Exact trig values — 30 45 60 degrees — Mathematics

Before this

SOH CAH TOA — sin, cos and tan introduced
Pythagoras' theorem · coming soon
Rationalising the denominator

This leads to

The sine rule
Area of a triangle using sine
Exact values and the unit circle · coming soon

A number the calculator can only show part of

Put your calculator in degree mode and type sin 30°. It shows 0.5. That answer is exact. Nothing is missing from it.

Now type sin 60°. The screen shows 0.8660254038. That is not the whole number. The digits go on for ever, and they never settle into a repeating pattern. The calculator simply stops at the edge of its screen.

Here is a question it cannot answer. Square the ten digits on the screen, and you get 0.750 000 000 03 (to 11 decimal places). So is $(sin 6 0^{\circ})^{2}$ exactly $\frac{3}{4}$ ? Or is it a number that only starts with 0.75? A calculator shows you more digits. It never shows you the reason. Only the reason can settle the question.

The first lecture on trigonometry promised that a later lecture would find exact values for 30°, 45° and 60° without a calculator. This is that lecture. You will need two shapes you already know, Pythagoras' theorem, and the three ratios. By the end, you will be able to write down sin 60° exactly, and you will know why it is that number and not another.

Before any building, test your own feeling for how the sine of an angle behaves.

Predict — what happens when the angle doubles?

sin 30° = 0.5. Without a calculator, what do you think sin 60° is? Choose one, then read the explanation.

This figure shows two lines of the same length, 2, at 30° and at 60° to the ground. The height of each end above the ground is drawn in.

Both lines have length 2. At 30°, the end is 1 above the ground, so sin 30° = 1 ÷ 2 = 0.5. At 60°, the end is √3 ≈ 1.73 above the ground, not 2. So sin 60° = √3 ÷ 2 ≈ 0.866, not 1. Doubling the angle does not double the height.

The figure in the answer shows why doubling fails. Picture a line of fixed length turning upward from the ground. At first, each extra degree lifts its end a lot. Near 90°, the end is almost at the top, and each extra degree lifts it only a little. So the height does not grow in proportion to the angle. The sine is a ratio of two sides, not a multiple of the angle.

Triangle one: half of an equilateral triangle

To find exact values, we need a right-angled triangle whose angles we know and whose sides we know exactly. Most triangles do not give us both. The trick is to start from a shape where everything is already known, and cut it.

Start with an equilateral triangle. Its three sides are equal, so its three angles are equal too. The angles of a triangle add up to 180°, so each one is 60°. Give each side the length 2. You will see in a moment why 2 is a better choice than 1.

Now draw a line from the top corner straight down to the middle of the base. This cuts the triangle into two halves. Each half is a right-angled triangle with angles 30°, 60° and 90°. The figure shows the left half, shaded.

The cut from A to the midpoint M of the base halves the angle at the top: 60° becomes 30° + 30°. The angle at B is not cut, so it stays 60°. In the shaded half, the side facing 30° is 1, the side facing 60° is √3, and the hypotenuse is 2.

Each claim in that figure has a reason, and every reason is a fact you already know. Step through them below. Read each reason before you go on.

Step through — building the 30°, 60°, 90° triangle

Press Next to add one step. Each step gives one fact and the reason it is true.

Each step gives one fact about the triangle in the figure, and the reason it is true.

Step 1. Triangle ABC has AB = BC = CA = 2, and each angle is 60°. Reason: equal sides give equal angles, and the three angles add up to 180°.
Step 2. M is the midpoint of BC, so BM = MC = 1. Reason: M halves the base of length 2. A side of 2 makes the halves whole numbers.
Step 3. Triangles ABM and ACM are congruent. Reason: AB = AC = 2, BM = CM = 1, and AM is shared. Three pairs of equal sides (SSS).
Step 4. ∠AMB = ∠AMC = 90°. Reason: the two angles are equal, because the triangles are congruent. Together they make a straight line, 180°. So each is 90°.
Step 5. ∠BAM = ∠CAM = 30°. Reason: congruent triangles have equal angles, so the cut splits the 60° at A into two equal parts.
Step 6. In triangle ABM, the angles are 60° at B, 30° at A and 90° at M. Reason: the angle at B was never cut. Check: 60° + 30° + 90° = 180°.
Step 7. AM = √3. Reason: Pythagoras in triangle ABM, where AB = 2 is the hypotenuse: AM² + 1² = 2², so AM² = 3.

Result: triangle ABM has sides 1, √3 and 2. The side 1 faces the 30° angle, √3 faces the 60° angle, and 2 faces the right angle.

This is the hard part of the lecture, so spend time here. The proof is short, but the order matters. The right angle at M is not assumed. It comes from the two congruent halves. The 30° is not a guess either. It is half of the top angle, because the two halves are mirror images. A common slip is to halve the angle at B instead. But the line does not pass through B, so the angle at B is untouched.

Notice a quick check on the sides. In any triangle, the shortest side faces the smallest angle and the longest side faces the largest angle. Here 1 < √3 < 2, because √3 ≈ 1.73. And 30° < 60° < 90°. The order matches.

Now read the ratios. From the 30° angle at A, the opposite side is BM = 1, the adjacent side is AM = √3, and the hypotenuse is 2.

sin 3 0^{\circ} = \frac{1}{2} cos 3 0^{\circ} = \frac{3}{2} tan 3 0^{\circ} = \frac{1}{3}

From the 60° angle at B, the two shorter sides swap roles. Now √3 is the opposite and 1 is the adjacent.

sin 6 0^{\circ} = \frac{3}{2} cos 6 0^{\circ} = \frac{1}{2} tan 6 0^{\circ} = 3

So the answer to the opening question is yes. $(sin 6 0^{\circ})^{2} = (\frac{3}{2})^{2} = \frac{3}{4}$ exactly. The tiny extra digits on the calculator came only from rounding.

Look at the two lists side by side. $sin 3 0^{\circ} = cos 6 0^{\circ}$ , and $cos 3 0^{\circ} = sin 6 0^{\circ}$ . This is not luck. The side opposite one acute angle is the side adjacent to the other. In the first lecture on trigonometry you met the rule $cos θ = sin (9 0^{\circ} - θ)$ . The angles 30° and 60° are complementary, so this triangle shows the rule with exact numbers.

Stop and think

Mei starts with an equilateral triangle of side 10 cm instead of 2. Will she get different exact values for sin 30° and sin 60°?

Show the answer

No. Her half-triangle has sides 5, $53$ and 10. Then $sin 3 0^{\circ} = \frac{5}{10} = \frac{1}{2}$ and $sin 6 0^{\circ} = \frac{5 3}{10} = \frac{3}{2}$ , the same as before. Every 30°, 60°, 90° triangle is the same shape, so the scale factor cancels in every ratio. This is the reason the ratios belong to the angle, from the first trigonometry lecture. A side of 2 is only a convenience: it makes the numbers small and whole.

Triangle two: half of a square

For 45°, the shape to cut is a square. Give each side the length 1. Draw one diagonal. It cuts the square into two right-angled triangles.

The diagonal is a line of symmetry of the square, so it halves each 90° corner into 45° and 45°. The shaded triangle has two sides of 1 and a hypotenuse of √2.

Why are the angles 45°? The diagonal is a line of symmetry of the square. So it cuts each right-angled corner into two equal parts, and 90° ÷ 2 = 45°. There is a second reason too. The triangle has two equal sides, so it is isosceles, and its two base angles are equal. They must add up to 180° − 90° = 90°, so each is 45°.

The hypotenuse comes from Pythagoras: $1^{2} + 1^{2} = 2$ , so the diagonal is $2$ . From either 45° angle, the opposite and the adjacent are both 1.

sin 4 5^{\circ} = \frac{1}{2} cos 4 5^{\circ} = \frac{1}{2} tan 4 5^{\circ} = 1

The sine and the cosine are equal, because the two shorter sides are equal. And the tangent is exactly 1, because the opposite equals the adjacent. This is the 45° check from the last lecture: below 45° the opposite is shorter than the adjacent, and above 45° it is longer.

You may have learned to rationalise the denominator, so that no surd is left on the bottom. Multiply the top and the bottom by $2$ :

\frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{2}

These are two ways of writing the same number, about 0.707. In the same way, $tan 3 0^{\circ} = \frac{1}{3} = \frac{3}{3}$ . Both forms are correct. Many textbooks and mark schemes print the rationalised form, so the table below gives both.

Two triangles, nine values

Exact values

Angle	sin	cos	tan
30°	$\frac{1}{2}$	$\frac{3}{2}$	$\frac{1}{3} = \frac{3}{3}$
45°	$\frac{1}{2} = \frac{2}{2}$	$\frac{1}{2} = \frac{2}{2}$	$1$
60°	$\frac{3}{2}$	$\frac{1}{2}$	$3$

Every value comes from two triangles: half of an equilateral triangle of side 2 (sides 1, √3, 2) and half of a square of side 1 (sides 1, 1, √2).

That is it. That is the whole idea. Do not learn this table as nine separate facts. If you forget a value in an exam, sketch the two triangles in the margin. That takes about twenty seconds, and it cannot be mixed up.

Once you understand where the values come from, the table also has patterns you can use as checks.

The sine column rises. Written with a 2 underneath, it is $\frac{1}{2}, \frac{2}{2}, \frac{3}{2}$ . As the angle grows, the opposite side grows.
The cosine column is the sine column upside down. This is $cos θ = sin (9 0^{\circ} - θ)$ again.
The tangent passes 1 at 45°. It is below 1 before 45° and above 1 after it. Also, $tan 3 0^{\circ} \times tan 6 0^{\circ} = \frac{1}{3} \times 3 = 1$ , because the opposite and the adjacent swap.

Use the patterns to check a value. Do not use them instead of the reason.

Worked examples

Worked Example 1 — An exact product

Find the exact value of $cos 3 0^{\circ} \times tan 6 0^{\circ}$ , without a calculator.

Think first. Replace each ratio by its exact value. Then multiply surds: $3 \times 3 = 3$ .

cos 3 0^{\circ} \times tan 6 0^{\circ} = \frac{3}{2} \times 3 = \frac{3}{2}

Check. With rounded decimals, $0.866 \times 1.732 = 1.499912$ . That is close to 1.5 but not equal to it. The rounding has hidden the true answer. The exact working shows it is exactly $\frac{3}{2}$ . ✓

Worked Example 2 — A ladder, in exact form

A ladder 5 m long leans against a vertical wall. It makes an angle of 60° with the level ground. Find, in exact form, how high up the wall it reaches and how far its foot is from the wall.

Think first. The ladder is the hypotenuse. From the 60° angle at the foot, the height is the opposite and the distance along the ground is the adjacent. So the height uses sine and the distance uses cosine.

h = 5 sin 6 0^{\circ} = 5 \times \frac{3}{2} = \frac{5 3}{2} m

d = 5 cos 6 0^{\circ} = 5 \times \frac{1}{2} = \frac{5}{2} m

As decimals, $h \approx 4.33$ m and $d = 2.5$ m exactly.

Check. Pythagoras with the exact sides: $\frac{75}{4} + \frac{25}{4} = \frac{100}{4} = 25 = 5^{2}$ . ✓ The angle is above 45°, so the height should be longer than the distance. It is. ✓

Worked Example 3 — Finding an angle without a calculator

A roof rises 4 m over a horizontal distance of $43$ m. Find the angle of the roof to the horizontal, without a calculator.

Think first. The rise is the opposite and the horizontal distance is the adjacent. So the tangent of the angle is their ratio. Then ask which angle has that tangent. This is the inverse question from the lecture on finding angles, but now the table answers it.

tan θ = \frac{4}{4 3} = \frac{1}{3} \Rightarrow θ = 3 0^{\circ}

Check. The rise, 4 m, is shorter than the run, $43 \approx 6.93$ m, so the angle must be below 45°. It is. ✓ The sides 4 and $43$ are in the ratio $1 : 3$ , exactly the two shorter sides of the half-equilateral triangle, with the 1 facing 30°. ✓

Worked Example 4 — Squares of sine and cosine

Show that $sin^{2} 4 5^{\circ} + cos^{2} 4 5^{\circ} = 1$ . Then test the same sum at 30°.

Think first. First, the notation. $sin^{2} 4 5^{\circ}$ means $(sin 4 5^{\circ})^{2}$ : find the sine, then square it. It does not mean the sine of $4 5^{2}$ .

sin^{2} 4 5^{\circ} + cos^{2} 4 5^{\circ} = (\frac{1}{2})^{2} + (\frac{1}{2})^{2} = \frac{1}{2} + \frac{1}{2} = 1

At 30°: $(\frac{1}{2})^{2} + (\frac{3}{2})^{2} = \frac{1}{4} + \frac{3}{4} = 1$ . The same answer.

Why? In any right-angled triangle with shorter sides $a$ and $b$ and hypotenuse $c$ , the sine and cosine of one angle are $\frac{a}{c}$ and $\frac{b}{c}$ . Their squares add up to $\frac{a ^{2} + b ^{2}}{c ^{2}}$ . Pythagoras says $a^{2} + b^{2} = c^{2}$ , so the sum is 1, for every angle. The result is Pythagoras' theorem divided by $c^{2}$ . A Pre-University lecture builds on it.

Check. With decimals at 60°: $0.86 6^{2} + 0. 5^{2} \approx 0.750 + 0.25 = 1.000$ . ✓

Three traps

Trap 1 — Doubling the angle, doubling the ratio

sin 60° is not 2 × sin 30°. Sine is a ratio of two sides, not a number multiplied by the angle. The same is true for cosine and tangent: $tan 6 0^{\circ} = 3 \approx 1.73$ , not $2 tan 3 0^{\circ} \approx 1.15$ . Always read the value from the triangle.

Trap 2 — Putting √3 opposite the 30° angle

In the half-equilateral triangle, the side 1 faces 30° and the side √3 faces 60°. Swapping them gives $sin 3 0^{\circ} = \frac{3}{2}$ , which is wrong. Check with the rule that the shortest side faces the smallest angle. The 30° angle is at the top of the triangle, where the cut was made, and it faces the short half of the base.

Trap 3 — Rounding too early, or writing "=" for a decimal

Write $sin 6 0^{\circ} = \frac{3}{2} \approx 0.866$ , never $sin 6 0^{\circ} = 0.866$ . Keep the exact form until the last line. If you round first, errors grow: $0.8 7^{2} = 0.7569$ , but $(\frac{3}{2})^{2} = 0.75$ exactly. Also take care with the fraction line: $\frac{3}{2}$ is not $\frac{3}{2}$ .

Try it yourself

Work each one on paper first, without a calculator. Open a hint only when you are stuck.

Find the exact value of $sin^{2} 6 0^{\circ} + cos 6 0^{\circ}$ .

Hint 1
Write down $sin 6 0^{\circ}$ and $cos 6 0^{\circ}$ from the half-equilateral triangle.

Hint 2
$sin^{2} 6 0^{\circ} = (\frac{3}{2})^{2} = \frac{3}{4}$ . Now add $cos 6 0^{\circ}$ .

Full answer
$sin^{2} 6 0^{\circ} + cos 6 0^{\circ} = \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$ . Check with decimals: $0.86 6^{2} + 0.5 \approx 0.750 + 0.5 = 1.25$ . ✓
A plank 6 m long rests on a step, making an angle of 30° with the level ground. Find, in exact form, the height of the top of the plank above the ground and the horizontal distance it covers.

Hint 1
The plank is the hypotenuse. From the 30° angle, the height is the opposite and the horizontal distance is the adjacent.

Hint 2
Height $= 6 sin 3 0^{\circ}$ . Horizontal distance $= 6 cos 3 0^{\circ}$ .

Full answer
Height $= 6 \times \frac{1}{2} = 3$ m. Horizontal distance $= 6 \times \frac{3}{2} = 33$ m $\approx 5.20$ m. Check: $3^{2} + (33)^{2} = 9 + 27 = 36 = 6^{2}$ . ✓ The angle is below 45°, so the height should be shorter than the horizontal distance. It is. ✓ The sides 3, $33$ , 6 are the triangle 1, √3, 2 scaled by 3.
A right-angled triangle has a hypotenuse of 10 cm and one shorter side of 5 cm. Without a calculator, find the third side in exact form and both acute angles.

Hint 1
Use Pythagoras for the third side. Then compare the sides with 1, √3, 2.

Hint 2
The angle facing the 5 cm side has sine $\frac{5}{10} = \frac{1}{2}$ . Which angle has that sine?

Full answer
Third side: $1 0^{2} - 5^{2} = 75 = 25 \times 3 = 53$ cm. The angle facing the 5 cm side has $sin θ = \frac{5}{10} = \frac{1}{2}$ , so $θ = 3 0^{\circ}$ . The other acute angle is $9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ . Check: the sides 5, $53$ , 10 are 1, √3, 2 scaled by 5, with the shortest side facing the smallest angle. ✓ And $tan 6 0^{\circ} = \frac{5 3}{5} = 3$ . ✓
A stretch. A floor tile is a regular hexagon with sides of 4 cm. A regular hexagon splits into six equilateral triangles that meet at its centre. Find, in exact form, (a) the distance across the tile between two opposite straight edges, and (b) the area of the tile.

Hint 1
Each of the six triangles is equilateral with side 4 cm. Find its height by cutting it in half.

Hint 2
The height of each triangle is $4 sin 6 0^{\circ} = 23$ cm. The distance across the flats is two of these heights, one on each side of the centre.

Full answer
(a) Each equilateral triangle of side 4 has height $4 sin 6 0^{\circ} = 4 \times \frac{3}{2} = 23$ cm. The distance between opposite edges runs from one edge, through the centre, to the other edge, so it is two heights: $43$ cm $\approx 6.93$ cm. (b) Each triangle has area $\frac{1}{2} \times 4 \times 23 = 43$ cm². Six of them make $243$ cm² $\approx 41.6$ cm². Check: the distance across the flats, 6.93 cm, is less than the distance between opposite corners, which is two sides, 8 cm. ✓ The area is less than that of a square of side $43$ cm, which is 48 cm². ✓

The bigger idea

Long before calculators, people needed tables of these ratios to work out positions of the stars. Around 150 CE, in Alexandria, Ptolemy wrote a book on astronomy that is known today as the Almagest. Its first part contains a table of chords: the lengths of straight lines across a circle, for every half degree. Ptolemy took the circle's diameter as 120 parts. A chord in that circle is the same as a sine in ours: the chord for a central angle of 120° is $120 \times sin 6 0^{\circ} = 603 \approx 103.92$ parts. (Sources: T. L. Heath, A History of Greek Mathematics, 1921, on the Almagest, Book I; "Ptolemy's Table of Chords", The Physics Hypertextbook.)

How did he fill a table of 360 lines? He began, as this lecture did, with angles whose values could be found exactly from regular shapes: the triangle, the square, the hexagon, and also the pentagon and the decagon. The pentagon gives another exact value, $sin 1 8^{\circ} = \frac{5 - 1}{4} \approx 0.309$ , which is beyond this course. From these starting values, he used geometry to find the chords of differences and of halves of angles. Only near the end, for the chord of 1°, did he have to estimate.

So a huge table grew outward from a few exact values, the same kind you found today. The next lecture takes trigonometry beyond the right angle, into triangles of any shape, with the sine rule. The exact values will still be useful there, because examiners often choose 30°, 45° and 60° so that answers come out exactly.

Key concepts in this lesson

Cutting an equilateral triangle of side 2 in half gives a right-angled triangle with sides 1, √3 and 2, where the side 1 faces 30° and the side √3 faces 60°.
Cutting a square of side 1 along a diagonal gives a right-angled triangle with sides 1, 1 and √2, and two angles of 45°.
Reading the three ratios from these two triangles gives every exact value for 30°, 45° and 60°, so the table can be rebuilt at any time instead of memorised.
The values are the same for every size of triangle, because triangles with the same angles are similar and the scale factor cancels in each ratio.
Exact forms such as $\frac{3}{2}$ are the true values, while decimals such as 0.866 are rounded descriptions of them and should be written with ≈.
Sine is a ratio of sides, not a multiple of the angle, so doubling an angle does not double its sine.