. These values must be recalled — they are not on the formula list, and in Paper 2 you have no calculator at all.
They also let you give an exact answer. $\sin 60 ° = \frac{\sqrt{3}}{2}$ is exact; $0.866$ is a rounded approximation.
When a question says 'give your answer in surd form' or 'give an exact answer', it is telling you to use these values rather than the calculator.
Here is the full table. Learn it, but also learn where it comes from — the next two sections show why, and a derivation you understand will outlast a table you crammed.
$\sin$ : $0 ° \to 0$ , $30 ° \to \frac{1}{2}$ , $45 ° \to \frac{\sqrt{2}}{2}$ , $60 ° \to \frac{\sqrt{3}}{2}$ , $90 ° \to 1$ .
$\cos$ : $0 ° \to 1$ , $30 ° \to \frac{\sqrt{3}}{2}$ , $45 ° \to \frac{\sqrt{2}}{2}$ , $60 ° \to \frac{1}{2}$ , $90 ° \to 0$ .
$\tan$ : $0 ° \to 0$ , $30 ° \to \frac{1}{\sqrt{3}}$ , $45 ° \to 1$ , $60 ° \to \sqrt{3}$ .
Notice the pattern: the cosine row is the sine row backwards. $\cos 30 ° = \sin 60 °$ , $\cos 60 ° = \sin 30 °$ , and so on. That halves what you have to remember.
Notice also that $\tan 90 °$ is missing from the objective. It is undefined — the tangent graph has an asymptote there (3.5).

Part 2 of 4

Where 45° comes from

Draw a right-angled triangle with both short sides equal to 1. The two acute angles must be equal, and since they sum to $90 °$ , each is $45 °$ .
Find the hypotenuse with Pythagoras.
$h = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$
Now read the ratios straight off. Opposite $= 1$ , adjacent $= 1$ , hypotenuse $= \sqrt{2}$ .
$\sin 45 ° = \frac{1}{\sqrt{2}} \cos 45 ° = \frac{1}{\sqrt{2}} \tan 45 ° = \frac{1}{1} = 1$
$\frac{1}{\sqrt{2}}$ and $\frac{\sqrt{2}}{2}$ are the same number — multiply top and bottom by $\sqrt{2}$ to see it. Either is accepted; the second is the rationalised form.
That $\sin 45 ° = \cos 45 °$ is no accident. The triangle is symmetrical, so the opposite and adjacent sides are the same length.

Part 3 of 4

Where 30° and 60° come from

Start with an equilateral triangle of side 2. All its angles are $60 °$ .
Cut it in half down the middle. That gives a right-angled triangle with hypotenuse 2, base 1, and angles of $30 °$ , $60 °$ and $90 °$ .
Find the height with Pythagoras.
$h = \sqrt{2^{2} - 1^{2}} = \sqrt{3}$
So the three sides are $1$ , $\sqrt{3}$ and $2$ — worth memorising as a set, because every value in this table comes out of it.
Two triangles generate the whole table.
From the $30 °$ angle: the opposite side is 1, the adjacent is $\sqrt{3}$ , the hypotenuse is 2.
$\sin 30 ° = \frac{1}{2} \cos 30 ° = \frac{\sqrt{3}}{2} \tan 30 ° = \frac{1}{\sqrt{3}}$
From the $60 °$ angle the opposite and adjacent swap over, since it is the other acute angle in the same triangle.
$\sin 60 ° = \frac{\sqrt{3}}{2} \cos 60 ° = \frac{1}{2} \tan 60 ° = \sqrt{3}$
That swap is exactly why the sine and cosine rows are reverses of each other. If you can sketch these two triangles, you can rebuild the entire table in under a minute.

Part 4 of 4

0° and 90°, and using the values

$0 °$ and $90 °$ do not come from a triangle — you cannot draw a triangle with a $0 °$ angle. Read them off the graphs from 3.1 instead.
The sine curve starts at 0 and peaks at 1 when $x = 90 °$ . So $\sin 0 ° = 0$ and $\sin 90 ° = 1$ .
The cosine curve starts at 1 and falls to 0 at $90 °$ . So $\cos 0 ° = 1$ and $\cos 90 ° = 0$ .
$\tan 0 ° = 0$ , and $\tan 90 °$ is undefined — which is why the objective stops at $60 °$ for tangent.
In use, substitute the exact value and keep it exact. For a triangle with hypotenuse 8 and angle $30 °$ : opposite $= 8 \sin 30 ° = 8 \times \frac{1}{2} = 4$ exactly.
Working with surds, keep them in surd form to the end. $10 \cos 30 ° = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3}$ — do not convert to $8.66$ unless the question asks for a decimal.
These values also let you recognise an angle from a ratio. If $\sin θ = \frac{\sqrt{3}}{2}$ , then $θ = 60 °$ — which is how the equations in 7.4 are solved without a calculator.

Worked examples

Example 1. A right-angled triangle has a hypotenuse of $12$ cm and an angle of $60 °$ . Work out the exact length of the side opposite that angle.

Show the working

Opposite and hypotenuse means sine.
$\sin 60 ° = \frac{x}{12}$
Rearrange to make $x$ the subject.
$x = 12 \sin 60 °$
Substitute the exact value $\sin 60 ° = \frac{\sqrt{3}}{2}$ rather than a decimal.
$x = 12 \times \frac{\sqrt{3}}{2}$
Simplify by cancelling the 2 into the 12.
$x = 6 \sqrt{3} cm$
Leave it as $6 \sqrt{3}$ . That is the exact answer, and converting to $10.4$ cm would throw away the exactness the question wanted.

Answer: $6 \sqrt{3}$ cm.

Example 2. In a right-angled triangle the side adjacent to an angle of $30 °$ is $5 \sqrt{3}$ cm. Work out the exact length of the opposite side.

Show the working

Opposite and adjacent means tangent.
$\tan 30 ° = \frac{x}{5 \sqrt{3}}$
Substitute the exact value $\tan 30 ° = \frac{1}{\sqrt{3}}$ .
$\frac{1}{\sqrt{3}} = \frac{x}{5 \sqrt{3}}$
Multiply both sides by $5 \sqrt{3}$ .
$x = \frac{5 \sqrt{3}}{\sqrt{3}}$
The $\sqrt{3}$ terms cancel exactly.
$x = 5 cm$
This is what surd form is for — the awkward roots cancel and leave a clean whole number that a decimal calculation would have hidden.

Answer: $5$ cm.

Example 3. Without using a calculator, find the exact value of $\sin 30 ° + \cos 60 ° + \tan 45 °$ .

Show the working

Write down each value from the table. $\sin 30 ° = \frac{1}{2}$ .
$\cos 60 ° = \frac{1}{2}$ — the same value, because cosine of $60 °$ equals sine of $30 °$ .
$\tan 45 ° = 1$ , from the isosceles triangle where opposite and adjacent are equal.
Add them.
$\frac{1}{2} + \frac{1}{2} + 1 = 2$
An exact whole number, with no surds and no rounding.

Answer: $2$ .

Where people go wrong

Mistake: Writing $\sin 30 ° = \frac{\sqrt{3}}{2}$ .

Instead: $\sin 30 ° = \frac{1}{2}$ ; it is $\cos 30 °$ that equals $\frac{\sqrt{3}}{2}$ .

The sine and cosine rows are reverses of each other, so the two are easy to swap. Sketch the half-equilateral triangle to check: the side opposite $30 °$ is the short one, of length 1.

Mistake: Giving a decimal when the question asks for an exact value.

Instead: Leave surds as surds: $\frac{\sqrt{3}}{2}$ , not $0.866$ .

'Exact' and 'in surd form' both mean no rounding is permitted, and a decimal loses the accuracy mark.

Mistake: Thinking $\frac{1}{\sqrt{2}}$ and $\frac{\sqrt{2}}{2}$ are different answers.

Instead: They are the same number, written two ways.

Multiplying top and bottom by $\sqrt{2}$ converts one into the other. Both are accepted.

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

Instead: It is undefined.

$\tan x = \sin x / \cos x$ and $\cos 90 ° = 0$ , so the tangent graph has an asymptote there.

Mistake: Reaching for the calculator on Paper 2.

Instead: Paper 2 is non-calculator — these values must be recalled.

The objective says 'know', which is exactly why the values are examined in the non-calculator paper.

Check yourself

Write down the exact values of $\sin 30 °$ , $\sin 45 °$ and $\sin 60 °$ .
Answer
$\frac{1}{2}$ , $\frac{\sqrt{2}}{2}$ (or $\frac{1}{\sqrt{2}}$ ), and $\frac{\sqrt{3}}{2}$ .
Write down the exact values of $\cos 0 °$ , $\cos 60 °$ and $\cos 90 °$ .
Answer
$1$ , $\frac{1}{2}$ and $0$ .
Write down the exact values of $\tan 30 °$ , $\tan 45 °$ and $\tan 60 °$ .
Answer
$\frac{1}{\sqrt{3}}$ , $1$ and $\sqrt{3}$ .
Which triangle gives the values for $30 °$ and $60 °$ , and what are its sides?
Answer
Half of an equilateral triangle of side 2, giving sides $1$ , $\sqrt{3}$ and $2$ .
Why is $\tan 90 °$ undefined?
Answer
$\tan 90 ° = \sin 90 ° / \cos 90 ° = 1 / 0$ , and division by zero is undefined.
Find the exact value of $\cos 30 ° \times \tan 60 °$ .
Answer
$\frac{\sqrt{3}}{2} \times \sqrt{3} = \frac{3}{2}$ .

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 $\cos 45 °$ . 1 mark No calculator

Mark scheme

B1 √2/2 oe, accept 1/√2

(b) Work out the exact value of $\sin 60 ° \times \cos 30 °$ . 2 marks No calculator

Mark scheme

M1 (√3/2) × (√3/2) seen
A1 3/4 oe

(c) Show that $\tan 60 ° \times \tan 30 ° = 1$ . 2 marks No calculator

Mark scheme

M1 √3 × 1/√3 seen
A1 reaches 1 with correct working; nfww

Question 2. A right-angled triangle has a hypotenuse of $20$ cm and one acute angle of $30 °$ . Do not use a calculator.

(a) Work out the exact length of the side opposite the $30 °$ angle. 2 marks No calculator

Mark scheme

M1 20 sin 30° or 20 × 1/2 seen
A1 10 cm

(b) Work out the exact length of the third side, giving your answer in surd form. 3 marks No calculator

Mark scheme

M1 20 cos 30° oe, or Pythagoras with 20 and their 10
M1 20 × √3/2 seen, or √(400 - 100)
A1 10√3 cm

Question 3. An equilateral triangle has sides of length $6$ cm. Do not use a calculator.

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

Mark scheme

M1 splits the triangle into two right-angled triangles with base 3 and hypotenuse 6
M1 6 sin 60° or √(36 - 9) seen
A1 reaches 3√3 with correct working; nfww

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

Mark scheme

M1 ½ × 6 × 3√3 oe; FT their height
A1 9√3 cm²

Words to know

Exact value: A value written without rounding, using fractions and surds rather than decimals.
Surd: A root that cannot be simplified to a rational number, such as $\sqrt{2}$ or $\sqrt{3}$ .
Rationalising the denominator: Rewriting $\frac{1}{\sqrt{2}}$ as $\frac{\sqrt{2}}{2}$ by multiplying top and bottom by $\sqrt{2}$ .
Undefined: Having no value, as for $\tan 90 °$ , where the cosine in the denominator is zero.

All Maths lessons Next: 7.4 Trigonometric functions