Trigonometry

Trigonometry is the branch of mathematics about the sides and angles of triangles and how they are related. Its name is Greek, trigonon “triangle” and metron “measure”. The Greek astronomer Hipparchus (about 190–120 BC) is called its father, because he made the first known table of what we now call trigonometric ratios, to work out where the Sun and Moon would be.

Trigonometry began with astronomy and with mapping the land, and it is still used wherever something can’t be measured directly: physics, navigation, engineering, surveying for bridges and roads, and finding the height of a building or a mountain from the ground.

Before we can use angles we must agree how to measure them. There are three systems: degrees, grades and radians.

2.1.1 Sexagesimal and centesimal systems

a. Sexagesimal system

When a line turns from its starting position all the way round once, it turns through 360 degrees, 360∘°. A quarter turn, a right angle, is 90∘°. Each degree is cut into 60 minutes (′) and each minute into 60 seconds (′′). Because everything is counted in sixties, this is the sexagesimal system, from the Latin for “sixtieth”.

1 complete turn	360∘°
1 right angle	90∘° (degrees)
1∘° (degree)	60′ (minutes)
1′ (minute)	60′′ (seconds)

b. Centesimal system

In the centesimal system (from the Latin for “hundredth”) a complete turn is 400 grades, 400ᵍ, so a right angle is exactly 100ᵍ. Each grade is cut into 100 centesimal minutes (′) and each of those into 100 centesimal seconds (′′). Everything goes in hundreds, like metric units.

1 complete turn	400ᵍ
1 right angle	100ᵍ (grades)
1ᵍ (grade)	100′ (minutes)
1′ (minute)	100′′ (seconds)

The same signs ′ and ′′ are used in both systems, so always look at whether the angle starts with ∘° or ᵍ: 30∘°20′ and 30ᵍ20′ are different angles, and their minutes are different sizes.

2.1.2 Conversion between sexagesimal and centesimal systems

One right angle is both 90∘° and 100ᵍ, so 90∘° = 100ᵍ. Everything else follows from that.

From degrees to grades	From grades to degrees
90∘° = 100ᵍ	100ᵍ = 90∘°
$1^\circ = \left(\frac{100}{90}\right)^{\mathrm{g}} = \left(\frac{10}{9}\right)^{\mathrm{g}}$	$1^{\mathrm{g}} = \left(\frac{90}{100}\right)^\circ = \left(\frac{9}{10}\right)^\circ$
$D^\circ = \left(\frac{10}{9} \times D\right)^{\mathrm{g}}$	$G^{\mathrm{g}} = \left(\frac{9}{10} \times G\right)^\circ$

A grade is a little smaller than a degree, so the same angle is always a bigger number of grades than of degrees.

When an angle has minutes and seconds, first make it a decimal in its own system, then convert, then turn the decimal part back into minutes and seconds.

Question 1

Fill in the blanks.

Question 2

Convert into sexagesimal seconds.

Question 3

Question 4

Convert into sexagesimal measure (degrees, minutes and seconds).

Question 5

Convert into centesimal measure (grades, minutes and seconds).

Question 6

Convert into degrees.

Question 7

Convert into grades.

Question 8

Convert into grades.

Question 9

In a right-angled triangle, one angle is 60∘°.

Question 10

Question 11

The three angles of a triangle are in the ratio 2 : 3 : 4.

Question 12

The angles of a triangle are in the ratio 5 : 7 : 8.

Question 13

The four angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6.

Question 14

In a triangle, one angle is 72∘°. The other two are in the ratio 1 : 3.

Question 15

Two angles of a triangle are in the ratio 3 : 4 and the third angle is 60ᵍ.

Question 16

One acute angle of a right-angled triangle is $\frac{3}{10}$ of a right angle.

Question 17

The sum of two angles is 100∘° and their difference is 20ᵍ.

Question 18

The sum of two angles is 45∘° and their difference is 30ᵍ.

2.1.3 System of circular measure

The degree and the grade are man-made: someone chose 360 or 400 parts. The circular measure uses the circle itself. In it, angles are measured in radians, written ᶜ (for “circular”) or rad.

Radian: one radian (1ᶜ) is the angle made at the center of a circle by an arc as long as the circle’s radius.

The whole circumference is $2\pi r$, which is $2\pi$ radius-lengths. So the angle of a complete turn is $2\pi$ radians: 360∘° = 2πᶜ, and a straight angle is 180∘° = πᶜ. Here $\pi \approx 3.14159$, often taken as $\frac{22}{7}$ in calculations.

2.1.4 Theorems on radian measure

Two facts about radians are proved as theorems.

Theorem 1: A radian is a constant angle. (It is the same angle in every circle.)

Proof. Take a circle with center O and radius OA = r. Mark B on the circle so that the arc AB = r; then ∠AOB = 1ᶜ, by the definition of a radian. Extend AO to C, so that AC is a diameter.

The arc ABC is half the circumference: $\frac{1}{2} \times 2\pi r = \pi r$. And ∠AOC is a straight angle, 180∘°.

In a circle, central angles are in the same ratio as their arcs:
$$\frac{\angle AOB}{\angle AOC} = \frac{\text{arc } AB}{\text{arc } ABC} \quad\text{so}\quad \frac{1^{\mathrm{c}}}{180^\circ} = \frac{r}{\pi r} = \frac{1}{\pi} \quad\text{and}\quad 1^{\mathrm{c}} = \frac{180^\circ}{\pi}.$$

The radius has canceled: $1^{\mathrm{c}} = \frac{180^\circ}{\pi}$ whatever the circle, and since 180∘° and $\pi$ are both constants, one radian is a constant angle. Proved.

Worked out, $1^{\mathrm{c}} = \frac{180^\circ}{\pi} \approx 57.2958^\circ \approx 57^\circ 17' 45''$. (With $\pi = \frac{22}{7}$, $1^{\mathrm{c}} = \frac{180 \times 7}{22} \approx 57.27^\circ$.)

2.1.5 The relationship among arc length, radius and central angle

Theorem 2: In a circle of radius r, an arc of length l makes a central angle
$$\theta = \left(\frac{l}{r}\right)^{\mathrm{c}}.$$

Proof. Take a circle with center O and radius OA = r. Mark B on it so that arc AB = r, so ∠AOB = 1ᶜ. Mark C so that the arc ABC = l, and let ∠AOC = θ.

Central angles are in the ratio of their arcs:
$$\frac{\angle AOC}{\angle AOB} = \frac{\text{arc } ABC}{\text{arc } AB} \quad\text{so}\quad \frac{\theta}{1^{\mathrm{c}}} = \frac{l}{r} \quad\text{and}\quad \theta = \left(\frac{l}{r}\right)^{\mathrm{c}}.$$
Proved. In words:
$$\text{central angle (in radians)} = \frac{\text{length of the arc}}{\text{radius of the circle}}, \qquad\text{or}\qquad l = r\theta.$$

This is why radians matter. In degrees the same fact needs a clumsy factor, $l = \frac{\theta}{360} \times 2\pi r$; in radians it is just l = rθ. Measuring the angle in radians is measuring it in radius-lengths of arc.

Converting to and from radians

From 180∘° = 200ᵍ = πᶜ:

$1^\circ = \left(\frac{\pi}{180}\right)^{\mathrm{c}}$	$1^{\mathrm{g}} = \left(\frac{\pi}{200}\right)^{\mathrm{c}}$	$1^{\mathrm{c}} = \left(\frac{180}{\pi}\right)^\circ = \left(\frac{200}{\pi}\right)^{\mathrm{g}}$

The map gave the Nyatapola’s latitude as 27∘°40′19′′ N. Latitude is the angle at the center of the Earth between the equator and the place, measured along a line of longitude: a central angle of a circle whose radius is the Earth’s radius, about 6,371 km. The distance from the equator is the arc, so l = rθ, with θ in radians.

First the angle in degrees: $27^\circ 40' 19'' = \left(27 + \frac{40}{60} + \frac{19}{3600}\right)^\circ = 27.67194^\circ$.

In radians: $27.67194 \times \frac{\pi}{180} = 0.48297^{\mathrm{c}}$.

The arc: $l = 6371 \times 0.48297 \approx 3077$ km.

So Bhaktapur is about 3,077 km north of the equator, measured along the Earth’s surface. (The Earth isn’t a perfect ball, so a surveyor would get a few kilometers more, but the method is the same.)

How far is one second of latitude on the ground? $1'' = \left(\frac{1}{3600}\right)^\circ = \frac{\pi}{180 \times 3600}$ radians, so
$$l = 6371000 \text{ m} \times \frac{\pi}{648000} \approx 30.9 \text{ m}.$$
A second of latitude is about 31 m: the length of a long school building. That is why a map app that gives seconds can tell one end of Durbar Square from the other, and why positions for very exact work are given in decimals of a second.

Question 1

Fill in the blanks.

Question 2

Question 3

Question 4

Question 5

Look back at the activities “Degrees, grades and radians side by side” and “Do arcs grow with their angles?” (Activities 2 and 3 of the textbook).

Question 6

Convert into radians.

Question 7

Convert into degrees.

Question 8

Convert into grades.

Question 9

A right-angled triangle ABC is given.

Question 10

One angle of a right-angled triangle is 60∘°.

Question 11

One acute angle of a right-angled triangle is 50∘°.

Question 12

Question 13

Find the angle between the hour hand and the minute hand of a clock at each time, in circular measure.

Question 14

Take $\pi = \frac{22}{7}$ in this question and the ones after it.

Question 15

Question 16

Question 17

A cow is tied to a peg with a rope 10 m long. With the rope kept tight, the cow walks round until the rope has turned through $\frac{7\pi^{\mathrm{c}}}{18}$.

Question 18

A goat is tied to a peg with a rope 14 m long. With the rope kept tight, it walks round until the rope has turned through 70∘° at the peg.

Question 19

Question 20

Dil walks along a circular path at 100 m a minute. In 36 seconds he turns through 56∘° about its center.

Question 21

In Grade 9 Math you met the sine, cosine and tangent of an angle in a right-angled triangle. Why do those ratios matter? Because they depend only on the angle, not on the size of the triangle: every right-angled triangle with a 30∘° angle has its side opposite 30∘° exactly half its hypotenuse, whether it is drawn in a notebook or laid out across a valley. So one measured angle and one measured length give all the others. That is how the surveyors found Everest’s height.

In a right-angled triangle the sides are linked by Pythagoras’ theorem, $h^2 = p^2 + b^2$. The ratios of the sides are linked too, by rules that hold for every angle. Those rules are the identities of this section.

2.2.1 The six trigonometric ratios

Ratio	Name	Short
$\frac{\text{opposite side}}{\text{hypotenuse}} = \frac{p}{h}$	sine	sin θ
$\frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{b}{h}$	cosine	cos θ
$\frac{\text{opposite side}}{\text{adjacent side}} = \frac{p}{b}$	tangent	tan θ
$\frac{\text{hypotenuse}}{\text{opposite side}} = \frac{h}{p}$	cosecant	$\operatorname{cosec}\theta$
$\frac{\text{hypotenuse}}{\text{adjacent side}} = \frac{h}{b}$	secant	$\sec\theta$
$\frac{\text{adjacent side}}{\text{opposite side}} = \frac{b}{p}$	cotangent	$\cot\theta$

The first three are the basic trigonometric ratios; the last three are their reciprocals (upside-downs).

The names are worth knowing. “Co-” is short for complement: the cosine of θ is the sine of its complement, 90∘° − θ, and the same for cotangent and cosecant (you will prove this in 2.4.3). “Tangent” and “secant” are the Latin for “touching” and “cutting”: on a circle of radius 1, they are the lengths of a line that touches the circle and a line that cuts through it.

2.2.2 Identities of trigonometric ratios

An identity is an equation that is true for every value of its letter, like $(a + b)^2 = a^2 + 2ab + b^2$. Trigonometric identities are true for every angle (where both sides are defined).

Reciprocal relations. What is $\sin\theta \times \operatorname{cosec}\theta$? From the definitions, $\frac{p}{h} \times \frac{h}{p} = 1$. In the same way $\cos\theta \times \sec\theta = \frac{b}{h} \times \frac{h}{b} = 1$ and $\tan\theta \times \cot\theta = \frac{p}{b} \times \frac{b}{p} = 1$.

$\sin\theta = \frac{1}{\operatorname{cosec}\theta}$	$\operatorname{cosec}\theta = \frac{1}{\sin\theta}$	$\sin\theta \times \operatorname{cosec}\theta = 1$
$\cos\theta = \frac{1}{\sec\theta}$	$\sec\theta = \frac{1}{\cos\theta}$	$\cos\theta \times \sec\theta = 1$
$\tan\theta = \frac{1}{\cot\theta}$	$\cot\theta = \frac{1}{\tan\theta}$	$\tan\theta \times \cot\theta = 1$

So sin and cosec, cos and sec, tan and cot are reciprocals of each other.

Quotient relations. Divide the top and the bottom of $\tan\theta = \frac{p}{b}$ by h:
$$\tan\theta = \frac{p/h}{b/h} = \frac{\sin\theta}{\cos\theta}.$$
Do the same with $\cot\theta = \frac{b}{p}$:
$$\cot\theta = \frac{b/h}{p/h} = \frac{\cos\theta}{\sin\theta}.$$

Quotient relations
$\tan\theta = \frac{\sin\theta}{\cos\theta}$	$\cot\theta = \frac{\cos\theta}{\sin\theta}$

2.2.3 Pythagoras relations

Take the right-angled triangle BET, with the right angle at E and the reference angle θ at T. Then p = BE, b = ET, h = BT, and by Pythagoras’ theorem $p^2 + b^2 = h^2$.

a. Prove that $\sin^2\theta + \cos^2\theta = 1$.
$BE^2 + ET^2 = BT^2$. Divide both sides by $BT^2$:
$$\frac{BE^2}{BT^2} + \frac{ET^2}{BT^2} = 1, \quad\text{so}\quad \left(\frac{BE}{BT}\right)^2 + \left(\frac{ET}{BT}\right)^2 = 1, \quad\text{so}\quad (\sin\theta)^2 + (\cos\theta)^2 = 1.$$
$\sin^2\theta + \cos^2\theta = 1$. Proved. ($\sin^2\theta$ means $(\sin\theta)^2$.)

b. Prove that $\sec^2\theta - \tan^2\theta = 1$.
$BE^2 + ET^2 = BT^2$. Divide both sides by $ET^2$:
$$\frac{BE^2}{ET^2} + 1 = \frac{BT^2}{ET^2}, \quad\text{so}\quad \tan^2\theta + 1 = \sec^2\theta,$$
since $\frac{BE}{ET} = \frac{p}{b} = \tan\theta$ and $\frac{BT}{ET} = \frac{h}{b} = \sec\theta$. So $\sec^2\theta - \tan^2\theta = 1$. Proved.

c. Prove that $\operatorname{cosec}^2\theta - \cot^2\theta = 1$.
$BE^2 + ET^2 = BT^2$. Divide both sides by $BE^2$:
$$1 + \frac{ET^2}{BE^2} = \frac{BT^2}{BE^2}, \quad\text{so}\quad 1 + \cot^2\theta = \operatorname{cosec}^2\theta,$$
since $\frac{ET}{BE} = \frac{b}{p} = \cot\theta$ and $\frac{BT}{BE} = \frac{h}{p} = \operatorname{cosec}\theta$. So $\operatorname{cosec}^2\theta - \cot^2\theta = 1$. Proved.

Each of the three can be turned round in two more ways:

Identity
$\sin^2\theta + \cos^2\theta = 1$	$\sin^2\theta = 1 - \cos^2\theta$	$\cos^2\theta = 1 - \sin^2\theta$
$\sec^2\theta - \tan^2\theta = 1$	$\sec^2\theta = 1 + \tan^2\theta$	$\tan^2\theta = \sec^2\theta - 1$
$\operatorname{cosec}^2\theta - \cot^2\theta = 1$	$\operatorname{cosec}^2\theta = 1 + \cot^2\theta$	$\cot^2\theta = \operatorname{cosec}^2\theta - 1$

Algebra with trigonometric ratios

A ratio such as sin A behaves like a letter in algebra: like terms are added, brackets multiplied out, expressions factorized, by the same rules. Only the identities are new.

To prove an identity, start from one side (usually the more complicated one) and change it, step by step, using the definitions and identities, until it becomes the other side. Never move terms across the equals sign, as if it were an equation to solve: that assumes what you are trying to prove.

Question 1

Fill in the blanks.

Question 2

Question 3

Question 4

Question 5

Simplify.

Question 6

Multiply.

Question 7

Factorize.

Question 8

Prove that:

Question 9

Prove that:

Question 10

Prove that:

2.2.4 Typical patterns of identities

Some identities don’t give way to the definitions alone. They need a trick from algebra: multiplying by a conjugate, $(a - b)(a + b) = a^2 - b^2$, or $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, or replacing a 1 by $\sec^2 x - \tan^2 x$. Study how each example below gets started: that first move is the part to learn.

Question 1

Prove that:

Question 2

Prove that:

Question 3

Prove that (θ and A acute):