Unit 1: Trigonometry
Introduction
Trigonometry is a branch of mathematics that studies relationships between the sides and angles
of triangles. Trigonometry is very important in real life and has many applications. For example,
it is used in Geography to measure the distance between landmarks, in Astronomy to measure the
distance of nearby stars, in Chemistry number theory, medical imaging, Electronics, Engineering,
Architecture, Oceanography, Seismology, Phonetics, Image compression, etc. Most importantly,
Calculus (which appears in Units 2, 3, and 4) is also based on Trigonometry and Algebra. Thus,
it is imperative to study Trigonometry. In this unit, we look at the two popular system of angle
measurement; degrees and radians. We will also introduce an important group of functions called
trigonometric functions. These functions are often used in applications involving relationships
among the sides and angles of triangles. Lastly, we will sketch graphs of trigonometric functions
as well as solve equations involving trigonometric expressions.
Learning Outcomes
On successful completion of the Unit you should be able to:
• State the two common used units of angles
• State the six trigonometric ratios
• Simplify trigonometric functions using addition, double angle, factor, and product formulae
• Prove trigonometric identities
• Sketch graphs of trigonometric functions
• Solve trigonometric equations
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Key Terms
Ensure that you understand the key terms or phrases used in this unit as listed below.
Degree measure, radian measure, standard angle, coterminal angles, cofunctions, reference angles,
special angles, trigonmetric functions, trigonometric equations, trigonometric graphs.
1.1
Angles
An angle consists of two rays in a plane with a common endpoint. The two rays are called the
sides of the angle and the common endpoint is called the vertex. If we position the angle so that
one side is horizontal and points to the right, then we call this side the initial side. The other side
will be called the terminal side. We imagine the initial side as being in a fixed position and the
terminal side being able to move as shown in Figure 1.1.
Figure 1.1: Terminal side, initial side, and vertex
A counterclockwise rotation generates a positive measure while a clockwise rotation generates a
negative measure. See Figure 1.2.
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Figure 1.2: Negative and positive angles
1.1.1
Standard Position
If we position an angle in the xy-plane with the vertex at the origin and one side along the positive
x-axis, then the angle is said to be in standard position. The side that lies along the positive
x-axis is called the initial side. The other side opens counter-clockwise and is called the terminal
side.
Figure 1.3: An angle in standard position
Angles in standard position whose terminal sides lie on the x-axis or y-axis are called quadrantal
angles. These include 0◦ , 90◦ , 180◦ , 270◦ , and 360◦ .
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Figure 1.4: Quadrantal angles
1.1.2
Coterminal Angles
Coterminal angles are angles which share the same initial side and terminal side. In other words,
they are angles in standard position (angles with the initial side on the positive x-axis) that have a
common terminal side. For example, 30◦ is coterminal with −330◦ . Also, 45◦ , −315◦ , and 405◦
are all coterminal.
Figure 1.5: Coterminal angles
Example 1.1
Find a positive and negative angle coterminal with a 55◦ angle.
Solution
Negative angle: 55◦ − 360◦ = −305◦ .
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Positive angle: 55◦ + 360◦ = 415◦
Thus, −305◦ and 415◦ angle are coterminal with 55◦ angle.
Example 1.2
Find the angles of the least possible measure coterminal with each angle.
(a) 908◦
(b) −75◦
Solution
There are two different approaches that can be employed to obtain the solution to the problem.
Method 1:
If the given angle is positive, then subtract 360◦ repeatedly until an angle between
0◦ and 360◦ is obtained.
Method 2:
If the angle is positive, then divide the angle by 360◦ . The remainder will be a
coterminal angle between 0◦ and 360◦ . Now, if the given angle is negative, then divide the angle
by 360◦ . The remainder gives a negative coterminal angle between −360◦ and 0◦ . Finally, add
360◦ to this angle to get a coterminal angle between 0◦ and 360◦ .
Employing Methods 1 and 2 lead to the following.
(a) 908◦ − 360◦ = 548◦
548◦ − 360◦ = 188◦
Therefore, 188◦ is coterminal with angle 908◦ .
(b) −75◦ + 360◦ = 285◦ .
Therefore, 285◦ is the coterminal with angle −75◦ .
A general expression can be obtained to generate all angles coterminal with any given angle. For
example, we can obtain any angle coterminal with 60◦ by adding an appropriate integer multiple
of 360◦ to 60◦ . Let η represent any integer. Then the expression
60◦ + η × 360◦
represents all such coterminal angles. For example when η = −1 then 60◦ − 360◦ = −300◦ . Thus,
60◦ and −300◦ are coterminal.
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Degree Measure
Consider the following unit circle(a circle whose radius is 1).
As θ = ∠AOP increases, AP increases as well. Thus, AP is proportional to θ. Then, we shall
say that ∠AOP has degree measure if
θ
Length of arc AP
=
Circumference
360◦
Making θ the subject of the formula yields
θ=
360◦ × Length of arc AP
2π
Note that Circumference = 2πr where r = 1. Also, the angle is measured in an anticlockwise
direction from A.
1.3
Radian Measure
Another way of measuring angles is to compare the length of an arc formed by the angles with
the radius of the circle. The unit used in this method is called the radian (sometimes denoted by
rad). An angle’s measurement in radians is numerically equal to the length of a corresponding arc
of a unit circle. A radian is a measure of an angle θ when drawn as a central angle, subtends an
arc whose length equals the length of the radius of the circle.
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We shall say that ∠AOP has radian measure θ if θ = Length of the arc AP. Since arc length AP
is θ (see Section 1.2), ∠AOP has a radian measure. Now, if ∠AOP is a right angle, then
Circumference
4
2π
=
4
π
=
2
θ =
Thus, 90◦ =
π
radians and 180◦ = π radians. Now dividing both sides by π we get
2
1 radian =
Example 1.3
180◦
π rads
= 57.296◦ =⇒ 1◦ =
π
180◦
Convert the following angles to radians.
(a) 135◦
(b) 210◦
Solution
(a) If 180◦ = π radians
then 135◦ = less
3π
135◦ × π
=
rads
Thus, we have
◦
180
4
(b) If 180◦ = π radians
then 210◦ = more
210◦ × π
7π
Hence,
=
rads
◦
180
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Note: When converting an angle from degrees to radians and from radians to degrees, multiply by
π
180◦
rads, respectively.
rads
and
180◦
π
Example 1.4
(a)
Convert the following angles into degrees.
3π
rads
2
(b) 3 rads
(c)
π
rads
4
Solution
(a)
3π 180◦
3π
rads =
×
= 270◦
2
2
π
(b) 3 rads = 3 ×
(c)
180◦
= 3 × 57.296◦ = 171.9◦
π
π
π 180◦
rads = ×
= 45◦
4
4
π
1.4
Arc Length
Recall that
θ
θ
Length of arc AP
=
=
Circumference
3600
2π
Thus,
θ × Circumference
2π
θ × 2πr
=
Since Circumference = 2πr
2π
= rθ
Length of arc AP =
Therefore, the Length of arc AP = rθ, where θ is in radians.
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Example 1.5
An arc of a cirle of radius 10cm subtends an angle of 14.3◦ at the centre. Find the
length of the arc.
Solution
We first convert 14.3◦ to radians. Thus, 14.3◦ = 14.3◦ ×
π
= 0.2496 rads. Therefore,
180◦
Length of the arc = rθ
= 10 × 0.2496
= 2.496cm.
Activity 1.1
1. Give an expression that generates all the angles coterminal with each angle. Let n represent
any integer.
(a) 30◦
(b) 45◦
(c) −90◦
2. Determine whether the following angles are coterminal.
(a) 30◦ , 330◦
(b) 125◦ , 845◦
(c)
π 7π
,
3 3
(d) −60◦ , 250◦
3. Convert the following angles to radians.
(a) 45◦
(b) −270◦
(c) 1400◦
4. Convert the following angles to degrees.
(a)
9π
4
(b)
−5π
6
(c) 4.25 rads
5. Find the length of an arc that subtends the central angle of 300 in a circle of radius 4 cm.
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1.5
Trigonometric Functions
Trigonometry is the study of the relations between the sides and angles of triangles. The word
”trigonometry” is derived from Greek words tri meaning ”three”, gono meaning ”angle”, and
metry meaning ”measure”. There are six functions that are the core of trigonometry. Out of the
six functions there are three primary functions namely:
• Sine (sin)
• Cosine (cos)
• Tangent (tan)
The other three are not used as often and can be derived from the three primary functions. Because
these functions can easily be derived, calculators and spreadsheets do not usually have them. These
functions are
• Secant (sec)
• Cosecant (csc or cosec)
• Cotangent (cot)
All six functions have three-letter abbreviations (shown in parentheses above).
1.5.1
Definition of Trigonometric Functions
For any acute angle θ we can construct a right-angled triangle with one of the angles being θ. The
three sides of this triangle are called the adjacent (denoted by adj which is the leg of the rightangled triangle that forms one side of the angle θ), the opposite (denoted by opp which is the leg
of the right-angled triangle that is opposite the angle θ, i.e. does not form part of the angle), and
the hypotenuse (denoted by hyp which is the longest side of the right-angled triangle).
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Figure 1.6: Right-angled triangle with names of its three sides
Using Figure 1.6, we define three basic trigonometric functions as follows:
Sine θ =
Side opposite angle θ
Hypotenuse
Cosine θ =
Side adjacent angle θ
Hypotenuse
Tangent θ =
Side opposite angle θ
Side adjacent angle θ
Now consider the following right-angled triangle in a unit circle.
Figure 1.7: Right-angled triangle in a unit circle
The six trigonometric functions are defined from Figure 1.7. Thus,
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x
x
= =x
r
1
y
y
sin θ = = = y
r
1
y
sin θ
tan θ = =
x
cos θ
1
1
sec θ =
=
cos θ
x
1
1
=
cosec θ =
sin θ
y
1
1
x
cos θ
cot θ =
= y = =
tan θ
y
sin θ
x
Cosine θ
= cos θ =
Sine θ
=
Tangent θ
=
Secant θ
=
Cosecant θ
=
Cotangent θ =
(1)
Note:
• (cos θ)2 = cos θ cos θ = cos2 θ 6= cos θ2 . In general, (cos θ)n = cosn θ. This fact applies to
all trigonometric functions.
• Secant, Cosecant, and Cotangent functions are simply reciprocals of Sine, Cosine, and Tangent functions, respectively.
1.5.2
The Pythagorean Identities
From the right-angled triangle in Figure 1.7 we see that
x2 + y 2 = r 2 = 1
(2)
and hence
(cos θ)2 + (sin θ)2 = 1
=⇒ cos2 θ + sin2 θ = 1
Now, dividing equation (3) by cos2 θ, we obtain
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cos2 θ sin2 θ
1
+
=
2
2
cos θ cos θ
cos2 θ
=⇒ 1 + tan2 θ = sec2 θ
(4)
Likewise, dividing equation (3) by sin2 θ yields
cos2 θ sin2 θ
1
+
=
2
2
sin θ sin θ
sin2 θ
=⇒ cot2 θ + 1 = csc2 θ
(5)
Equations (3), (4), and (5) form the Pythagorean identities which are useful in simplifying
trigonometric expressions.
1.5.3
Trigonometric Identities
An identity is a relationship that reduces (or has already been reduced) to an equation of the form
x = x i.e. the quantity on the left is exactly equal to the quantity on the right. For example, 7 = 7
is an identity.
Our objective is to prove some trigonometric identities using well-known trigonometric formulae.
Example 1.6
Prove the following trigonometric identities.
(a) sin θ tan θ = sec θ − cos θ
(b) (cot θ − csc θ)2 =
1 − cos θ
1 + cos θ
(c) tan2 θ − sin2 θ = tan2 θ sin2 θ
(d)
sin x cos x
+
=1
csc x sec x
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Solution
When proving trigonometric identities, the technique is to simplify either the left-hand side (LHS)
or the right-hand side (RHS) of the identity, but not both and show that the side we started with is
equal to the other side. Note that there is no particular method that is employed when simplifying
trigonometric identities apart from one’s knowledge of trigonometric formulae. It is also important
to master the skill of knowing when to apply a particular formula. With this in mind, we proceed
as follows.
(a) We start with the LHS and show that is is equal to the RHS. Thus,
LHS = sin θ tan θ
sin θ
= sin θ ×
cos θ
2
sin θ
=
cos θ
1 − cos2 θ
=
cos θ
1
cos2 θ
=
−
cos θ
cos θ
tan θ =
since
sin2 θ = 1 − cos2 θ
= sec θ − cos θ = RHS
since
If we start with the RHS, we proceed as follows.
RHS = sec θ − cos θ
1
=
− cos θ
cos θ
1 − cos2 θ
=
cos θ
sin2 θ
sin θ sin θ
=
=
cos θ
cos θ
sin θ
= sin θ ×
cos θ
= sin θ tan θ = LHS
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sin θ
cos θ
since
1
= sec θ
cos θ
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(b) Starting with the LHS yields
LHS = (cot θ − csc θ)2
2
cos θ
1
=
−
sin θ
sin θ
2
cos θ − 1
(cos θ − 1)2
=
=
sin θ
sin2 θ
(cos θ − 1)(cos θ − 1)
=
1 − cos2 θ
(cos θ − 1)(cos θ − 1)
difference of two squares on (1 − cos2 θ)
=
(1 − cos θ)(1 + cos θ)
−(1 − cos θ)(cos θ − 1)
=
(1 − cos θ)(1 + cos θ)
1 − cos θ
−(cos θ − 1)
=
= RHS
=
1 + cos θ
1 + cos θ
(c)
RHS = tan2 θ sin2 θ = tan2 θ(1 − cos2 θ)
= tan2 θ − tan2 θ cos2 θ
sin2 θ
2
= tan θ −
× cos2 θ
2
cos θ
= tan2 θ − sin2 θ = LHS
If we start with the LHS, then this can be worked as follows;
LHS = tan2 θ − sin2 θ
= (tan θ − sin θ)(tan θ + sin θ)
difference of two squares
sin θ
sin θ
=
− sin θ
+ sin θ
cos θ
cos θ
sin θ − sin θ cos θ
sin θ + sin θ cos θ
=
cos θ
cos θ
sin θ (1 − cos θ) sin θ (1 + cos θ)
=
×
cos θ
cos θ
2
2
2
sin θ (1 − cos θ)
sin θ
=
=
× sin2 θ
2
cos θ
cos2 θ
= tan2 θ sin2 θ = RHS
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(d) Consider the LHS. Then
sin x cos x
+
cscx
secx
sin x cos x
=
+ 1
1
LHS =
sin x
2
cos x
2
= sin x + cos x
= 1 = RHS
Activity 1.2
Prove the following trigonometric identities.
csc2 ϑ
(a)
= cot2 ϑ
2
1 + tan ϑ
(d)
1
1
+
= 2 sec2 θ
1 − sin θ 1 + sin θ
(b) cot x + tan x = csc x sec x
(e)
sin θ + cos θ
= 1 + tan θ
cos θ
(c) (cos φ + sin φ)2 + (cos φ − sin φ)2 = 2
(f) sin2 y − cos2 y = sin4 y − cos4 y
1.5.4
Quadrants and Signs of Trigonometric Functions
The plane is partitioned by the x-and y-axis into four quadrants. The trigonometric functions are
positive or negative depending on the quadrant they lie. We label the quadrants I, II, III, and IV as
shown in Figure 1.8.
In the definitions of the trigonometric functions in equations (1), r > 0 is the distance from the
origin to the point (x, y). If a point (x, y) is chosen in quadrant I, then both x and y will be positive and hence, the values of all the six trigonometric functions will be positive in quadrant I. In
quadrant II, a point (x, y) has x < 0 and y > 0. Thus, for angles in quadrant II, the values of the
sine and cosecant functions are positive while the other four angles take negative values. Similar
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Figure 1.8: Quadrants of a plane
results can be obtained for the other quadrants as summarized in Table 1.1 and Figure 1.9.
Table 1.1: Signs of values of tringonometric functions
θ in quadrant
I
II
III
IV
Example 1.7
sin θ
+
+
−
−
cos θ
+
−
−
+
tan θ
+
−
+
−
cot θ
+
−
+
−
sec θ
+
−
−
+
csc θ
+
+
−
−
Identify the quadrant (or quadrants) of any angle θ that satisfies sin θ > 0 and
tan θ < 0.
Solution
Since sin θ > 0 in quadrants I and II, while tan θ < 0 in quadrants II and IV, then both conditions
are satisfied in quadrant II only.
1.5.5
The Range of Trigonometric Functions
Recall that the domain of a function is the set of all possible input values while the range is the set
of all output values of a function. We want to find the range of the sine and cosine functions. Now,
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Figure 1.9: Signs of trigonometric functions in all quadrants
suppose that P (x, y) is a point lying on the terminal side of an angle θ (in standard position).
Figure 1.10: Angle θ in standard position
From Figure 1.10, we have
x2 + y 2 = r 2 .
(6)
It follows from (6) that x2 ≤ r2 which implies that
− r ≤ x ≤ r.
(7)
− r ≤ y ≤ r.
(8)
Likewise,
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Dividing inequalities (7) and (8) through by r > 0 leads to
−1≤
x
≤1
r
(9)
−1 ≤
y
≤1
r
(10)
Based on definitions of cos θ and sin θ as provided in (1.7), inequalities (10) and (10) become
since cos θ =
− 1 ≤ cos θ ≤ 1
(11)
−1 ≤ sin θ ≤ 1
(12)
x
y
, sin θ = , and r = 1. Therefore, the range of both cos θ and sin θ is the interval
r
r
[−1, 1].
The range of secant and cosecant functions can be derived from the inequalities above. Thus, the
range of secant and cosecant functions is the same and is given by
(−∞, −1] ∪ [1, ∞)
x
y
and cot θ = . This implies that the output value of these
y
x
functions depends on the values of x and y that we pick. Thus, the range of the tangent and
Note that by definition tan θ =
cotangent function is the set (−∞, +∞).
Example 1.8
(a) cos θ = 4
Determine whether or not the following statements are practical.
(b) cot θ = 250.5
(c) sin θ = 0.7
Solution
We use our knowledge of the range of trigonometric functions to answer this question.
(a) This is not possible because cos θ cannot be more than 1.
(b) This is possible because cot θ can assume any real number.
(c) This is possible because sin θ = 0.7 is withing the range of the sine function.
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Cofunctions
Consider the following right-angled triangle.
Figure 1.11: Right-angled triangle with angle θ and ϕ
Recall that the sum of angles in a triangle adds up to 180◦ . Since one of the angles in Figure
1.11 is a right-angle, that is, 90◦ , then the sum of angles ϕ and θ is 90◦ . These angles are called
complementary angles. In other words, two angles are complementary when their sum is 90◦ or
π
. Thus,
2
◦
ϕ + θ = 90 =⇒
In triangle (1.11), sin θ =
ϕ = 90◦ − θ
θ = 90◦ − ϕ
(13)
3
3
and cos ϕ = . Thus,
5
5
sin θ = cos(90◦ − θ)
sin θ = cos ϕ =⇒
cos ϕ = sin(90◦ − ϕ)
(14)
From (14), we can conclude that the cosine of an acute angle is equal to the sine of its complement
and the sine of an acute angle is equal to the cosine of its complement.
The above results lead to three sets of cofunctions identities namely:
(a) Sine and Cosine are Cofunctions
sin θ = cos(900 − θ)
or
cos θ = sin(900 − θ) or
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sin θ = cos
π
−θ
2
π
cos θ = sin
−θ
2
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(b) Secant and Cosecant are Cofunctions
0
sec θ = csc(90 − θ)
or
csc θ = sec(900 − θ) or
π
−θ
2
π
csc θ = sec
−θ
2
sec θ = csc
(c) Tangent and Cotangent are Cofunctions
tan θ = cot(900 − θ)
or
cot θ = tan(900 − θ) or
tan θ = cot
π
−θ
2
π
cot θ = tan
−θ
2
Note that the first two letters of the word complement ”co” defines cofunctions.
Example 1.9
Find the value of x for which sec x = csc 25◦ .
Solution
Recall that secant and cosecant are cofunctions as such the value of x will be the complement of
25◦ . Thus, the value of x = 90◦ − 25◦ = 65◦ .
1.6
Values of Trigonometric Functions of Quadrantal Angles
Consider the following unit circle.
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Recall that,
y
=y
r
x
cos θ =
=x
r
y
tan θ =
x
sin θ =
As θ approaches 0, r coincides with x, i.e., x = r = 1, and y approaches 0. Therefore,
sin 0 = y = 0,
cos 0 = x = 1,
0
y
= = 0.
tan 0 =
x
1
Similarly, as θ approaches 90◦ , r coincides with y, i.e., y = r = 1 and x approaches 0. Therefore,
sin 90◦ = y = 1,
cos 90◦ = x = 0,
tan 90◦
is undefined.
Table 1.2 summarises the values of trigonometric functions of quadrantal angles which are obtained using the definitions of trigonometric functions.
Table 1.2: Values of trigonometric functions of quadrantal angles
θ
0◦ π
90◦
2
◦
180
(π) 3π
270◦
2
◦
360 (2π)
sin θ
0
cos θ
1
tan θ
0
cot θ
undefined
sec θ
1
csc θ
undefined
1
0
undefined
0
undefined
1
0
−1
0
undefined
−1
undefined
−1
0
undefined
0
undefined
−1
0
1
0
undefined
1
undefined
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Trigonometric Ratios of Special Angles
There are a very few angles that have relatively ”neat” trigonometric values, involving at worst,
one square root. These angles are called special angles. Because of their relatively simple values,
these angles occur often in trigonometry and it is imperative to know their values. These angles
π
π
π
are: (45◦ ), (30◦ ), and (60◦ ).
4
6
3
1.7.1
Functional Values for π/4 (45◦ )
Consider an isosceles right-angled triangle (Figure 1.12) whose two equal sides have a length of 1
unit.
Figure 1.12: Isosceles right-angled triangle
√
Note that the length of the hypotenuse ( 2) is determined by applying the Pythagoras theorem.
From Figure 1.12 we have
√
√
π
1
2
π
1
1
sin = √ =
=⇒ csc =
2
π = √1 =
4
2
4
sin 4
2
2
√
√
π
1
2
π
1
1
cos = √ =
=⇒ sec =
=
=
2
√1
4
2
4
sin π4
2
2
tan
π
1
π
1
1
= = 1 =⇒ cot =
=1
π =
4
1
4
tan 4
1
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Functional Values for π/6 (30◦ ) and π/3 (60◦ )
Consider an equilateral triangle having sides of length 2 units. See Figure 1.13 below.
Figure 1.13: Equilateral triangle with a = 2
The median from vertex A (also works with any of the three vertices) to the opposite side BC
bisects angle, A, that is, it divides the angle into two equal angles of 30◦ each. Note that bisecting
one angle of this equilateral triangle leads to two right-angled triangles, each of which has angles
of 30◦ , 60◦ , and 90◦ . See Figure 1.14 below.
Figure 1.14: Equilateral triangle with a = 2
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From Figure 1.14, we obtain
π
1
π
2
=
=⇒ csc = = 2
6
2
6
1
√
√
π
3
π
2
2 3
cos =
=⇒ sec = √ =
6
2
6
3
3
√
√
3
3 √
π
1
π
tan = √ =
=⇒ cot =
= 3
6
3
6
1
3
sin
and
√
√
π
2
π
3
2 3
=⇒ csc = √ =
sin =
3
2
3
3
3
π
1
π
2
=
=⇒ sec = = 2
3
2
3
1
√
√
π
3 √
1
3
π
tan =
= 3 =⇒ cot = √ =
3
1
3
3
3
cos
A summary for the trigonometric ratios for special angle is given in Table 1.3.
Table 1.3: Trigonometric ratios for special angles
1.8
θ
sin θ
π
or 30◦
6
π
or 45◦
4
π
or 60◦
3
1
√2
2
√2
3
2
cos θ
√
3
√2
2
2
1
2
tan θ
√
3
3
cot θ
√
3
1
√
3
1
√
3
3
sec θ
√
2 3
3
√
2
2
csc θ
2
√
2
√
2 3
3
Negative Angles
Angles are measured in anticlockwise direction. The clockwise direction gives negative angles.
Now, consider the following unit circle.
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Let P and Q be points on the unit circle with coordinates (x, y) and (x, −y), respectively. From
the figure we see that
cos (−θ) = x = cos θ
sin (−θ) = −y = − sin θ
−y
tan (−θ) =
= − tan θ
x
since sin θ = y
since tan θ =
y
x
In general,
cos (−θ) = cos θ
sin (−θ) = − sin θ
tan (−θ) = − tan θ
(15)
From the results in (15), we can conclude that a cosine function is an even function where as sine
and tangent functions are odd functions.
Example 1.10
π
(a) tan −
3
Evaluate the following.
2π
(b) cos −
3
Solution
We use the formulae in (15).
Trigonometry and Elementary Calculus
π
(c) cos −
6
π
(d) sin −
4
27
(a)
(b)
(c)
(d)
π
π √
tan −
= − tan
=− 3
3
3
π 2π
2π
1
cos −
= cos
= − cos
=−
3
3
3
2
π
π √3
cos −
= cos
=
6
6
2
√
π π
2
= − sin
=−
sin −
4
4
2
1.9
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(cosine is negative in quadrant II)
Reference Angles
Every non-quadrantal angle which is in standard position is associated with a positive acute angle
called a reference angle. A reference angle for an angle θ, denoted θ0 , is the positive acute
angle that is formed by the terminal side of angle θ and the x-axis. θ and θ0 are the same in the
first quadrant. If it happens that θ is negative or it is more than 360◦ , then its reference angle
can be found by first finding its coterminal angle that is between 0◦ and 360◦ , and then using an
appropriate formulae as shown in Figure 1.15.
Figure 1.15: Reference angle and its formula in each quadrant
Note that every angle is measured from the positive side of the x-axis to the angle’s terminal side.
Example 1.11
(a) 250◦
Find the reference angle for the following angles.
(b) 1600◦
(c) −230◦
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Solution
(a) 250◦ is in quadrant III as such its reference angle is θ − 180◦ = 250◦ − 180◦ = 70◦
(b) Since 1600◦ is more than 360◦ , we first find its coterminal angle. Thus,
1600◦ − 360◦ = 1240◦
1240◦ − 360◦ = 880◦
880◦ − 360◦ = 520◦
520◦ − 360◦ = 160◦
Therefore, 1600◦ is coterminal with 160◦ which is in the second quadrant. Hence, its reference
angle is
180◦ − 160◦ = 20◦ .
(c) Since −230◦ is negative, we find its coterminal angle first. Thus, −230◦ is coterminal with
−230◦ + 360◦ = 130◦ which is in quadrant II and its reference angle is 180◦ − 130◦ = 50◦ .
Note that reference angles are very useful in trigonometry especially when we want to find trigonometric function values (e.g. sine or cosine or tangent, etc.) of any arbitrary angle. Thus, for any
given angle, we first find its reference angle and then find its value. Lastly, we assign an appropriate sign to the value depending on the quadrant the reference angle lies.
Example 1.12
(a) cos (−240◦ )
Without using a calculator, find the exact value of the following.
(b) tan 675◦
(c) sin
5π
6
Solution
(a) −240◦ is coterminal with an angle of −240◦ + 360◦ = 120◦ which is in quadrant II. Thus,
the reference angle is 180◦ − 120◦ = 60◦ . Since cosine is negative in quadrant II, then
1
cos (−240◦ ) = − cos 60◦ = − .
2
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(b) 675◦ is coterminal with angle 675◦ − 360◦ = 315◦ which is in fourth quadrant. Hence, its
reference angle is 360◦ − 315◦ = 45◦ . But tangent is negative in quadrant IV and therefore,
tan 675◦ = − tan 45◦ = −1.
(c)
5π
5π
π
is in quadrant II and its reference angle is π −
= . Since sine is positive in quadrant
6
6
6
II, then
5π
π
1
sin
= sin = .
6
6
2
Activity 1.3
1. Find the reference angle for the following angles.
17π
2π
(c) −650◦
(d)
3
36
2. Find, where possible, the exact values of all six trigonometric functions for the following
values of θ. Leave your answer in surd form where possible.
(a) 1900◦
(b)
(a) 420◦
1.10
(b)
4π
3
(c) 135◦
Addition Formulae
It is useful in many applications to have formulae for the values of expressions such as sin (A+B),
cos (A+B), and tan (A+B). In this section, we derive formulae for these important expressions.
1.10.1
Addition Formula for Cosine
We begin by deriving the addition angle formula for cosine which will then be used to find the
difference angle formula for cosine.
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Consider the following figure.
Notice that
∠ROP = ∠QOS = A + B
Hence,
chord P R = chord QS
(16)
Furthermore,
P = (cos (A + B), sin (A + B))
Q = (cos B, sin B)
S = (cos (−A), sin (−A)) = (cos A, − sin A)
By (16) we see that P R = QS.
Now, using the distance formula, we have
P R2 = [cos (A + B) − 1]2 + [sin (A + B) − 0]2
= cos2 (A + B) − 2 cos(A + B) + 1 + sin2 (A + B)
= 2 − 2 cos (A + B)
since
cos2 (A + B) + sin2 (A + B) = 1
Likewise,
QS 2 = [cos B − cos (−A)]2 + [sin B − sin (−A)]2
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(17)
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But cos (−A) = cos A and sin (−A) = − sin A. Therefore, formula (17) becomes
QS 2 = [cos B − cos A]2 + [sin2 B + sin A]2
= cos2 B − 2 cos B cos A + cos2 A + sin2 B + 2 sin B sin A + sin2 A
= (cos2 B + sin2 B) + (cos2 A + sin2 A) − 2 cos B cos A + 2 sin B sin A
= 1 + 1 − 2 cos B cos A + 2 sin B sin A
= 2 + 2 sin B sin A − 2 cos B cos A
Since P R2 = QS 2 , we obtain
2 − 2 cos (A + B) = 2 + 2 sin B sin A − 2 cos B cos A
Making cos (A + B) the subject of formula yields
cos (A + B) = cos B cos A − sin B sin A
(18)
Also, since cos (−B) = cos B and sin (−B) = − sin B, then changing B to −B in equation
(18) leads to
cos (A − B) = cos (−B) cos A − sin (−B) sin A
= cos B cos A + sin B sin A
The addition formula for cosine is summarised below.
Example 1.13
(a) cos 75◦
cos (A + B) = cos A cos B − sin A sin B
(19)
cos (A − B) = cos A cos B + sin A sin B
(20)
Find the exact value of the following.
(b) cos
π
12
Solution
We rewrite the given angle as a sum or difference of two angles.
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(a)
cos 75◦ = cos(45◦ + 30◦ )
= cos 45◦ √
cos30◦ − sin 45◦ sin 30◦
1
3 1
1
= √ ×
− ×√
2
2
2
2
√
√
3
1
3−1
= √ − √ = √
2 2 2 2
2 2
(b) Likewise,
1.10.2
π
π π
= − . Thus,
12
3
4
π
π π cos
= cos
−
12
3
4
π π π π = cos
cos
+ sin
sin
3
3
4
√4
1
1
1
3
×√ +
×√
=
2
2
2
2
√
√
3
1+ 3
1
= √ + √ = √
2 2 2 2
2 2
Addition Formula for Sine
Now, let us consider sin (A + B).
Recall from Section 1.5.6 that sin A = cos (90◦ − A) and cos A = sin (90◦ − A). Thus,
sin(A + B) = cos [90◦ − (A + B)]
= cos[(90◦ − A) − B]
expand using equation (20)
= cos (90◦ − A) cos B + sin (90◦ − A) sin B
= sin A cos B + cos A sin B
Similarly, it can be easily shown that sin (A − B) = cos B sin A − sin B cos A and we obtain the
following addition formula for sine.
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Example 1.14
(a) csc 105◦
sin (A + B) = sin A cos B + cos A sin B
(21)
sin (A − B) = sin A cos B − cos A sin B
(22)
Evaluate the following leaving your answer in surd form.
(b) sin
2π
3
(c) sin
π
12
Solution
We rewrite the given angle as a sum or difference of two angles.
(a) 105◦ = 60◦ + 45◦ . Since csc A =
1
, we first find the value of sin 105◦ . Thus,
sin A
sin 105◦ = sin (60◦ + 45◦ )
expand using equation (21)
◦
◦
◦
◦
= sin
√ 60 cos 45 + cos 60 sin 45
3
1
1
1
=
×√ + ×√
2
2 2
2
√
3+1
√
=
2 2
(b)
2π
π π
= + and hence
3
3
3
π π 2π
sin
= sin
+
3
3
3
π π π π cos
+ cos
sin
= sin
3 √
3
3
√ 3
3 1 1
3
=
× + ×
2 2 √2
√2
√
3
3
2 3
=
+
=
4
4
√4
3
=
2
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π
π π
(c)
= − . Therefore
12
3
4
π
π π sin
= sin
−
12
3 4 π
π π π
= sin
cos
− cos
sin
4
3
4
√ 3
3
1
1
1
=
×√ − ×√
2
2 2
2
√
3−1
√
=
2 2
1.10.3
Addition Formula for Tangent
To find the addition formula for the tangent function we combine the results of the addition formulae for sine and cosine functions.
Recall that tan A =
sin A
. Therefore,
cos A
tan (A + B) =
sin (A + B)
cos (A + B)
sin A cos B + cos A sin B
=
cos A cos B − sin A sin B
(23)
Now, dividing the numerator and denominator of (23) by cos A cos B, we get
sin A cos B
cos A sin B
+
tan (A + B) = cos A cos B cos A cos B
cos A cos B
sin A sin B
−
cos A cos B cos A cos B
tan A + tan B
=
1 − tan A tan B
(24)
To find the formula for tan (A − B) we modify (24) and use the fact that tangent is an odd
function, that is, tan (−A) = − tan A. Thus, we have
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tan (A − B) = tan [A + (−B)]
expand using (24)
=
tan A + tan (−B)
1 − tan A tan (−B)
=
tan A − tan B
1 + tan A tan B
(25)
The addition formula for tangent is summarized as follows.
tan (A + B) =
tan A + tan B
1 − tan A tan B
(26)
tan (A − B) =
tan A − tan B
1 + tan A tan B
(27)
Example 1.15 Find the exact value of tan
7π
.
12
Solution
π π
7π
= + . Therefore,
12
3
4
tan
7π
12
π
+
3
4
π π tan
+ tan
3 4π =
π
1 − tan
tan
3
4
√
3+1
√
=
1− 3
= tan
π
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Activity 1.4
1. Use the appropriate addition formulae to evaluate the following. Leave your answer in surd
form.
π
4π
(a) sec 105◦
(c) cot
(e) sin
12
3
11π
7π
(b) tan
(f) csc
2
◦
(d)
cos
75
6
4
2. Prove the following identities.
(a) 2 cos2 x = 1 + cos 2x
(b) sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A
(c) sin (π − x) = sin x
1.11
Double Angle and Half Angle Formulae
1.11.1
Double Angle Formula for Cosine
We know that
cos (A + B) = cos A cos B − sin A sin B
(28)
Now setting A = B, equation (28) becomes
cos (A + A) = cos A cos A − sin A sin A
=⇒ cos 2A = cos2 A − sin2 A
Recall that cos2 A + sin2 A = 1. Then it follows that
cos2 A = 1 − sin2 A
Trigonometry and Elementary Calculus
and
sin2 A = 1 − cos2 A.
(29)
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Substituting cos2 A = 1 − sin2 A in (29) yields
cos 2A = cos2 A − sin2 A = 1 − sin2 A − sin2 A = 1 − 2 sin2 A
(30)
Similarly, replacing sin2 A with 1 − cos2 A in (29) leads to
cos 2A = cos2 A − sin2 A = cos2 A − (1 − cos2 A) = 2 cos2 A − 1
(31)
In summary, the double angle formulae for cosine are given below.

 cos2 A − sin2 A
1 − 2 sin2 A
cos 2A =

2 cos2 A − 1
Example 1.16
(32)
1
Find the value of cos 2A if cos A = − .
2
Solution
We apply the double angle formula for cosine and write cos 2A in terms of cos A. Thus, we
proceed as follows.
cos 2A = 2 cos2 A − 1
2
1
= 2× −
−1
2
1
1
= 2× −1=−
4
2
1.11.2
Double Angle Formula for Sine
Recall that
sin(A + B) = sin A cos B + cos A sin B
(33)
Now, letting A = B in (33) we obtain
sin(A + A) = sin A cos A + cos A sin A
=⇒ sin 2A = 2 sin A cos A
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Thus, the double angle formula for sine is given by
sin 2A = 2 sin A cos A
Example 1.17
(34)
Show that 2 sin2 x = tan x sin 2x.
Solution
We simplify the RHS and show that it is equal to the LHS.
RHS = tan x sin 2x
expand sin 2x using (34)
= tan x (2 sin x cos x)
sin x
= 2
sin x cos x
cos x
= 2 sin2 x = LHS
1.11.3
Double Angle Formula for Tangent
Recall that
tan (A + B) =
tan A + tan B
1 − tan A tan B
(35)
Setting A = B, in (35) yields
tan A + tan A
1 − tan A tan A
2 tan A
=⇒ tan 2A =
1 − tan2 A
tan (A + A) =
Therefore, the double angle formula for tangent is given by
tan 2A =
1.11.4
2 tan A
1 − tan2 A
(36)
Half Angle Formulae
The next three identities are useful for simplifying certain expressions involving powers of trigonometric functions.
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Recall that:
cos 2A = 2 cos2 A − 1 =⇒
2
cos 2A = 1 − 2 sin A =⇒
cos 2A + 1
2
1 − cos 2A
2
sin A =
2
cos2 A =
(37)
Using (37), we can find the formula for tan2 A as follows.
tan2 A =
1
(1 − cos 2A)
sin2 A
1 − cos 2A
2
=
=
1
2
cos A
1 + cos 2A
(cos 2A + 1)
2
The half-angle formulae are found by replacing A with
(38)
1
A in equations (37) and (38). Thus, we
2
obtain
1 − cos A
1
sin2 A =
2
2
1
1
+
cos
A
cos2 A =
2
2
1 − cos A
1
tan2 A =
2
1 + cos A
Example 1.18
Show that tan θ =
(39)
sin 2θ
.
cos 2θ + 1
Solution
We simplify the RHS and show that it is equal to the LHS. Thus,
RHS =
=
sin 2θ
cos 2θ + 1
expand sin 2θ and cos 2θ
2 sin θ cos θ
2 cos2 θ − 1 + 1
2 sin θ cos θ
2 cos θ cos θ
sin θ
=
cos θ
= tan θ = LHS
=
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Activity 1.5
1. Prove the following identities.
(a) sin 4A = 4 sin A cos A cos 2A
(b) cos 3θ = 4 cos3 θ − 3 cos θ
[Hint: sin 4A = sin (2A + 2A)]
[Hint: cos 3θ = cos (2θ + θ)]
(c) sin 3θ = 3 sin θ − 4 sin3 θ
2. Let sin θ =
4
3
and cos θ = . Find
5
5
1
(c) tan 2θ
(e) tan2 θ
2
2 1
θ
(d)
cos
(f) csc 2θ
(b) cos 2θ
2
θ
θ
1
3. Find the exact value of sin
and cos
given that cos θ = − .
2
2
8
(a) sin 2θ
1.12
Product and Factor Formulae
We have already derived formulae for sin (A ± B) and cos (A ± B) in terms of the functions
sin A, sin B, cos A, and cos B. The motivation for this was the fact that
sin (A + B) 6= sin A + sin B.
In this section we want to go further and derive formulae for the expressions sin A ± sin B, and
cos A ± cos B.
Product Formulae
Recall that by (21) and (22) we have
sin (A + B) = sin A cos B + cos A sin B
(40)
sin (A − B) = sin A cos B − cos A sin B
(41)
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Adding equations (40) and (41), we obtain
sin (A + B) + sin (A − B) = 2 sin A cos B
(42)
Likewise, subtracting equations (40) and (41), we have
sin (A + B) − sin (A − B) = 2 cos A sin B
(43)
Now, dividing both sides of equations (42) and (43) by 2 yields the following product formulae.
sin A cos B =
sin (A + B) + sin (A − B)
2
(44)
cos A sin B =
sin (A + B) − sin (A − B)
2
(45)
Now, let
A+B = u
(46)
A−B = v
(47)
Adding and subtracting (46) and (47), we obtain
2A = u + v
and
2B = u − v,
u+v
2
and
B=
or
A=
1.12.1
u−v
.
2
(48)
Sine Factor Formulae
Substituting (48) into (42) and (43), we obtain the sine factor formulae
u+v
u−v
cos
2
2
u−v
u+v
sin u − sin v = 2 sin
cos
2
2
sin u + sin v = 2 sin
(49)
(50)
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Product Formulae
Similarly, recall that by (19) and (20), we have
cos (A + B) = cos A cos B − sin A sin B
(51)
cos (A − B) = cos A cos B + sin A sin B
(52)
Adding and subtracting (51) and (52) lead to two more product formulae
cos (A + B) + cos (A − B) = 2 cos A cos B
(53)
cos (A + B) − cos (A − B) = −2 sin A sin B.
(54)
and
Thus, dividing both sides of (53) and (54) by 2 gives
1.12.2
cos A cos B =
cos (A + B) + cos (A − B)
2
(55)
sin A sin B =
cos (A − B) − cos (A + B)
2
(56)
Cosine Factor Formulae
Making the same substitutions as before we get the cosine factor formulae
u−v
u+v
cos
2
2
u+v
u−v
cos u − cos v = −2 sin
sin
2
2
cos u + cos v = 2 cos
(57)
(58)
Note: Do not attempt to memorise the formulae in this Section rather we recommend that you
understand how they are derived.
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Example 1.19
(a) Without using a calculator, evaluate the following leaving your answer in surd form where
possible.
(i) sin 75◦ + sin 15◦
(ii) cos 105◦ − cos 15◦
(b) Find the exact value of sin 52.5◦ cos 7.5◦ .
(c) Prove that sin2 x − sin2 y = sin (x + y) sin (x − y).
Solution
(a)
(i) We apply formula (49) and proceed as follows.
75◦ + 15◦
75◦ − 15◦
cos
2
2
90◦
60◦
2 sin
cos
2
2
◦
2 sin√
45 cos√30◦
2
3
2×
×
2
√ 2
6
2
sin 75◦ + sin 15◦ = 2 sin
=
=
=
=
(ii) We use formula (58). Thus,
105◦ + 15◦
105◦ − 15◦
sin
2
2
120◦
90◦
−2 sin
sin
2
2
◦
◦
−2 sin√
60 sin√45
3
2
−2 ×
×
2
√ 2
6
−
2
cos 105◦ − cos 15◦ = −2 sin
=
=
=
=
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(b) Applying formula (44) we obtain
sin (52.5◦ + 7.5◦ ) + sin (52.5◦ − 7.5◦ )
2
sin 60◦ + sin 45◦
=
2
√
√
3
2
+
2
= 2
2
√
√
3+ 2
√
√
3
+
2
2
=
=
2
4
sin 52.5◦ cos 7.5◦ =
(c) We simplify the LHS and show that it is equal to RHS.
LHS = sin2 x − sin2 y
apply difference of two squares
= (sin x + sin y)(sin x − sin y)
expand using (49) and (50)
x+y
x−y
x+y
x−y
=
2 sin
cos
cos
2 sin
2
2
2
2
x+y
x−y
x+y
x−y
cos
cos
=
2 sin
2 sin
simplify using (34)
2
2
2
2
x−y
x+y
sin 2
= sin 2
2
2
= sin (x + y) sin (x − y) = RHS
Exercise: Prove this identity from RHS to LHS
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Activity 1.6
1. Write the expression cos 3x cos 5x as a sum of two trigonometric functions.
2. Express the following as a product of sine and cosine.
(a) sin 5x + sin 3x
(b) cos 2t − cos 3t
3. Evaluate the following leaving your answer in surd form.
(a) sin 75◦ − sin 15◦
(b) cos 75◦ cos 15◦
4. Prove the following identities.
(a)
sin 4x + sin 6x
= cot x
cos 4x − cos 6x
(b)
sin x + 2 sin 2x + sin 3x
= tan 2x
cos x + 2 cos 2x + cos 3x
Periodicity
A function f is periodic if there exists a positive real number c such that
f (x + c) = f (x)
for every x in the domain of f . The least such positive number real number c, if it exists, is called
the period of f .
Sine, cosine, and tangent functions are periodic functions. For sine and cosine functions, the
period is 2π and for tangent it is π.
A period of say 2π means that the same pattern is repeated in intervals of length 2π along the
x − axis (in mathematics and trigonometry in particular).
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Graphs of Trigonometric Functions
In this section, we discuss the graphs of elementary trigonometric functions; sin x, cos x, and
tan x. Since we wish to sketch the graphs of these functions on an xy-coordinate system, we shall
consider the equations of the form y = sin x, y = cos x, and y = tan x.
Note that x denotes a real number or radian measure of an angle (and not degrees).
1.13.1
The graph of y = sin x
It is not difficult to sketch the graph of the sine function, i.e. the equation y = sin x.
Since −1 ≤ sin x ≤ 1 for every real number x, the graph lies entirely between the horizontal
lines y = 1 and y = −1. That is, range is [−1, 1].
Since the sine function is periodic with period 2π, it is sufficient to determine the graph for
0 ≤ x ≤ 2π, because the same pattern is repeated in intervals of length 2π along the x-axis.
Note: We count a peak (∩) and a trough (∪) as a wave.
The manner in which sin x varies in interval [0, 2π] is summarized as follows.
Variation of x
0 to π2
π
to π
2
π to 3π
2
3π
to
2π
2
Variation of sin x
0 to 1 (increases - sin π2 = 1)
1 to 0 (decreases - sin π = 0)
0 to −1 (sin x is negative, continues to decrease)
−1 to 0 (increases)
Using the Table above and some intermediate values of x (that we may calculate) we may plot the
points (x, y) and draw a smooth curve through them. See Figure 1.16.
Note: We refer to the part of the graph corresponding to the interval [0, 2π] as a sine wave.
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Figure 1.16: The graph of y = sin x
1.13.2
The graph y = cos x
The graph of the cosine function can be found in a similar manner by observing how cos x varies
as x varies from 0 to 2π. See the table below.
Variation of x
0 to π2
π
to π
2
3π
π to
2
3π
to
2π
2
Variation of cos x
1 to 0 (decreases; cos π2 = 0)
0 to −1 (continues to decrease and becomes negative; cos π = −1)
−1 to 0 (increases; cos 3π
= 0)
2
0 to 1 (increases; cos 2π = 1)
Figure 1.17: The graph of y = cos x
Alternatively, notice that the shape of the graph y = cos x in Figure 1.17 is the same as that of
π
the graph of the sin x only that it has been moved to the left by . Thus, sin x is obtained by
2
π
moving the graph cos x to the right by . The following Figure shows the graphs of sin x and
2
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cos x plotted on the same axes.
1.13.3
The graph of y = tan x
3π
π
and
.
2
2
Similarly, the minimum value of tan x is −∞, i.e. tan x decreases without bound as x approaches
π
3π
− and − .
2
2
The maximum value of tan x is ∞, i.e. tan x increases without bound as x approaches
3π
sin x
π
and x =
since tan x =
and cos x = 0 when
The value of tan x is undefined x =
2
2
cos x
π
3π
x = and x =
.
2
2
π 3π
3π 5π
The same pattern is repeated in the open intervals
,
,
,
, that is, after a period
2 2
2 2
of π. The graph of tanx crosses the x-axis when sin x = 0, i.e. at x = 0, x = π, and x = 2π
(=⇒ tan x = 0).
The vertical lines x =
π
3π
and x =
are called asymptotes.
2
2
Figure 1.18: The graph of y = tan x
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Activity 1.7
Study the graphs of cot x, sec x, and csc x.
1.13.4
Shifting Graphs
Let c be a constant. Vertical and horizontal shifts of the graph y = f (x) are represented as follows;
(i) Vertical shift of c units upwards: y = f (x) + c.
(ii) Vertical shift c units downwards: y = f (x) − c.
(iii) Horizontal shift c units to the left: y = f (x + c).
(iv) Horizontal shift c units to the right: y = f (x − c).
Example 1.20
On the same axes, sketch the graphs of the following.
(a) y = sin x + 2
and y = sin x − 3.
(b) y = cos x − 1
and y = cos x + 1.
Solution
(a) y = sin x + 2 is the graph of y = sin x but shifted upwards by 2 units.
y = sin x − 3 is the graph of y = sin x but shifted downwards by 3 units.
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(b) y = cos x − 1 is the graph of y = cos x but shifted downwards by 1 unit.
y = cos x + 1 is the graph of y = cos x but shifted upwards by 1 unit.
1.14
Graphs of the form y = a sin (bx + c)
We want to consider graphs of the form
y = a sin (bx + c)
for real numbers a, b, and c.
We shall also discuss graphs of similar equations that involve different trigonometric functions.
We will use information about the graphs of the trigonometric functions discussed earlier (other
than plotting many points).
Let us start by considering a special case in which c = 0, b = 1, and a > 0. Thus, we wish to
sketch graphs with the equation of the form y = a sin x where a is called the amplitude of the
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graph (i.e. how high the graph is).
Example 1.21
Sketch the graphs of y = 3 sin x, y =
1
sin x, and y = sin x on the same axes.
2
Solution
1
The amplitudes of the graphs are 3, , and 1, respectively. However, they all have a period of 2π.
2
Below are the graphs.
Note: To obtain the graph of y = a sin x, we multiply the y-coordinates of points of y = sin x by
a.
Example 1.22
Sketch the graph of y = −3 cos x.
Solution
The graph of y = −3 cos x is just like the graph of y = cos x, except that the values will vary
between −3 and 3 and the graph is turned over.
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Remark: If f (x) = sin bx and b 6= 0, then the period of f is
2π if and only if x ranges from 0 to
Example 1.23
2π
. In this case, bx ranges from 0 to
|b|
2π
.
|b|
Sketch the graph of
(a) y = sin 2x
(b) y = cos 2x
Solution
(a) We identify b = 2.
2π
2π
=
= π. Thus, there is exactly one sine wave of amplitude
b
2
1 corresponding to the interval [0, π]. We proceed by dividing the interval [0, π] into 4 equal
π−0
π
parts. Each part has length
= . We now plot the graph by replicating the sine wave
4
4
with period π.
So, y = sin 2x has period
(b) The period is
2π
π
= π and each part is . Below is the graph.
2
4
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Sketch the graph y = 3 sin 4x.
Example 1.24
Solution
The amplitude is 3 and the period is
2π
π
= .
4
2
h πi
Thus, one sine wave is completed over the interval 0, .
2
π
−0
π
= . The graph is plotted below.
Note that each part is 2
4
8
Activity 1.8
Sketch the graphs of the following functions;
(a) y = 4 cos
x
2
(b) y = − sin 4x
(c) y = 2 + tan x
(d) y = −6 cos 2x
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We conclude our discussion on graphs by considering the general situation
f (x) = a sin (bx + c).
We have already observed that the amplitude is |a|. We also know that the sine wave for the graph
of y = sin (bx + c) is obtained if bx + c varies from 0 to 2π, that is, if bx ranges from −c to 2π − c.
In turn, the variation of bx is obtained by letting x range from
−c
b
Recall that
to
(2π − c)
2π c
=
− .
b
b
b
2π
is the period. Now,
b
• If
−c
−c
> 0, this amounts to shifting the graph of y = a sin bx to the right by
units.
b
b
• If
−c
< 0, the shift is to the left.
b
−c
is called the phase shift associated with the function. Do not memorize the
b
general formula for finding the phase shift. In any specific problem the interval that contains one
The number
complete sine wave can be found by solving the two equations
bx + c = 0
and
bx + c = 2π
for x. See the examples below.
Example 1.25
Sketch the graph of
π
(a) f (x) = 2 sin 2x −
2
π
(b) y = cos x +
3
Solution
(a) From the function, we have amplitude = 2,
Trigonometry and Elementary Calculus
b = 2,
c=−
π
2π
and period =
= π.
2
2
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π
The interval containing exactly one sine wave is obtained by letting 2x −
range from 0 to
2
2π. Thus, the end points of the interval can be found by solving the equations
2x −
π
=0
2
2x −
and
π
= 2π.
2
Solving the equations we obtain
2x =
π
π
=⇒ x =
2
4
2x = 2π +
and
π
5π
5π
=
=⇒ x =
.
2
2
4
π 5π
Hence, one sine wave will occur in the interval
,
. Note that from this interval we can
4 4
5π
π
determine the period by finding the difference between the endpoints, that is,
− = π.
4
4
π
Also, each part is . The graph of the function is plotted below.
4
(b) The graph will complete one cycle when x +
π
varies from 0 to 2π. That is, the end points
3
are
x+
π
−π
= 0 =⇒ x =
3
3
and
x+
So, the graph will complete one cycle in the interval
π
5π
= 2π =⇒ x =
.
3
3
−π 5π
,
.
3 3
2π
2π
2π
π
=
= 2π. Therefore, each part is
= .
b
1
4
2
π
π
The y-intercept is found when x = 0, that is, y = cos 0 +
= cos = 0.5. The graph of
3
3
the function is shown below.
Amplitude = 1, and period =
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Solutions of Trigonometric Equations
In this section, we discuss how to solve various trigonometric equations.
Definition 1.1
A trigonometric equation is an equation that contains trigonometric expres-
sions.
Usually these equations can be reduced to one of the following forms: cos θ = a, sin θ = a, or
tan θ = a.
We wish to determine an angle θ that gives us a. Sometimes this exercise is not easy. Generally,
there may be many solutions and sometimes no solution. Solutions of trigonometric equations
may be expressed in terms of either real numbers or angles. If a trigonometric equation is not an
identity, then techniques similar to those used for algebraic equations may be employed to find the
solutions, for instance, factoring, collecting like terms, expanding etc.
Recall that, we usually solve for sin x, cos θ, tan y and so on, and then find x, θ and y.
1.15.1
The Equation cos θ = a
Suppose that |a| ≤ 1 or −1 ≤ a ≤ 1, so that the equation has solutions.
The line y = a cuts the cosine curve in an infinite number of positions. Hence, there is an infinite
number of solutions to the equation.
If θ1 is a solution, then θ1 + 2π is also a solution since cosine has a period of 2π. Thus, cos θ1 = a
and cos (θ1 + 2π) = a.
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In general, all the solutions are given by;
n ∈ Z.
θ1 + 2nπ,
Since cosine is an even function, that is, cos (−θ1 ) = cos θ1 , it implies that −θ1 is also a solution.
Hence, −θ1 + 2π is also a solution.
Thus, the general solution to the equation cos θ = a is given by
±θ1 + 2nπ,
Example 1.26
n∈Z
1
Find the general solution of the equation cos x = .
2
Solution
We want to find the value of x whose cosine is
1
π
. Thus, we know that x =
will satisfy the
2
3
equation. So the general solution is
±
Example 1.27
π
+ 2nπ, n ∈ Z.
3
Find all values of θ in the interval [0, 2π] that satisfy the equation cos 3θ = 1.
Solution
We see that if 3θ = 0, then the equation will be satisfied. So, the general solution is
3θ = ±0 + 2nπ
=⇒
θ=
2nπ
, , n ∈ Z.
3
Now, if
n = 0,
n = 1,
n = 2,
n = 3,
θ=0
2π
θ=
3
4π
θ=
3
θ = 2π
Therefore, the values of θ that will satisfy the equation are: 0,
2π 4π
,
and 2π.
3 3
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1.15.2
The Equation sin θ = a
Suppose θ1 is a solution to the equation sin θ = a. Then θ1 + 2π is also a solution since sine has
a period of 2π. Hence, θ1 + 2nπ, n ∈ Z is a solution.
Recall that sin θ = sin (π − θ). Thus, π − θ1 is also a solution to the equation sin θ = a so is
(π − θ1 ) + 2nπ, n ∈ Z.
Therefore, the general solution to the equation sin θ = a is given by
θ1 + 2nπ, n ∈ Z
(π − θ1 ) + 2nπ, n ∈ Z
or
Example 1.28
Find the general solution of the following equations.
1
2
(b) sin2 x − sin x = 2
(a) sin x =
Solution
(a) We know that
sin
π 6
=
1
2
=⇒
x=
π
.
6
Therefore, the general solution is
π
+ 2nπ
6
π
x = + 2nπ
6
x=
i.e
or
π−
π
+ 2nπ, n ∈ Z
6
5π
+ 2nπ, n ∈ Z
6
or
(b) This equation is different from the equation above because we do not know the value of x
that will satisfy the equation. Thus, we must simplify the equation first.
Now, let sin x = y. Substituting in the equation yields
y2 − y − 2 = 0
(y − 2)(y + 1) = 0
=⇒ y = 2
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y = −1.
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Replacing y with sin x yields
sin x = 2 or
sin x = −1.
Since sin x = 2 is outside the range of the sine function, we discard it. Thus, we consider the
equation sin x = −1.
We know that sin
3π
3π
= −1 =⇒ x =
.
2
2
Thus, the general solution is
x=
1.15.3
3π
+ 2nπ
2
or
−
π
+ 2nπ, n ∈ Z.
2
The Equation tan θ = a
If θ1 is a solution (between −π/2 and π/2) to the equation tan θ = a, then θ1 + π is also a
solution (since tangent has a period of π).
So, the general solution to the equation tan θ = a is given by
θ1 + nπ, n ∈ Z
Example 1.29
Find the general solution of the equation tan4 x + 7 = 4 sec2 x.
Solution
Recall that 1 + tanx = sec2 x. Thus, the equation becomes
tan4 x + 7 = 4 (1 + tan2 x)
So,
tan4 x − 4 (1 + tan2 x) + 7 = 0
=⇒ tan4 x − 4 − 4 tan2 x + 7 = 0
=⇒ tan4 x − 4 tan2 x + 3 = 0
(59)
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Let y = tanx , then (59) becomes
y 2 − 4y + 3 = 0
=⇒ (y − 3)(y − 1) = 0
Therefore,
y=3
y−1
or
So,
tan2 x = 3
√
=⇒ tan x = ± 3
or
tan2 x = −1
or
tan x = ±1
√
√
π
π
= ± 3, then x = ± satisfies the equation tan x = ± 3.
3
3
π
Similarly, x = ± satisfies the equation tan x = ±1.
4
Since tan ±
Therefore, the general solution is
x=±
π
+ nπ
3
or
±
π
+ nπ, n ∈ Z.
4
Activity 1.9
Find all solutions of the following equations.
(a) (cos θ − 1)(sin θ + 1) = 0
(c) cos3 x = cos x
(b) cot2 x − 3 = 0
(d) cos2 x + cos x = sin2 x
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Summary
In this Unit, we have looked at the two systems of measuring angles. Also, the unit has comprehensively covered trigonometric functions which are used in applications that involve relationships
among the sides and angles of a triangle. We went further to derive trigonometric identities as well
as addition, product, and factor formulae. Functional values for the trigonometric functions for different angles, including quadrantal and special angles, have also been extensively discussed. The
last two topics cover graphs of trigonometric functions and solutions of trigonometric equations.
Suggestions for Further Reading
Ewer, J.P.G. (1994). Algebra, Trigonometry and Calculus. Science Teachers Association of
Malawi, Zomba (unpublished).
McKeague, C.P. & Turner, M.D. (2012). Trigonometry, 7th edition. Brooks Cole, Boston.
Larson, R. (2013). Trigonometry, 9th edition. Brooks Cole, Boston.
Unit Test
1. Convert
(a) 450◦ to radians.
(b) 0.3π radians to degrees [approximate to the nearest degree].
2. Evaluate the following without using a calculator. Leave your answer in surd form.
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4π
3
(b) cos 165◦ − cos 75◦
5π
12
(d) cos 75◦ sin 15◦
(a) sin
(c) sec
3. By using the addition formulae, evaluate exactly:
(a) tan 15◦
(b) cos 75◦
(c) sec 105◦
4. Prove the following trigonometric identities.
(c) (sin θ − cos θ)2 = 1 − sin 2θ
x
x
(d) sin x = 2 sin cos
2
2
(a) tan x sin x + cos x = sec x
sin 3t + sin 5t
(b)
= cot t
cos 3t − cos 5t
5. Suppose A + B =
π
1 − tan A
. Show that tan B =
.
4
1 + tan A
6. Sketch graphs of the following functions
(a) y = tan x − 3
(b) y = 1 + cos 2x
(c) y = 3 sin
π
6
+
x
2
7. Find the general solution of the following equations in the interval 0 < x < 2π.
(a) cos 2x + cos x = 0
(b) sin 3x − sin x = 0
(c) tan θ + sec θ = 1
Answers to Unit Activities
ACTIVITY 1.1
1. (a) 30◦ + n × 360◦
2. (a) Not coterminal
3. (a)
π
4
4. (a) 405◦
(b) 45◦ + n × 360◦
(b) Coterminal
(b) −
3π
2
(b) −150◦
5. 2.094cm
Trigonometry and Elementary Calculus
(c) −90◦ + n × 360◦
(c) Coterminal
(c)
(d) Not coterminal
70π
9
(c) 243.5◦
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ACTIVITY 1.3
1. (a) 80◦
θ
420◦
2.
4π
3
135◦
(b)
sin
√θ
3
2
√
3
−
√2
2
2
cos θ
1
2
1
−
√2
2
−
2
π
3
(c) 70◦
tan θ
√
3
csc θ
2
√
3
2
−√
3
2
√
2
√
3
−1
sec θ
2
−2
2
−√
2
(d)
17π
36
cot θ
1
√
3
1
√
3
−1
ACTIVITY 1.4
4
√
(a) √
2− 6
√
3
(b) −
3
(c)
√
√
3+2
(e) −
√
( 3 − 1)2
(d)
8
3
2
√
(f) − 2
ACTIVITY 1.5
24
25
7
(b)
25
θ
3
3. sin
= ;
2
4
2. (a)
24
7
9
(d)
10
√
θ
7
cos
=
2
2
1
9
25
(f)
24
(c)
(e)
ACTIVITY 1.6
1. cos 8x + cos 2x
2. (a) 2 sin 4x cos x
(b) 2 sin
√
2
3. (a)
2
(b)
5t
t
sin
2
2
1
4
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ACTIVITY 1.8
(a)
(b)
(c)
(d)
ACTIVITY 1.9
(a) θ = 2nπ
or
3π
+ 2nπ, n ∈ Z
2
π
5π
+ nπ or
+ nπ, n ∈ Z
6
6
π
(c) x = ± + 2nπ or ± π + 2nπ or ± 2π + 2nπ, n ∈ Z
2
π
(d) x = ± + 2nπ or ± π + 2nπ, n ∈ Z
3
(b) x =
Trigonometry and Elementary Calculus