A Quick Survey of Trigonometry: Part I

K Ahmed and PR Hewitt

Trigonometry is all about triangles. A “trigon” is a figure with 3 sides; “metric” refersto measurement. We will begin our study of trigonometry classically, in the context of righttriangles (that is, triangles which contain a right angle). We will then show how to representtrig functions as coordinates of points on the unit circle (that is, a circle of radius 1). Wewill then derive many important identities for the trig functions, and apply these to studytheir limits and derivatives. Finally we will look at the inverses of the trig functions.

1. Ancient geometry

In section 1 we will take a general look at the fundamental ideas needed to study trigono-metric functions. We will start our study of a brief review of similar triangles. We will thendefine trigonometric ratios, and apply them to solve triangles. Finally, we will see the famousPythagoras Theorem in its trigonometric form, and use it to determine information aboutany triangle.

1.1. Similar triangles. One of the beautiful and fundamental results from ancientplane geometry is the following:

Theorem 1. If two triangles have equal angles then their correspondingsides are in constant proportion.

This is illustrated below.

α

α

ββ

γ

γ

A1

A2B1

B2

C 1

C 2

T

T

1

2

Figure 1. Corresponding sides are

in constant proportion: A1/A2 = B1/B2 = C1/C2.

In this picture we see two triangles, T1 and T2, with equal angles, labeled with the Greekletters α (“alpha”), β (“beta”), and γ (“gamma”). We label the sides in the traditionalway, with the side opposite an angle using the corresponding Latin letter. In symbols, thetheorem above states that

A1

A2

=B1

B2

=C1

C2

. (1)

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Another way to say this is that T1 is just a scaled version of T2, where the scale factor s isthe common ratio of the corresponding sides. We say that T1 and T2 are similar triangleswhen this happens.

When we have similar triangles we can apply a little algebra to derive lots of otherequations between ratios. For example:

A1

B1

=A2

B2

, andA1

C1

=A2

C2

, andB1

C1

=B2

C2

, . . . (2)

These equations say that the ratios of two sides in a given triangle depend only on theangles in that triangle, and not on the size of the triangle. This fact is the basis for creatingtrigonometric functions , which are functions where the input is an angle and the outputis a ratio.

Example 1. Pictured below is a schematic representation of an imageprojected through a lens. Distances are given in centimeters. Explain whythe two right triangles T1 and T2 are similar. Find the scale factor, and thedistances x1 and y1.

α 1

x1

y1

α 2

β 1

γ 2

β 2

T2

T1

γ 1

27

35

3

Figure 2. An image projected through a lens.

Solution: Angles γ1 and γ2 are each right angles, and angles α1 andα2 are vertical angles , formed at the intersection of two straight lines.Vertical angles are equal. Angles β1 and β2 are equal because they are com-plementary to the equal angles α1 and α2. Hence, by theorem 1, trianglesT1 and T2 are similar.

Scale factor: 27/3 = 9. Distance x1: x1/35 = 9 cm, so x1 = 9 ·35 = 315 cm. Distance y1: using Pythagoras theorem we have y1 =√

3152 + 272 ≈ 17.088 cm, to 3 decimal places. (If you have forgottenPythagoras Theorem, don’t despair: we will review it in the next section.)

Problem 1. A shadow 5 m long is cast by a light pole 6 m high shiningon a person 2 m tall. Why are the two triangles similar? Find the scalefactor and the distance x from the person to the base of the light pole.

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2

5x

Figure 3. A shadow cast by a light shining on a person.

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More details, a little history, and bit of trivia. α, β, and γ are the first three letters of the Greek alphabet. Ourword “alphabet” is just a shortening of “alpha-beta-gamma”, which was used the way we would say “ABCs”. The Romans gottheir alphabet from the Greeks. The Roman empire spread the alphabet to Europe. The Greeks learned the alphabet from thePhoenicians, a Semitic people.

(The Semites also include Arabs, Jews, and many other people of the Middle East. In the biblical tradition, these are thepeople descended from Noah’s son Shem.)

The symbols of the Semitic alphabet were originally simplified versions of Egyptian hieroglyphics. This great inventiondates from the Egypt’s “Middle Kingdom”, roughly 4000 years ago, when many Semites lived under Egyptian rule.

1.2. Right triangles. In a planar triangle the 3 angles sum to a straight angle — oras Euclid would have said it, “two right angles”. You can see this by cutting out a papertriangle then tearing off the angles and fitting them together.

α

β

γ

βα γ

Figure 4. α + β + γ = straight angle = two right angles

We focus on right triangles, which are triangles in which one of the angles is a rightangle. If we have labeled the triangle so that γ is the right angle then α and β are comple-mentary angles , which is to say that α+β is a right angle. The angles α and β are acuteangles , meaning that each is less than a right angle. The side opposite the right angle iscalled the hypotenuse , while the sides opposite the acute angles are called the legs .

Let T be any right triangle with an acute angle equal to α.

α

βC

B

A

Figure 5. C = hypotenuseA = opposite leg to α = adjacent leg to β

B = adjacent leg to α = opposite leg to β

The ratio of the opposite side A to the hypotenuse C depends only on α, not on T . We callthis ratio the sine of the angle α, and abbreviate it sin α:

sin α =opposite leg

hypotenuse=

A

C. (3)

The cosine of α, abbreviated cos α, is defined to be the ratio of the adjacent leg to thehypotenuse:

cos α =adjacent leg

hypotenuse=

B

C. (4)

Another way to say this is that the cosine of an angle is the sine of the complementary angle.3

There are four other possible ratios, and they also get names:

tangent α = tan α =opposite leg

adjacent leg=

A

B=

sin α

cos α(5)

cotangent α = cot α =adjacent leg

opposite leg=

B

A=

cos α

sin α(6)

secant α = sec α =hypotenuse

adjacent leg=

C

B=

1

cos α(7)

cosecant α = csc α =hypotenuse

opposite leg=

C

A=

1

sin α(8)

In each case, the “co”-function of an angle equals the function of the complementary angle.

More details, a little history, and bit of trivia. Trigonometry is different on a sphere. For example, the sum of theangles in a triangle is always more than two right angles. For example, start at the north pole; travel due south to the equator;turn right and travel due west one quarter of the way around the earth; then turn right and return to the north pole, travelingdue north. You have just completed a triangle with three right angles! In fact, the area of a spherical triangle is proportionalto the difference between the sum of its angles and a straight angle.

By the way, what do we mean by a triangle on a sphere? Doesn’t a triangle have to have straight sides? In the plane atriangle can have straight sides, but on a sphere the shortest distance between two points is not a straight line, but an arc of a“great circle”. A great circle is the intersection of the sphere and a plane which passes through the center of the sphere.

If you walk due north or due south then you are walking along a great circle. But unless you are at the equator, walkingdue east or west is not walking along a great circle. The east-west “lines” are called lines of latitude, and they are cut by planesparallel to the equator. If you walk one mile due west from some point in the US, for example, then you are less than one milefrom where you started. You can walk a shorter path home by turning slightly northeast, then walking along a great circlewhich eventually bends a bit north of the line of latitude.

equator

great circle

line of latitude

Figure 6. Following a line of latitude is longer than following a great circle.

You may not be able to see the difference in the crude picture above, but you can check it out on a globe, using a piece ofstring. The difference is certainly noticeable on long-range flights. Failure to map a route along a great circle can cost a lot offuel.

Ordinary plane trigonometry can be used to develop spherical trigonometry, since trig functions are intimately tied tocircles, as we shall see. The subject was thoroughly developed by Islamic scholars long before the invention of airplanes.Muslims around the world pray five times a day, each time facing Mecca. The direction they face is called qibla , and follows agreat circle. Qibla is determined very precisely at each mosque. Here in Toledo, qibla is fairly close to northeast, even thoughMecca is at a much more southern latitude.

1.3. Solving triangles. In the ancient world (in fact, until a few hundred years ago)trigonometry was used exclusively for astronomical calculations. The standard problemsolved using trigonometry was to determine all of the sides and angles in a right trianglegiven only part of the information. This was known as “solving a triangle”.

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Example 2. To solve the right triangle below we must find all missinginformation about its angles and sides, given that the angle β = 27.4◦ andthe side A = 5 m.

Solution: First, let’s find angle α. Remember, α + β + γ = 180◦. So,α = 180− 90− 27.4 = 62.6◦.

C

γ=90

B

α

β=27.4

A=5

Figure 7. Sketching a right triangle with A = 5 and α = 27.4.

To find the missing sides, we reflect on the trigonometric ratios. Note:Always use acute angles to calculate with trigonometric ratios.

In this case, let’s use the given angle β = 27.4◦ and the given side A = 5.What is the relationship between β and A? A is adjacent to β. So, we canset up the ratios

cos(27.4◦) = A/C

tan(27.4◦) = B/A

cos(27.4◦) = 5/C

C = 5/ cos(27.4) = 5.632

tan(27.4◦) = B/5

B = 5 tan(27.4) = 2.592

(Remember, set your calculator in degree mode when the angles are givenin degrees. Otherwise, use radian mode.)

Problem 2. Sketch and solve the following right triangles. (Let γ be theright angle.)

i. α = 42.9◦, and A = 155.ii. β = 31.9◦, and A = 1.74.

More details, a little history, and bit of trivia. Back in the day, a large part of trig class was spent learning howto use such tables. In fact, some old timers probably remember having to memorize many of these values! This was especiallytrue of civil engineers and others who worked in the field. Nowadays all you have to do is push the “sin x” or “cos x” buttonsto compute the trig functions. How easy is that? But there is a subtlety, having to do with units of measurement for angles.We will return to this point after learning about radian measure.

Although you don’t need to memorize tables of trig values, you should memorize the definitions of the trig functions and

their basic identities. There will be a one-page summary sheet after I post part II of these notes. Do yourself a favor: study it

and commit as much of it to memory as you can.

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1.4. Pythagoras theorem. One of the more famous theorems in all of mathematics isabout right triangles.

Theorem 2. The square of the hypotenuse of a right triangle equals thesum of the squares of the legs.

If the hypotenuse is C and the legs are A and B then this theorem says that

C2 = A2 + B2. (9)

This has come to be known as “Pythagoras Theorem”, although historians do not findmuch evidence that the theorem or its proof originated with Pythagoras, or even any of hisfollowers. We will bow to tradition and also call it Pythagoras Theorem.

If we divide both sides of Pythagoras Theorem by C2, we get the most fundamentaltrigonometric identity:

C2

C2=

A2

C2+

B2

C2

1 =

(A

C

)2

+

(B

C

)2

Since A/C = sin α and B/C = cos α, we can write this last line as

1 = sin2 α + cos2 α. (10)

A word about notation: when we write something like sin2 α we really should insteadwrite (sin α)2, to be consistent with our usual notation for exponents and order of operations.The trigonometric functions are the only functions where we use this funny notation. Again,this has become traditional, and we are loathe to challenge tradition. At any rate, be careful:sin2 α = (sin α)2 6= sin α2 = sin(α2).

More details, a little history, and bit of trivia. “Pythagoras” Theorem was known in ancient Egypt and ancientIraq long before Pythagoras. We do not have any record of their proofs or explanation for this theorem. They may not havehad any. However, there are very ancient proofs from other cultures. In China this theorem was called the “Kou-Ku” Theorem— which means something like the “Short-Long” Theorem. They had a very simple and very visual proof.

An algebraic quantity x2 is called a “square” because it gives the area of a square of side x. Thus, the Kou-Ku Theoremcan be interpreted geometrically to say that if we use the three sides of a right triangle to make three squares, then the sum ofthe areas of the smaller two squares is exactly the area of the larger square.

A

B

C

Figure 8. The two smaller squares togetherequal the area of the largest square.

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If we make a red square of side A + B and place around its inside edge four blue right triangles with legs A and B. Wewill see exposed a red square of side C. If we re-arrange the four blue triangles we see that the left over (red) area consists ofsquares of sides A and B. Hence

C2 = red area = A2 + B2.

B

C

A

BC

A

A B

B

CC

A

B A

A

B A

A

BB

Figure 9. A square of side A + B contains four right triangles and a square of side C.It also contains the same four triangles and squares of sides A and B.

There are many other proofs of this theorem. In Euclid’s Elements the proof shows how to cut up the large square intotwo pieces which have the same areas as the two smaller squares:

A

B

C

Figure 10. A square of side C is cut into piecesequaling the two squares of sides A and B.

This proof is based on the simple fact that the area of a parallelogram depends only on the length of one side and the distanceto the opposite side:

h

b b b

Figure 11. All of these parallelograms have the same area.

A good trivia question asks who is the only US President to have published a mathematical paper. Many people guessThomas Jefferson, since he was such a learned, inquisitive man. Oddly, very few people guess any of the recent Presidents. Thecorrect answer is James Garfield, who is otherwise known only as one of the four Presidents to be assassinated. Garfield’s proofwas somewhat similar to the ancient Chinese proof, but used a trapezoid and some algebra:

B

C

A

BC

A

Figure 12. Garfield’s proof: 12(A + B)2 = 1

2C2 + AB.

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1.5. The law of cosines. Pythagoras Theorem has an important generalization. If A,B, and C are the sides of any triangle, and if γ is the angle opposite C, then

C2 = A2 + B2 − 2AB cos γ. (11)

Suppose our triangle looks like this:

C

γ

B

A

Figure 13. Typical non-right triangle.

Drop a perpendicular to A from the vertex at top, so that we have two right triangles: onewith sides A− E, D, and C; the other with sides E, D, and B.

D

C

γ

A − E E

B

Figure 14. Dropping a perpendicular to A.

If we use Pythagoras Theorem twice and the definition of cosine as a ratio (4) we find that

C2 = (A− E)2 + D2 (Pythagoras Theorem)

= (A− E)2 + B2 − E2 (Pythagoras Theorem)

= A2 − 2AE + E2 + B2 − E2

= A2 + B2 − 2AE

= A2 + B2 − 2AB cos γ (definition of cosine)

Example 3. Use law of cosines to solve the triangle given the sidesA = 17, B = 10, and C = 12.

γ

B=10

C=12

A=17

α

β

Figure 15. Solving a triangle with given side lengths.

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Solution: First we use the law of cosines to find angle α:

A2 = B2 + C2 − 2BC cos(α)

cos(α) =B2 + C2 − A2

2BC

cos(α) =102 + 122 − 172

2 · 10 · 12= −0.1875

α = arccos(−0.1875) = 100.807◦

Notice that to find an angle given its cosine, we use the inverse functionof cosine, which we write as arccos but is sometimes written cos−1. Theinverse-cosine button on your calculator is probably labeled cos−1. However,we will avoid the notation cos−1 since it is too easily confused with 1/ cos,which is completely different. Similarly, when sine of an angle is given, weuse the inverse function of sine, arcsin, etc. We will study inverse functionsin section 4. For now, you should simply use the inverse functions on yourcalculator.

Next, we use the law of cosines to find angle β:

B2 = A2 + C2 − 2AC cos(β)

cos(β) =A2 + C2 −B2

2AC

cos(β) =172 + 122 − 102

2 · 17 · 12= 0.8162

β = arccos(0.8162) = 35.296◦

Finally, the angle γ can be found by using the fact that α+β+γ = 180◦.Hence γ = 180− 100.807− 35.296 = 43.897◦.

Problem 3. Sketch the triangles given the information below, then solvethem for the missing information.

i. α = 85◦, γ = 30◦, A = 3.985, and C = 2.ii. α = 133◦, B = 12, and C = 15.

Section summary. The main points we have learned in the first section:

• The laws of proportional triangles: formulas (1)–(2).• The definitions of the trig functions as ratios: formulas (3)–(8).• The technique of “solving triangles”: example 2.• Pythagoras Theorem and the Law of Cosines: formulas (9) and (11).

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2. Circular functions

What’s in store for section 2?

• We will represent sine and cosine as coordinates on the unit circle.• We will introduce radians, the natural way to measure angles.• We will look at the trig values at some special angles.

2.1. The unit circle. If we set x = cos α and y = sin α then the fundamental iden-tity (10) tells us that the point P with coordinates x and y lies on the unit circle .

yα

α= sin

x = cos α

P(x,y)

O

1

Figure 16. Each point on the unit circle x2 + y2 = 1

determines a right triangle with hypotenuse equal to 1.

This is the circle of radius 1 centered at the origin, with equation x2 + y2 = 1. In fact, thepoint (x, y) and the origin (0, 0) are the endpoints of the hypotenuse, of length 1, in a righttriangle with acute angle α, adjacent leg x, and opposite leg y.

Definition 1. Cosine and sine of an angle α are the x- and y-coordinatesof the point P on the unit circle such that the line OP makes an angle equalto α with the positive x-axis.

This representation defines sine and cosine for all angles, even obtuse ones, even negativeones! We measure the angle starting at the point (1, 0) on the positive x-axis, movingcounterclockwise for positive angles, and clockwise for negative angles.

(1,0)

(0,−1)

(0,1)

(−1,0)

positive

angles

negative

angles

Figure 17. Positive and negative directions on the unit circle.

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For example, cos 0 = 1 and sin 0 = 0. One positive right angle takes us to the point (0, 1).That is, the cosine of a right angle equals 0, and the sine of a right angle equals 1. Continuingalong, after two right angles (a straight angle) we come to the point (−1, 0); after three wecome to (0,−1); and four brings us back full circle, to the point (1, 0).

Acute angles are in the first quadrant. Their negatives are in the fourth. Angles betweena right angle and a straight angle are in the second.

Since x and y take on both positive and negative values, the value of sine and cosine canbe positive or negative. Sine is positive in the first and second quadrant. Cosine is positivein the first and fourth. Both are negative in the third quadrant.

More details, a little history, and bit of trivia. Although the foundations of trigonometry lie in plane geometry, theancient Greeks did not use the trig functions that we do. Instead they looked at chord subtended (“stretched”) by a circulararc. Rather than compute the ratios sine and cosine, they computed and tabulated the chord as a function of the angle.

We can see why they were interested in the relation of a chord and an arc when we remember the application they made oftrigonometry: astronomy. They imagined the stars and planets as circling the earth, and used trig to predict their movements.

However, the algebra involved in using the chord function directly is not very neat. Lots of factors of 2 and 1/2 pop upall over the place. The modern approach to trig was developed first in medieval India. They realized that the function whichmeasures the half-chord of twice the angle is easier to work with than the chord of an angle. This eliminates all of the messyfactors.

yα

α y

= sin α

2y = chord of 2α

Figure 18. The sine of an angle equals the half-chord of twice the angle.

By the way, the Sanskrit word for half-chord is “jya-ardha”. Through the centuries this became shortened to just “jya”.The pronunciation of this word eventually became “jiva”. When the great Islamic scholars of Baghdad studied Indian astronomyand trigonometry, they borrowed this word as “jiba”.

In written Arabic and other Semitic languages, vowels are often omitted, since they can usually be inferred from context.A few centuries later Europeans studied the Arabic trig texts, and they misread the word “jiba” as “jaib”, since they didn’tunderstand the context. This translates to Latin as “sinus”, which originally meant “bay”, but came to mean “hollow” or“cavity”. For example, the Romans used it to refer to the sinus in their nose, which is where we get that term. It is also theroot of the word “sinewy”. As Berlinghoff and Gouvea point out in Math through the Ages, the graph of the sine function doesindeed have a rather sinewy character!

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2.2. Radian measure. The representation of the trig functions on the unit circle leadsto a natural measure for angles, called radians.

Definition 2. The radian measure of an angle is the length of the arc ofthe unit circle spanned by the angle.

P(x,y)

Q(1,0)

α

1

Figure 19. The radian measure of α is

the length of the arc from Q(1, 0) to P (x, y).

Since the radius of the unit circle is 1, the circumference of the entire circle is 2π. Onequarter of this length is π/2, which is thus the radian measure of a right angle. Similarly, astraight angle has radian measure π.

Example 4. An angle measures 43◦. How large is the angle in radianmeasure? Another angle measures 1.3 radians, what is its degree equiva-lence?

Solution: We use the fact that 180◦ = π to convert between degreesand radian measures of an angle. To convert from degree to radian, wemultiply the degrees by the conversion factor π/180◦. To convert fromradian to degree measure, multiply the radian by the conversion factor180◦/π. Hence

43◦ = 43 · π

180◦= 0.7505 radian

1.3 radian = 1.3 · 180◦

π= 74.485◦

Problem 4. Use the appropriate conversion factor to convert the follow-ing angles measures.

i. Convert 122◦ to radians.ii. Convert 0.23 radian to degrees.

iii. Convert 6.12 radian to degrees.

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Example 5. The point where x = y is a good illustration of the rela-tionship of radian measure to cosine and sine values. This is the point 1/8of the way around the entire circle, and so the radian measure of this angleis 2π/8, or π/4. On the other hand, if x = y then the fundamental trigidentity (10) tells us that

1 = x2 + y2 = x2 + x2 = 2x2,

hence x =√

1/2. That is, when x = y the radian measure is π/4 and

cos(π/4) = sin(π/4) =√

1/2 = 1/√

2 =√

2/2.

Example 6. Another familiar triangle from elementary geometry is the“30-60-90” triangle:

A

C

B

α

β

Figure 20. A triangle with acute angles

α = 60◦ = π/3 radian, β = 30◦ = π/6 radian.

One of the things we should recall from elementary geometry is that in sucha triangle, B is one-half C. In other words,

cos(π/3) = sin(π/6) = B/C = 1/2.

Pythagoras Theorem tells us that

A2 = C2 −B2 = C2 − C2/4 = 34C2.

Hence A =√

32

C, and

cos(π/6) = sin(π/3) = A/C =√

3/2.

We summarize the special acute angles and their sine and cosine valuesin the picture below. You should memorize these special values.

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Figure 21. Sine and cosine of some special angles.

Problem 5. Fill in all of the missing entries in the table below, andlocate the angles on the unit circle. Give exact values, not decimals. Donot use a calculator.

α in degrees α in radians sin α cos α tan α

0◦

π/6√

2/2 1

60◦

1 0√

3/2 −√

3

3π/4

1/2 −√

3/2

π

210◦

−√

2/2 1

5π/4

−1/2 −√

3/2

3π/2

300◦

√2/2 −1

11π/6

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If we need the length of an arc in a circle of a different radius, then we simply scaleappropriately:

Theorem 3. The length of an arc spanned by an angle α in a circle ofradius r is rα, where α is measured in radians.

P

α

r

Q

s = r α

Figure 22. The length of the arc from Q to P is rα.

More details, a little history, and bit of trivia. We have mentioned that trig was a tool for astronomers. Why didthe ancient world need astronomy? Calendars.

Much of the way we keep time dates back ancient Iraq, the world’s first civilization. The ancient Iraqis divided the dayand the night into 12 hours each, for a total of 24 each calendar day. They divided each hour into 60 minutes and each minuteinto 60 seconds. Note that 60 is five twelves. Why all these twelves and sixties? Around the world people found it convenientto count either in tens or twelves: ten because of the number of fingers, twelve because it is evenly divided by lots of factors.Probably some time in the pre-history of ancient Iraq two counting systems merged — one based on ten and one based ontwelve. Since 60 is the least common multiple of 10 and 12, counting in sixties makes it relatively easy to translate from onebase to another.

All of that is just speculation. It is known that the Iraqis had a sophisticated base-60 numeration system. They used aplace-value system, just like ours. When we write 237 and 723 we recognize these as the names of different numbers, becausethe digits 2, 3 and 7 appear in different places in the two numerals. The ancient Iraqis had the same system, except that theplaces represented not increasing powers of 10 — ones, tens, hundreds, thousands, . . . — but increasing powers of 60. Thinkabout it: students in ancient Iraq would have to learn multiplication tables up to 59 times 59!

This system continued to be used by astronomers across much of the world, well into modern times. But the system nevercaught on with anyone else, including mathematicians and scientists (except, of course, when they were doing trigonometry andastronomy). Most of the world used very primitive numeration system, more like Roman numerals than like our system.

The place-value system we use now was developed somewhere in the Hindu world, possibly in Cambodia over 1000 yearsago, when it was an Indian kingdom with increasing trade with China. Very possibly the Hindu system was influenced by theChinese merchants’ counting board, which in effect was a base-100 place-value system.

At any rate, as with trig and astronomy, Europe learned modern numeration from the Arabs, who in turn learned it fromthe Hindus. Hence we now call this the Hindu-Arabic system.

Back to ancient Iraq: they also divided a circle into 360 degrees, probably because this was six sixties, and also fifteentwenty-fours, and possibly also because it was very close to the number of days in a year. So, in ancient Iraq, a right anglewould have been 90 degrees. It is a sorry fact of life that some good ideas — like the place-value system — take a long timeto catch on, and sometimes never do; while many bad ideas — like Roman numerals, measuring angles in degrees, and incometaxes — refuse to go away.

The point of all of this historical rambling is to show how arbitrary the use of degrees is. For the most part you shouldalways use the more natural radian measure, rather than degrees, unless a problem requires you to do otherwise. This isespecially true when you use trig functions in calculus: the formulas you learn will not work if you use degrees. In particular,unless you are consciously using degrees, make sure your calculator is in radian mode.

You can find much more history of the calendar in the fun book Calendar, by David Duncan. I cannot resist one moregem from that book, which again shows how arbitrary decisions made eons ago affect us even today. I am referring specificallyto the days of the week. Why are there seven? And where did we get their names?

Ancient Iraqis noticed that seven of the stars wander. The word “planet” comes from the Greek word for “wandering”.They worshiped these seven as gods, in descending order of apparent “height” in the heavens: Saturn, Jupiter, Mars, the Sun,Venus, Mercury, and the Moon. (They had different names, of course.) Their priests rotated through a series of round-the-clockprayers, each hour being devoted to the next god in succession.

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the sun

saturn

mars

venus

mercury

the moon jupiter

Figure 23. Order of worship in ancient Iraq.

So the first hour of the first day was devoted to Saturn, and it was natural to call this “Saturn-day”. After three full prayercycles on a Saturday, there are three hours left, which are devoted to Saturn, Jupiter, and Mars. Thus, the first hour of thenext day is devoted to the Sun, and so it is natural to call this “Sun-day”. Three more full prayer cycles, with three hours leftover, and they arrived at the first hour of the next day, devoted to the Moon. Hence this became “Moon-day”. The remainingdays become “Mars-day”, “Mercury-day”, “Jupiter-day”, and “Venus-day”, before returning again to “Saturn-day”.

This scheme was later brought to Rome by Caesar, who learned it when he conquered Egypt. He used Roman names(Saturn, Sol, Luna, Mars, Mercury, Jove, Venus) and we see many of these names, somewhat altered, in modern Latinatelanguages. In French: lundi, mardi, mercredi, jeudi, etc. In Spanish: lunes, martes, miercoles, jueves, etc.

The English names are a funny combination of Latinate and Germanic. For example, the Germanic equivalents of thegods Mars, Mercury, Jove, and Venus are Tiuw, Wodin, Thor, and Frida. Why “Saturday” persisted in English is yet anotherarbitrary twist of fate.

Section summary. The main points we have learned in the second section:

• Sine and cosine are coordinates on the unit circle: definition 1.• The natural way to measure angles, using radians: definition 2.• The values of the trig functions at some special angles.

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3. Identities

We already have a fundamental identity for sine and cosine:

sin2 α + cos2 α = 1. (12)

If we divide both sides of this equation by cos2 α

sin2 α

cos2 α+

cos2 α

cos2 α=

1

cos2 αthen we get a second one:

tan2 α + 1 = sec2 α. (13)

In section 3 we will look at the periodic nature of trigonometric functions and developtrigonometric identities that are essential to solving problems using trigonometric functions.

3.1. Symmetry and periodicity. The symmetry of the circle leads to many usefultrig identities. First of all, adding 2π to an angle takes all the way round and back to thesame point. Hence the trig functions are periodic:

sin(α + 2π) = sin α, cos(α + 2π) = cos α. (14)

We can repeatedly add or subtract any multiple of 2π to α and not change the values of sineor cosine:

sin(α + 2nπ) = sin α, cos(α + 2nπ) = cos α, for n = 0,±1,±2,±3, . . . . (15)

If we reflect the unit circle across the x-axis then the angle α becomes −α, and the point(x, y) becomes (x,−y):

(x,y)

α

−α

(x,−y)

Figure 24. Reflecting across the x-axis:

cos(−α) = cos α, sin(−α) = − sin α.

We find thatcos(−α) = cos α, sin(−α) = − sin α. (16)

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If instead we reflect across the y-axis

(x,y)

α

(−x,y)

π−α

Figure 25. Reflecting across the x-axis:

cos(π − α) = − cos α, sin(π − α) = sin α.

then we find that

cos(π − α) = − cos α, sin(π − α) = sin α. (17)

Increasing the angle α by π takes us to the point where both coordinates are negated:

(x,y)

α

(−x,−y)

π

Figure 26. Increasing α by π:

cos(α + π) = − cos α, sin(α + π) = sin α.

cos(π − α) = − cos α, sin(π − α) = sin α. (18)

These identities, together with the periodicity of sine and cosine, tell us that tangent andcotangent repeat every π:

tan(α + π) =sin(α + π)

cos(α + π)=− sin α

− cos α= tan α.

Similarly for cotangent.

tan(α + nπ) = tan α, cot(α + nπ) = cot α, for n = 0,±1,±2,±3, . . . . (19)

Since π/2 is a right angle, the complement of α is π/2 − α. Our rule for co-functionsreads:

sin(π/2− α) = cos α, tan(π/2− α) = cot α, sec(π/2− α) = csc α. (20)18

Example 7. Use the special values (illustrated in figure 21 or tabulatedin problem 5) and the identities (14)–(20) to find the exact values of thefollowing. Justify your answers by citing by number the formulas you use.Do not use a calculator.

i. cos(16π3

)

ii. tan(16π3

)Solution:

i. cos(16π

3) = cos(4π +

4π

3) (formula (14))

= cos(4π

3) = cos(π +

π

3) (formula (18))

= − cos(π

3) = −1

2(figure 21)

ii. tan(16π

3) = tan(5π +

π

3) (formula (19))

= tan(π

3) =

√3 (figure 21)

Problem 6. Use the special values (illustrated in figure 21 or tabulatedin problem 5) and the identities (14)–(20) to match the values on the leftwith their equals on the right. (Do not use a calculator.)

sin(7π/3) sin(3π/4)sin(π/6) cos(π/6)

cos(11π/4) cos(3π/4)cos(7π/4) − cos(2π/3)

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3.2. The addition and subtraction formulas. All of the identities of the previoussection are really special cases of what are called the addition and subtraction formulas:

cos(α± β) = cos α cos β ∓ sin α sin β (21)

sin(α± β) = sin α cos β ± cos α sin β (22)

We show how the definitions of the trig functions as ratios (3)–(8) can be used to derivethe formula for sin(α− β). Consider the various right triangles in the following picture:

d

f

g

1

α − βπ/2−β

π/2−β

β

β

e

Figure 27. Cosine of a difference.

We are seeking a formula d = sin(α − β). Since g = cos α and f/g = tan β we find thatf = cos α tan β. Since e + f = sin α we find that e = sin α − cos α tan β. Finally, sinced/e = cos β we find that

sin(α− β) = d = e cos β

= (sin α− cos α tan β) · cos β

= sin α cos β − cos α · sin β

cos β· cos β

= sin α cos β − cos α sin β.

The other three identities are similar.We can now derive addition and subtraction formulas for tangent:

tan(α± β) =sin(α± β)

cos(α± β)

=sin α cos β ± cos α sin β

cos α cos β ∓ sin α sin β

If we now divide numerator and denominator by cos α cos β and simplify we find that

tan(α± β) =tan α± tan β

1∓ tan α tan β. (23)

Example 8. Use the addition and subtraction formulas (21), (22),and (23) and the table of special values to find the exact values of thefollowing.

i. sin(7π12

)ii. tan( π

12)

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Solution:

i. sin(7π

12) = sin(

π

3+

π

4)

= sin(π

3) cos(

π

4) + cos(

π

3) sin(

π

4) (formula (22))

=

√3

2·√

2

2+

1

2·√

2

2(figure 21)

=1 +

√6

4

ii. tan(π

12) = tan(

π

3− π

4)

=tan(π

3)− tan(π

4)

1 + tan(π3) tan(π

4)

(formula (23))

=

√3− 1

1 +√

3 · 1(figure 21)

Problem 7. Suppose that angles α and β are in quadrant II, and thatsin α = 0.2 and cos β = −0.7. Use the fundamental trig identity (12) tofind cos α, sin β, tan α, and tan β. (Remember that the quadrant tells youthe sign of the square roots.)

Next use the addition and subtraction formulas (21), (22), and (23) tofind sin(α+β), sin(α−β), cos(α+β), cos(α−β), tan(α+β), and tan(α−β).

(You may use your calculator on this problem.)

More details, a little history, and bit of trivia. For millenia, trig identities were used to simplify the algorithm formultiplying numbers with many digits. The process was known as prosthaphaeresis. It was based on the identity

sin α sin β =1

2(cos(α− β)− cos(α + β)),

which is easily derived from the addition and subtraction formulas.Let’s look at an example. Say you wanted to multiply 0.98273463 by 0.23964771 — by hand! You could use the usual

process you learned in grade school, but this is tedious and error prone. Here is a quicker way: look in a table of trig values forangles α and β such that sin α = 0.98273463 and sin β = 0.23964771. You will find that α = 1.384703435 and β = 0.2420029709.Compute their sum and difference — by hand! This step is not trivial, but is a lot easier than multiplication. The result:α + β = 1.626706406, α − β = 1.142700464. Now look up the cosines of these numbers in your table. Result: cos(α − β) =0.4151392522, cos(α + β) = −0.05588095519. Now subtract these values, and divide by 2. Final result: 0.2355101037.

The trade-off: two additions, one subtraction, one division by 2, and four table look-ups in place of one multiplication. Ifyou are not sure this is worth it try doing several multiplications the grade-school way and see how long it takes you and howmany errors you make.

In the sixteenth century a brilliant but very strange Scottish mathematician named John Napier invented a better way:logarithms. (Even if it weren’t a better way, it is certainly easier to say “logarithm” than “prosthaphaeresis”.) He used theprinciples of calculus to devise a function ln(x), called the “natural logarithm”, with the property that

ln(xy) = ln(x) + ln(y).

That is, Napier’s logarithm turns multiplication into addition. So, to multiply 0.98273463 by 0.23964771 using logarithms yousimply look up their logs in a suitable table; add the results; then look up the “inverse log” in the same table. The greatmathematician Laplace said that Napier’s invention, “by shortening the labors, doubled the life of the astronomer”.

By the way, the inverse of the logarithm is the familiar exponential (to the base e). Euler later discovered a link from theexponential back to the trig functions: if i =

√−1 then

eiα = cos α + i sin α.

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3.3. Double angles and half angles. If we take α = β in the addition formulas (21)–(23) then we get double-angle formulas :

cos(2α) = cos2 α− sin2 α = 2 cos2 α− 1 = 1− 2 sin2 α (24)

sin(2α) = 2 sin α cos α (25)

tan(2α) =2 tan α

1− tan2 α(26)

If we replace α by α/2 in the double-angle formula (24) we get half-angle formulas :

cos(α) = 2 cos2(α/2)− 1 = 1− 2 sin2(α/2),

hence

cos(α/2) = ±√

12(1 + cos α) (27)

sin(α/2) = ±√

12(1− cos α) (28)

The sign on this formula has to be determined in each case by noting which quadrant α/2is in.

If we divide equation (28) by equation (27) we get a half-angle formula for tangent:

tan(α/2) = ±√

1− cos α

1 + cos α.

But we can be more precise. Since

1− cos α

1 + cos α=

1− cos2 α

(1 + cos α)2=

sin2 α

(1 + cos α)2

and similarly

1− cos α

1 + cos α=

(1− cos α)

1− cos2 α=

(1− cos α)2

sin2 α

we find that

tan(α/2) =sin α

1 + cos α=

1− cos α

sin α. (29)

In these formulas there is no ambiguity of signs, as you can verify by considering the quad-rants where sine and tangent are positive.

Example 9. Suppose that angle α is in quadrant I, β is in quadrant IV,sin α = 0.2, and cos β = 0.7. Use the fundamental trig identity (12) tofind cos α, sin β, tan α, and tan β. Next use the double-angle formulas andhalf-angle formulas to find sin(α/2), sin(2α), cos(β/2), cos(2β), tan(2α),and tan(β

2). (Keep 4 decimal places.)

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Solution: In quadrant I both sine and cosine are positive. In quadrant IIcosine is positive but sine is negative. Hence

cos α =√

1− 0.22 = 0.9798

sin β = −√

1− 0.72 = −0.7141

tan α = 0.2/0.9798 = 0.2041

tan β = −0.7/0.7141 = −0.9802

If 0 < α < π/2 then 0 < α/2 < π/4. In particular, α/2 is also in quadrant I.If 0 > β > −π/2 then 0 > β/2 > −π/4. In particular, β/2 is also inquadrant IV. Hence

sin(α/2) =√

12(1− cos α) = 0.1005

sin(2α) = 2 sin α cos α = 0.3919

cos(β/2) =√

12(1 + cos β) = 0.7071

cos(2β) = 2(cos β)2 − 1 = −0.02

tan(2α) =2 tan α

1− tan2 α= 0.4259

tan(β/2) =sin β

1 + cos β= −0.4201

Problem 8. Suppose that angle α and β are in quadrant II, and thatsin α = 0.4 and cos β = −0.3. Use the fundamental trig identity (12) tofind cos α, sin β, tan α, and tan β. Next use the double-angle formulas andhalf-angle formulas to find sin(α/2), sin(2α), cos(β/2), cos(2β), tan(2α),and tan(β

2). (Keep 4 decimal places.)

Section summary. The main identities we have learned in the third section:

• Fundamental formulas (12) and (13).• Periodicity formulas (14), (19).• Symmetry formulas (16), (17), (18), and (20).• Addition and subtraction formulas (21), (22), and (23).• Double-angle formulas (24), (25), and (26).• Half-angle formulas (27), (28), and (29).

Toledo OH USA

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