Download A Quick Survey of Trigonometry: Part I

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” refers
to measurement. We will begin our study of trigonometry classically, in the context of right
triangles (that is, triangles which contain a right angle). We will then show how to represent
trig functions as coordinates of points on the unit circle (that is, a circle of radius 1). We
will then derive many important identities for the trig functions, and apply these to study
their 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 trigonometric functions. We will start our study of a brief review of similar triangles. We will then
define trigonometric ratios, and apply them to solve triangles. Finally, we will see the famous
Pythagoras Theorem in its trigonometric form, and use it to determine information about
any triangle.
1.1. Similar triangles. One of the beautiful and fundamental results from ancient
plane geometry is the following:
Theorem 1. If two triangles have equal angles then their corresponding
sides are in constant proportion.
This is illustrated below.
C1
β
β
C2
α
A1
T1
α
B1
T2
A2
γ
B2
γ
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 Greek
letters α (“alpha”), β (“beta”), and γ (“gamma”). We label the sides in the traditional
way, with the side opposite an angle using the corresponding Latin letter. In symbols, the
theorem above states that
B1
C1
A1
=
=
.
(1)
A2
B2
C2
1
Another way to say this is that T1 is just a scaled version of T2 , where the scale factor s is
the common ratio of the corresponding sides. We say that T1 and T2 are similar triangles
when this happens.
When we have similar triangles we can apply a little algebra to derive lots of other
equations between ratios. For example:
A2
A1
A2
B1
B2
A1
=
, and
=
, and
=
, ...
(2)
B1
B2
C1
C2
C1
C2
These equations say that the ratios of two sides in a given triangle depend only on the
angles in that triangle, and not on the size of the triangle. This fact is the basis for creating
trigonometric functions, which are functions where the input is an angle and the output
is a ratio.
Example 1. Pictured below is a schematic representation of an image
projected through a lens. Distances are given in centimeters. Explain why
the two right triangles T1 and T2 are similar. Find the scale factor, and the
distances x1 and y1 .
β1
y1
27
T1
γ1
γ2
35
α1
α2
x1
T2
3
β2
Figure 2. An image projected through a lens.
Solution: Angles γ1 and γ2 are each right angles, and angles α1 and
α2 are vertical angles, formed at the intersection of two straight lines.
Vertical angles are equal. Angles β1 and β2 are equal because they are complementary to the equal angles α1 and α2 . Hence, by theorem 1, triangles
T1 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 forgotten
Pythagoras 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 shining
on a person 2 m tall. Why are the two triangles similar? Find the scale
factor and the distance x from the person to the base of the light pole.
6
2
x
5
Figure 3. A shadow cast by a light shining on a person.
2
More details, a little history, and bit of trivia. α, β, and γ are the first three letters of the Greek alphabet. Our
word “alphabet” is just a shortening of “alpha-beta-gamma”, which was used the way we would say “ABCs”. The Romans got
their alphabet from the Greeks. The Roman empire spread the alphabet to Europe. The Greeks learned the alphabet from the
Phoenicians, a Semitic people.
(The Semites also include Arabs, Jews, and many other people of the Middle East. In the biblical tradition, these are the
people descended from Noah’s son Shem.)
The symbols of the Semitic alphabet were originally simplified versions of Egyptian hieroglyphics. This great invention
dates 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 — or
as Euclid would have said it, “two right angles”. You can see this by cutting out a paper
triangle then tearing off the angles and fitting them together.
β
α
β
α
γ
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 right
angle. If we have labeled the triangle so that γ is the right angle then α and β are complementary angles, which is to say that α + β is a right angle. The angles α and β are acute
angles, meaning that each is less than a right angle. The side opposite the right angle is
called 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.
β
C
A
α
B
Figure 5. C = hypotenuse
A = 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 call
this ratio the sine of the angle α, and abbreviate it sin α:
sin α =
A
opposite leg
= .
hypotenuse
C
(3)
The cosine of α, abbreviated cos α, is defined to be the ratio of the adjacent leg to the
hypotenuse:
adjacent leg
B
cos α =
= .
(4)
hypotenuse
C
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:
opposite leg
adjacent leg
adjacent leg
cotangent α = cot α =
opposite leg
hypotenuse
secant α = sec α =
adjacent leg
hypotenuse
cosecant α = csc α =
opposite leg
tangent α = tan α =
A
B
B
=
A
C
=
B
C
=
A
=
sin α
cos α
cos α
=
sin α
1
=
cos α
1
=
sin α
=
(5)
(6)
(7)
(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 the
angles 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, traveling
due north. You have just completed a triangle with three right angles! In fact, the area of a spherical triangle is proportional
to 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 a
triangle 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, walking
due east or west is not walking along a great circle. The east-west “lines” are called lines of latitude, and they are cut by planes
parallel to the equator. If you walk one mile due west from some point in the US, for example, then you are less than one mile
from where you started. You can walk a shorter path home by turning slightly northeast, then walking along a great circle
which eventually bends a bit north of the line of latitude.
great circle
line of latitude
equator
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 of
string. The difference is certainly noticeable on long-range flights. Failure to map a route along a great circle can cost a lot of
fuel.
Ordinary plane trigonometry can be used to develop spherical trigonometry, since trig functions are intimately tied to
circles, 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 a
great circle. Qibla is determined very precisely at each mosque. Here in Toledo, qibla is fairly close to northeast, even though
Mecca 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 problem
solved using trigonometry was to determine all of the sides and angles in a right triangle
given only part of the information. This was known as “solving a triangle”.
4
Example 2. To solve the right triangle below we must find all missing
information about its angles and sides, given that the angle β = 27.4◦ and
the side A = 5 m.
Solution: First, let’s find angle α. Remember, α + β + γ = 180◦ . So,
α = 180 − 90 − 27.4 = 62.6◦ .
α
C
B
γ=90
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 can
set 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 given
in degrees. Otherwise, use radian mode.)
Problem 2. Sketch and solve the following right triangles. (Let γ be the
right 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 how
to use such tables. In fact, some old timers probably remember having to memorize many of these values! This was especially
true of civil engineers and others who worked in the field. Nowadays all you have to do is push the “sin x” or “cos x” buttons
to 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.
5
1.4. Pythagoras theorem. One of the more famous theorems in all of mathematics is
about right triangles.
Theorem 2. The square of the hypotenuse of a right triangle equals the
sum of the squares of the legs.
If the hypotenuse is C and the legs are A and B then this theorem says that
C 2 = A2 + B 2 .
(9)
This has come to be known as “Pythagoras Theorem”, although historians do not find
much evidence that the theorem or its proof originated with Pythagoras, or even any of his
followers. We will bow to tradition and also call it Pythagoras Theorem.
If we divide both sides of Pythagoras Theorem by C 2 , we get the most fundamental
trigonometric identity:
A2 B 2
C2
=
+
C2
C2 C2
2 2
A
B
1=
+
C
C
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 instead
write (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 ancient
Iraq long before Pythagoras. We do not have any record of their proofs or explanation for this theorem. They may not have
had 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 Theorem
can be interpreted geometrically to say that if we use the three sides of a right triangle to make three squares, then the sum of
the areas of the smaller two squares is exactly the area of the larger square.
C
A
B
Figure 8. The two smaller squares together
equal the area of the largest square.
6
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. We
will 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 of
squares of sides A and B. Hence
C 2 = red area = A2 + B 2 .
B
B
A
A
A
B
C
C
B
B
A
A
C
C
A
B
B
B
A
A
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 into
two pieces which have the same areas as the two smaller squares:
C
A
B
Figure 10. A square of side C is cut into pieces
equaling 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 distance
to 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 guess
Thomas Jefferson, since he was such a learned, inquisitive man. Oddly, very few people guess any of the recent Presidents. The
correct answer is James Garfield, who is otherwise known only as one of the four Presidents to be assassinated. Garfield’s proof
was somewhat similar to the ancient Chinese proof, but used a trapezoid and some algebra:
C
C
A
B
B
A
Figure 12. Garfield’s proof:
7
1
(A
2
+ B)2 =
1 2
C
2
+ AB.
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
C 2 = A2 + B 2 − 2AB cos γ.
(11)
Suppose our triangle looks like this:
C
B
γ
A
Figure 13. Typical non-right triangle.
Drop a perpendicular to A from the vertex at top, so that we have two right triangles: one
with sides A − E, D, and C; the other with sides E, D, and B.
C
B
D
γ
A−E
E
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
C 2 = (A − E)2 + D2
2
(Pythagoras Theorem)
2
= (A − E) + B − E
2
2
2
(Pythagoras Theorem)
2
= A − 2AE + E + B − E
2
= A2 + B 2 − 2AE
= A2 + B 2 − 2AB cos γ
(definition of cosine)
Example 3.
Use law of cosines to solve the triangle given the sides
A = 17, B = 10, and C = 12.
β
A=17
C=12
γ
α
B=10
Figure 15. Solving a triangle with given side lengths.
8
Solution: First we use the law of cosines to find angle α:
A2 = B 2 + C 2 − 2BC cos(α)
B 2 + C 2 − A2
2BC
2
2
10 + 12 − 172
= −0.1875
cos(α) =
2 · 10 · 12
α = arccos(−0.1875) = 100.807◦
cos(α) =
Notice that to find an angle given its cosine, we use the inverse function
of cosine, which we write as arccos but is sometimes written cos−1 . The
inverse-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, we
use the inverse function of sine, arcsin, etc. We will study inverse functions
in section 4. For now, you should simply use the inverse functions on your
calculator.
Next, we use the law of cosines to find angle β:
B 2 = A2 + C 2 − 2AC cos(β)
A2 + C 2 − B 2
2AC
2
2
17 + 12 − 102
cos(β) =
= 0.8162
2 · 17 · 12
β = arccos(0.8162) = 35.296◦
cos(β) =
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 solve
them 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).
9
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 identity (10) tells us that the point P with coordinates x and y lies on the unit circle.
P(x,y)
1
y = sin α
α
O
x = cos α
Figure 16. Each point on the unit circle x2 + y 2 = 1
determines a right triangle with hypotenuse equal to 1.
This is the circle of radius 1 centered at the origin, with equation x2 + y 2 = 1. In fact, the
point (x, y) and the origin (0, 0) are the endpoints of the hypotenuse, of length 1, in a right
triangle 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-coordinates
of the point P on the unit circle such that the line OP makes an angle equal
to α with the positive x-axis.
This representation defines sine and cosine for all angles, even obtuse ones, even negative
ones! We measure the angle starting at the point (1, 0) on the positive x-axis, moving
counterclockwise for positive angles, and clockwise for negative angles.
(0,1)
positive
angles
(1,0)
(−1,0)
negative
angles
(0,−1)
Figure 17. Positive and negative directions on the unit circle.
10
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. Continuing
along, after two right angles (a straight angle) we come to the point (−1, 0); after three we
come 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 between
a 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 can
be positive or negative. Sine is positive in the first and second quadrant. Cosine is positive
in 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, the
ancient Greeks did not use the trig functions that we do. Instead they looked at chord subtended (“stretched”) by a circular
arc. 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 of
trigonometry: 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 up
all over the place. The modern approach to trig was developed first in medieval India. They realized that the function which
measures the half-chord of twice the angle is easier to work with than the chord of an angle. This eliminates all of the messy
factors.
y = sin α
α
α
2y = chord of 2α
y
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 astronomy
and 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’t
understand 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 the
root of the word “sinewy”. As Berlinghoff and Gouvêa point out in Math through the Ages, the graph of the sine function does
indeed have a rather sinewy character!
11
2.2. Radian measure. The representation of the trig functions on the unit circle leads
to a natural measure for angles, called radians.
Definition 2. The radian measure of an angle is the length of the arc of
the unit circle spanned by the angle.
P(x,y)
1
α
Q(1,0)
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. One
quarter of this length is π/2, which is thus the radian measure of a right angle. Similarly, a
straight angle has radian measure π.
Example 4. An angle measures 43◦ . How large is the angle in radian
measure? Another angle measures 1.3 radians, what is its degree equivalence?
Solution: We use the fact that 180◦ = π to convert between degrees
and radian measures of an angle. To convert from degree to radian, we
multiply the degrees by the conversion factor π/180◦ . To convert from
radian to degree measure, multiply the radian by the conversion factor
180◦ /π. Hence
π
43◦ = 43 ·
= 0.7505 radian
180◦
180◦
1.3 radian = 1.3 ·
= 74.485◦
π
Problem 4. Use the appropriate conversion factor to convert the following angles measures.
i. Convert 122◦ to radians.
ii. Convert 0.23 radian to degrees.
iii. Convert 6.12 radian to degrees.
12
Example 5. The point where x = y is a good illustration of the relationship of radian measure to cosine and sine values. This is the point 1/8
of the way around the entire circle, and so the radian measure of this angle
is 2π/8, or π/4. On the other hand, if x = y then the fundamental trig
identity (10) tells us that
1 = x2 + y 2 = x2 + x2 = 2x2 ,
hence x =
p
1/2. That is, when x = y the radian measure is π/4 and
cos(π/4) = sin(π/4) =
p
√
√
1/2 = 1/ 2 = 2/2.
Example 6. Another familiar triangle from elementary geometry is the
“30-60-90” triangle:
β
C
A
α
B
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 such
a 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 = C 2 − B 2 = C 2 − C 2 /4 = 34 C 2 .
√
Hence A =
3
C,
2
and
cos(π/6) = sin(π/3) = A/C =
√
3/2.
We summarize the special acute angles and their sine and cosine values
in the picture below. You should memorize these special values.
13
Figure 21. Sine and cosine of some special angles.
Problem 5.
Fill in all of the missing entries in the table below, and
locate the angles on the unit circle. Give exact values, not decimals. Do
not 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
3/2
3π/4
1/2
√
− 3/2
π
210◦
√
− 2/2
1
5π/4
−1/2
√
− 3/2
3π/2
300◦
√
11π/6
14
2/2
−1
If we need the length of an arc in a circle of a different radius, then we simply scale
appropriately:
Theorem 3. The length of an arc spanned by an angle α in a circle of
radius r is rα, where α is measured in radians.
P
r
s=r α
α
Q
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 did
the 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 day
and the night into 12 hours each, for a total of 24 each calendar day. They divided each hour into 60 minutes and each minute
into 60 seconds. Note that 60 is five twelves. Why all these twelves and sixties? Around the world people found it convenient
to 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 on
twelve. Since 60 is the least common multiple of 10 and 12, counting in sixties makes it relatively easy to translate from one
base to another.
All of that is just speculation. It is known that the Iraqis had a sophisticated base-60 numeration system. They used a
place-value system, just like ours. When we write 237 and 723 we recognize these as the names of different numbers, because
the digits 2, 3 and 7 appear in different places in the two numerals. The ancient Iraqis had the same system, except that the
places represented not increasing powers of 10 — ones, tens, hundreds, thousands, . . . — but increasing powers of 60. Think
about 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 never
caught on with anyone else, including mathematicians and scientists (except, of course, when they were doing trigonometry and
astronomy). 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 years
ago, when it was an Indian kingdom with increasing trade with China. Very possibly the Hindu system was influenced by the
Chinese 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 from
the 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 fifteen
twenty-fours, and possibly also because it was very close to the number of days in a year. So, in ancient Iraq, a right angle
would have been 90 degrees. It is a sorry fact of life that some good ideas — like the place-value system — take a long time
to catch on, and sometimes never do; while many bad ideas — like Roman numerals, measuring angles in degrees, and income
taxes — 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 should
always use the more natural radian measure, rather than degrees, unless a problem requires you to do otherwise. This is
especially 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 more
gem from that book, which again shows how arbitrary decisions made eons ago affect us even today. I am referring specifically
to 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-clock
prayers, each hour being devoted to the next god in succession.
15
saturn
jupiter
the moon
mercury
mars
venus
the sun
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 prayer
cycles on a Saturday, there are three hours left, which are devoted to Saturn, Jupiter, and Mars. Thus, the first hour of the
next day is devoted to the Sun, and so it is natural to call this “Sun-day”. Three more full prayer cycles, with three hours left
over, and they arrived at the first hour of the next day, devoted to the Moon. Hence this became “Moon-day”. The remaining
days 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 Latinate
languages. 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 the
gods Mars, Mercury, Jove, and Venus are Tiuw, Wodin, Thor, and Frida. Why “Saturday” persisted in English is yet another
arbitrary 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.
16
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
1
+
=
2
2
cos α cos α
cos2 α
then 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 develop
trigonometric identities that are essential to solving problems using trigonometric functions.
3.1. Symmetry and periodicity. The symmetry of the circle leads to many useful
trig identities. First of all, adding 2π to an angle takes all the way round and back to the
same 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 sine
or 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 that
cos(−α) = cos α, sin(−α) = − sin α.
17
(16)
If instead we reflect across the y-axis
π−α
(−x,y)
(x,y)
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)
π
α
(−x,−y)
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 and
cotangent repeat every π:
tan(α + π) =
sin(α + π)
− sin α
=
= tan α.
cos(α + π)
− cos 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-functions
reads:
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 tabulated
in problem 5) and the identities (14)–(20) to find the exact values of the
following. Justify your answers by citing by number the formulas you use.
Do not use a calculator.
)
i. cos( 16π
3
ii. tan( 16π
)
3
Solution:
16π
4π
) = cos(4π +
)
(formula (14))
i. cos(
3
3
4π
π
(formula (18))
= cos( ) = cos(π + )
3
3
π
1
= − cos( ) = −
(figure 21)
3
2
16π
π
ii. tan(
) = tan(5π + )
(formula (19))
3
3
√
π
= tan( ) = 3
(figure 21)
3
Problem 6. Use the special values (illustrated in figure 21 or tabulated
in problem 5) and the identities (14)–(20) to match the values on the left
with their equals on the right. (Do not use a calculator.)
sin(7π/3)
sin(π/6)
cos(11π/4)
cos(7π/4)
sin(3π/4)
cos(π/6)
cos(3π/4)
− cos(2π/3)
19
3.2. The addition and subtraction formulas. All of the identities of the previous
section 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 derive
the formula for sin(α − β). Consider the various right triangles in the following picture:
β
d
1
e
π/2−β
α−β
π/2−β
f
β
g
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 that
f = cos α tan β. Since e + f = sin α we find that e = sin α − cos α tan β. Finally, since
d/e = cos β we find that
sin(α − β) = d = e cos β
= (sin α − cos α tan β) · cos β
sin β
= sin α cos β − cos α ·
· 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:
sin(α ± β)
tan(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 ± tan β
tan(α ± β) =
.
(23)
1 ∓ tan α tan β
Example 8.
Use the addition and subtraction formulas (21), (22),
and (23) and the table of special values to find the exact values of the
following.
i. sin( 7π
)
12
π
ii. tan( 12
)
20
Solution:
7π
π π
i. sin( ) = sin( + )
12
3
4
π
π
π
π
= sin( ) cos( ) + cos( ) sin( )
(formula (22))
3
4
3
4
√ √
√
3
2 1
2
=
·
+ ·
(figure 21)
2 √2
2 2
1+ 6
=
4
π
π π
ii. tan( ) = tan( − )
12
3
4
tan( π3 ) − tan( π4 )
=
(formula (23))
1 + tan( π3 ) tan( π4 )
√
3−1
√
(figure 21)
=
1+ 3·1
Problem 7. Suppose that angles α and β are in quadrant II, and that
sin α = 0.2 and cos β = −0.7. Use the fundamental trig identity (12) to
find cos α, sin β, tan α, and tan β. (Remember that the quadrant tells you
the sign of the square roots.)
Next use the addition and subtraction formulas (21), (22), and (23) to
find 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 for
multiplying numbers with many digits. The process was known as prosthaphaeresis. It was based on the identity
sin α sin β =
1
(cos(α − β) − cos(α + β)),
2
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 for
angles 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. If
you are not sure this is worth it try doing several multiplications the grade-school way and see how long it takes you and how
many 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 the
principles 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 you
simply look up their logs in a suitable table; add the results; then look up the “inverse log” in the same table. The great
mathematician 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 the
exponential back to the trig functions: if i = −1 then
eiα = cos α + i sin α.
21
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
2 tan α
tan(2α) =
1 − tan2 α
(25)
(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
q
cos(α/2) = ± 12 (1 + cos α)
q
sin(α/2) = ± 12 (1 − cos α)
(27)
(28)
The sign on this formula has to be determined in each case by noting which quadrant α/2
is in.
If we divide equation (28) by equation (27) we get a half-angle formula for tangent:
r
1 − cos α
tan(α/2) = ±
.
1 + cos α
But we can be more precise. Since
1 − cos2 α
sin2 α
1 − cos α
=
=
1 + cos α
(1 + cos α)2
(1 + cos α)2
and similarly
1 − cos α
(1 − cos α)
(1 − cos α)2
=
=
1 + cos α
1 − cos2 α
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 quadrants 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) to
find cos α, sin β, tan α, and tan β. Next use the double-angle formulas and
half-angle formulas to find sin(α/2), sin(2α), cos(β/2), cos(2β), tan(2α),
and tan( β2 ). (Keep 4 decimal places.)
22
Solution: In quadrant I both sine and cosine are positive. In quadrant II
cosine 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 in
quadrant IV. Hence
q
sin(α/2) = 12 (1 − cos α) = 0.1005
sin(2α) = 2 sin α cos α = 0.3919
q
cos(β/2) = 12 (1 + cos β) = 0.7071
cos(2β) = 2(cos β)2 − 1 = −0.02
2 tan α
= 0.4259
tan(2α) =
1 − tan2 α
sin β
= −0.4201
tan(β/2) =
1 + cos β
Problem 8. Suppose that angle α and β are in quadrant II, and that
sin α = 0.4 and cos β = −0.3. Use the fundamental trig identity (12) to
find cos α, sin β, tan α, and tan β. Next use the double-angle formulas and
half-angle formulas to find sin(α/2), sin(2α), cos(β/2), cos(2β), tan(2α),
and tan( β2 ). (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
23