Objectives

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Mathematics C30
Module 1
Lesson 1
Mathematics C30
Trigonometric Ratios and Functions
1
Lesson 1
Mathematics C30
2
Lesson 1
Trigonometric Ratios and Functions
Introduction
The Greek words for triangle (trigonon) and measure (metron) form the basis for the word
trigonometry. It is, therefore, quite common to use letters of the Greek alphabet to denote
the measures of angles.
In this first lesson, the trigonometric ratios and their inverses are defined for acute angles,
and then these definitions are extended to general angles. The development of the hand
held calculator eliminated the need for extensive tables of trigonometric ratios, and
calculations have become much quicker. For this course, a scientific calculator is a
requirement. A scientific calculator is one that contains the sin cos tan functions.
Graphing calculators also have these functions; however, a graphing calculator will not be a
requirement for this course.
Greek Alphabet
alpha


beta

gamma

delta
epsilon



zeta
eta
theta





iota
kappa
lambda
mu
nu
xi
omicron
pi





rho
sigma
tau
upsilon
phi
chi, khi
psi






omega

Mathematics C30
3
Lesson 1
Mathematics C30
4
Lesson 1
Objectives
After completing this lesson you will be able to
Mathematics C30
•
determine the trigonometric ratios of acute angles measured
in degrees.
•
determine the trigonometric ratios of general angles
measured in degrees.
•
determine the solutions to a simple trigonometric equation.
•
express the trigonometric function of any angle as a
trigonometric function of an acute angle.
5
Lesson 1
Mathematics C30
6
Lesson 1
1.1 Trigonometric Ratios of Acute Angles Measured in
Degrees
An angle is the union of two rays with a common end point.
The two rays of the angle are sometimes called arms and the point where the rays meet is
called the vertex of the angle.
B
arm
0
vertex
arm
A
The name of the above angle is angle AOB or  AOB . If no other rays are involved this
could be simply named O .
It is convenient to think of an angle being generated by rotating one ray about the vertex
O to form the initial ray and the terminal ray.
Positive Angle
R
al
in
rm
Te
O
ay
Initial Ray
Negative Angle
B
In
O
A
Fig. 1
y
Ra
l
a
iti
A
Terminal Ray B
Fig. 2
The angle in Figure 1 is defined as a positive angle because the initial ray has to rotate in
a counterclockwise direction to reach the position of the terminal ray.
In Figure 2, the angle is negative because the initial ray has to rotate in a clockwise
direction to reach the position of the terminal ray.
An angle is in standard position when it is drawn on the coordinate axes, its initial ray
extends horizontally to the right from the origin (in other words its initial ray is the
positive portion of the x-axis), and its vertex is at the origin.
Mathematics C30
7
Lesson 1
y
y
x
x
Posit ive angle in st an dard
posit ion.
Negat ive an gle in st an dard
posit ion.
Angles in standard position which have the same terminal ray are called coterminal
angles
There are infinitely many angles coterminal with a given angle. However, the angle in
standard position with the smallest, positive measure is called the principal angle.
The diagram below shows that the angles in standard position with measures of 45°,
 315  , and 405° are coterminal. The principal angle of each angle is 45°.
y
y
45°
y
–315°
x
x
x
405°
If the principal angle has a measure of 60° then all the other angles coterminal with it can
be written 60   n 360  , where n is any integer or n  I



If n  0  the measure is 60°
If n  1  the measure is 420°
If n  1  the measure is  300  .
The principal angle of a  240  angle is a 120° angle. The principal angle of an 840° angle is
a 120° angle since 840  2360   120 .
In the formula 60   n 360  , the first measure must always be the measure of the
principal angle.
For example, it is incorrect to write  50   n 360  . This should be written 310   n 360  .
Positive and negative angles will be discussed and used in a later lesson. This lesson will
continue with positive angles only.
Mathematics C30
8
Lesson 1
Trigonometric Ratios
B1
B

A
C
C1
The triangles ABC and AB 1 C1 are similar since the corresponding angles are equal in
measure. Consequently the ratios of the sides are equal.
BC B 1 C1 BC B 1 C1 AC AC 1
.

,

,

AC
AC 1 AB
AB 1 AB AB 1
The ratios depend only on the size of the angles and not on the size of the triangle.
The relationships between the angles and the ratios of the sides of a right-angled triangle
are called the trigonometric ratios or trigonometric functions. These ratios are the
foundation of trigonometry.
y
 is a Greek letter
pronounced (thay-da).
A
hypotenuse
opposite side

O
B
x
adjacent side
In the diagram,  is an acute angle in standard position. AB is any perpendicular drawn
to the initial ray OB. Hence, OA is the hypotenuse, AB is the side opposite to angle  ,
and OB is the side adjacent to angle  .
Mathematics C30
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Lesson 1
The trigonometric ratios of angle  with reference to the sides of this right-angled triangle
are defined in the following table.
A
Trigonometric
Ratio
Abbreviation
Definition

O
sine 
sin 
cosine 
cos 
tangent 
tan 
cotangent 
cot 
secant 
sec 
cosecant 
csc 
opposite side
hypotenuse
adjacent side
hypotenuse
opposite side
adjacent side
adjacent side
opposite side
hypotenuse
adjacent side
hypotenuse
opposite side
B
AB
OA
OB
OA
AB
OB
OB
AB
OA
OB
OA
AB
For any other right triangle with angle  , regardless of size, the ratios will be the same.
Notice that the cotangent, secant, and cosecant ratios are reciprocals of the tangent,
cosine, and sine ratios.
1
tan 
1
sec  
cos 
1
csc  
sin 
cot  
Other ratios can also be shown as true.
tan  
sin 
cos 
 AB

sin   OA
Proof:

cos   OB

 OA
cos 
Similarly, cot  
.
sin 
Mathematics C30


  AB  OA  AB .
 OA OB OB


10
Lesson 1
Example 1
For the right-angled triangle ABC, state the six
trigonometric ratios, first of C and then of A.

A
5
Any one leg of a triangle is opposite to one angle
but adjacent to another angle. For example, AB is
opposite C but adjacent to A.
C
3
B
4
Solution:
A
5
side opposit e
C
3
C
4
side adjacen t C
sin C 
cos C 
tan C 
cot C 
sec C 
csc C 
A
B
5
4
side opposit e A
C
opp. 3

hyp. 5
adj. 4

hyp. 5
opp. 3

adj. 4
1
4

tan C 3
1
5

cos C 4
1
5

sin C 3
3 side adjacen t
A
sin A 
cos A 
tan A 
cot A 
sec A 
csc A 
When any two sides of a right triangle are given, all the
trigonometric ratios can be found by first using the
Pythagorean Theorem to find the third side.
B
opp. 4

hyp. 5
adj. 3

hyp. 5
opp. 4

adj. 3
1
3

tan A 4
1
5

cos A 3
1
5

sin A 4
2
2
a +b =c
2
c
a
b
Mathematics C30
11
Lesson 1
Example 2
B
For right triangle ABC with right angle at C, if tan A 
cos A, and sin B.
5
, find sin A,
2
Solution:

tan A 
A
C
opposite
5

adjacent
2
Label the triangle so that the sides give this ratio.
Calculate side c.
B
c
c 2  a 2  b2
a=5
c 2  5 2  22
A
c 2  29
b=2
C
c  29
Determine the ratios of the sides.
sin A 
cos A 
sin B 
5
29
2
29
2
29

5 29
29

2 29
29

2 29
29
Remember to rationalize the denominator when evaluating trigonometric functions.
Mathematics C30
12
Lesson 1
Example 3
If  is an angle in standard position and P(3, 7) is a point on the terminal ray, find
all the trigonometric ratios of  .
Solution:
Draw a diagram.
y
P(3, 7)
r
7

3
x
When a perpendicular is drawn from P to the x-axis, a right triangle is formed with
sides 3 and 7. The hypotenuse is found by use of the Pythagorean Theorem.
Calculate the hypotenuse.
r 2  32  72
r  58
Determine the ratios of the sides.
7
sin  
58
3
cos  
tan  
58

7 58
58
csc  
58
7

3 58
58
sec  
58
3
7
3
cot  
3
7
Trigonometric Functions
Each acute angle  corresponds to a right triangle with  as one of its angles. From the
sides of the right triangle the trigonometric ratios can be determined.
Mathematics C30
13
Lesson 1
Example 4
Using a protractor and ruler, find tan 35° and csc 35°.
Solution:
Method
A
B

Draw your own accurate triangle.

Draw angle AOC = 35° with a protractor.

Let B be any point on OA. Draw BD
perpendicular to OC. Measure the sides of the
right-angled triangle BOD.
OB
BD
OD
=
=
=
35°
O
___ cm
___ cm
___ cm
tan 35 
opp. BD


adj. OD
Round to two decimal places.
=
Determine the tan 35°.
Determine the csc 35°.
csc 35 
D
C
hyp. OB


opp. BD
Round to two decimal places.
=
csc 35° = reciprocal of sin 35°
The above graphical method gives an excellent answer, if all the measurements are made
carefully. It is, of course, a slow method. Hence, to save time and work, Tables of
Trigonometric Ratios have been calculated by methods employing advanced mathematics.
These tables have now been replaced by scientific calculators. The values given by tables
or the calculator are not exact but are infinite decimals rounded to a certain number of
decimal places.
Mathematics C30
14
Lesson 1
Now use your calculator to check the solutions from Example 4.
 Tan 35° = 0.7002075
1
1

 1.7434468 .
 Csc 35° =
sin 35 0.5735764
Scientific calculator instructions will not be given in this course. You will be
reminded to use your calculator at certain times. If you cannot do this correctly,
contact your instructor.
Example 5
Given that sin  = 0.4000, use a scientific calculator to find  in degrees. Then find
tan  and cos  .
Solution:
Enter 0.4 followed by a sine inverse function.
 = 23.578178 degrees.
tan 23.578178 = 0.4364358
cos 23.578178 = 0.9165151.
You should be familiar with the following procedures on the calculator.
•
If an angle is given in degrees, find any of the six trigonometric ratios of that angle.
•
Conversely, find the angle if a trigonometric ratio of that angle is given.
Special Angles
The angles whose measures are 30°, 45°, or 60° are special because their trigonometric
ratios can be found in exact form without the use of a calculator. Exact form does not
include a decimal rounded to a certain number of decimal places, but may be, for example,
4, 0.5, 2 , 3 , etc. These angles occur in many situations and are often called common
angles.
Mathematics C30
15
Lesson 1
Angles of 30° and 60°
Draw an equilateral triangle whose sides are each equal to 2 units. From the vertex, draw
a perpendicular to the base. This perpendicular produces two 30° - 60° - 90° triangles, and
the perpendicular bisects the base of the equilateral triangle.
Calculate the length of the perpendicular.
A
AC 2  BC 2  AB 2
30° 30°
AC 2  1  4
2
3
AC 2  3
AC  3
60°
B
1
1
C
From this special right triangle, the following exact trigonometric ratios can be
determined.
BC
AB
AC
cos 30 
AB
BC
tan 30 
AC
AC
cot 30 
BC
sin 30 



1
2
3
2
1
3

3
3
BC
1
3


AC
3
3
AB
sec 60 
2
BC
cot 60 
 3
AB
2
2 3


AC
3
3
AB
csc 30 
2
BC
sec 30 
Mathematics C30
AC
3

AB
2
BC 1
cos 60 

AB 2
AC
tan 60 
 3
BC
sin 60 
16
csc 60 
AB
2
2 3


AC
3
3
Lesson 1
An Angle of 45°
Draw a square with each side one unit in length. The diagonal forms the hypotenuse of a
triangle with 45° angles. Calculate the length of the hypotenuse.
A
AB 2  BC 2  AC 2
2
AB 2  1  1
1
AB 2  2
AB  2
B
1
C
From this special right triangle, the following exact trigonometric ratios can be
determined.
sin 45 
cos 45 
tan 45 
cot 45 
sec 45 
csc 45 
AC
1
2


AB
2
2
BC
AB
AC
BC
BC
AC
AB
BC
AB
AC

1
2

2
2
1
1
 2
 2
Rather than memorize the values of these ratios of the common angles, it is best
to learn how to label the equilateral triangle or square each time exact values
are required.
Mathematics C30
17
Lesson 1
Example 6
Evaluate by giving the value in simplest exact form.
sin 60  cos 45  tan 2 30
Solution:
tan 2 30  tan 30
2
Note:
Evaluate each trigonometric function.
3 1  3 
sin 60  cos 45  tan 30 


2
2  3 
2
2

3
2 3


2
2 9

3
2 1


2
2 3

3 3 3 2 2


6
6
6

3 3 3 2 2
6
Exercise 1.1
1.
Give the measure of one positive angle and one negative angle, each coterminal with
the given angle. State the measure of the principal angle.
a.
b.
c.
d.
2.
15°
 60
 390
1605°
Write the formula for the measures of all angles coterminal with the given angle.
a.
b.
c.
d.
30°
 30
 210
 725
Mathematics C30
18
Lesson 1
y
3.
B
A
6
4
x
Find the trigonometric ratios of A.
Find the trigonometric ratios of B.
a.
b.
1
.
4
4.
Find the remaining five trigonometric ratios of an acute angle A if sin A 
5.
Find the remaining five trigonometric ratios of an acute angle A if cos A 
6.
The terminal side of an angle  in standard position passes through the point
P 3, 3 3 . Give the exact values of sin  and cos  .
7.
Use your calculator to find each ratio and round to 4 decimal places.

a.
b.
c.
d.
8.
3
.
2

sin 89°
csc 89°
tan 10°
sec 47.5°
Use your calculator to find the measure of the angle for each given trigonometric
ratio. Round the measure to the nearest tenth of a degree.
a.
b.
c.
d.
e.
sin  = 0.8290
csc  = 10.4226
sec  = 10.4226
tan  = 1.0000
cot  = 2.1350
Mathematics C30
19
Lesson 1
9.
Find the lengths of the indicated unknown sides.
a.
b.
B
12
c
C
30 30
y
60° b
18
A
10.
x
Evaluate by giving the value in simplest exact form.
a.
b.
c.
d.
e.
f.
tan 30° + 2 sin 30°
cos 45° · tan 60° · csc 30°
2
tan 2 30  sec 2 30
Note: tan 2 30  tan 30
sin 2 45  cos 2 45
tan 45  cot 2 45  csc 2 45
2 cos 2 30  1  cos 60
1.2 Trigonometric Functions of General Angles
The trigonometric ratios of angles of any measure can be defined in the same way as the
ratios of acute angles if the angle is placed in standard position.
Let P(x, y) be any point on the terminal arm of angle A. The segment from P to the origin
is called the radius vector and is always a positive length.
y
OP  r  x 2  y 2
r
O
Mathematics C30
20
A
x
P(x, y)
y
x
Lesson 1
The coordinate x and y may be positive or negative, depending on the quadrant.
y
y
P(x, y)
P(x, y)
r
A
x (positive)
y (positive)
x
A
y (negative)
x
x (negative)
y
x (negative) A
r
y (positive)
y
A
x
x (positive)
x
r
r
y (negative)
P(x, y)
P(x, y)
Definitions: Trigonometric functions of an angle A in
any quadrant, where P (x, y) is a point on the
terminal side of A.
r
y
r
sec A 
x
x
cot A 
y
y
r
x
cos A 
r
y
tan A 
x
csc A 
sin A 
Notice that if angle A is acute, the angle is in the first quadrant and the definitions
coincide with those you learned for acute angles and all the ratios are positive. For angles
in other quadrants, some ratios are negative.
Mathematics C30
21
Lesson 1
Example 1
Evaluate the six trigonometric functions for angle A whose terminal ray passes
through the point  2,  4 .
Solution:
y
Determine the value of r.
r 2  x 2  y2
A
x
r  20  2 5
Evaluate the trigonometric functions.
sin A 
r
y
4
2
2 5



r 2 5
5
5
(–2, –4)
x
2
5


r 2 5
5
y 4
tan A  
2
x 2
cos A 
csc A 
r 2 5  5


y
4
2
r 2 5

 5
x
2
x 2
1
cot A  

y 4
2
sec A 
Example 2
Find the exact values of the six trigonometric ratios of a 315° angle.
Solution:
y
Notice that the radius vector forms a 45° angle with the
x-axis. Therefore, a triangle with sides of length 1 and
hypotenuse 2 can be formed.
315°
45°
x
2
P(1, –1)
Mathematics C30
22
Lesson 1
Apply the definitions of the trigonometric functions.
sin 315 
cos 315 
1
2
1
2
tan 315  1


2
2
csc 315   2
2
2
sec 315  2
cot 315  1
Any angle in standard position whose terminal side coincides with either the x or y axis is
called a quadrantal angle. The next two examples show how the trigonometric ratios of
quadrantal angles are determined.
Example 3
Evaluate the six trigonometric functions of the quadrantal angle whose measure is
180°.
Solution:

Let P = (x, y) be any point on the terminal side of the
angle. In particular, let P   2, 0 . The radius vector
has length 2.
y
r
180°
x
P(–2, 0)
Apply the definition of the trigonometric functions.
y 0
 0
r 2
x 2
cos 180  
 1
r
2
y
0
tan 180  
0
x 2
sin 180 
r 2
(undefined)

y 0
r
2
sec 180  
 1
x 2
x 2
(undefined)
cot 180  
y
0
csc 180 
Note: Any other radius vector length would have given the same answers. You may
repeat the example using P   6, 0 .
Mathematics C30
23
Lesson 1
Example 4
Evaluate sin, cos, and tan of the quadrantal angle 270°.
Solution:
Let P(x, y) be any point on the terminal side of the angle. In
particular, let P  0,  3 . The radius vector has length 3.
Evaluate for sin, cos and tan of 270°.
270°
x
r
y 3

 1
r
3
x 0
cos 270    0
r 3
y 3
tan 270  
(undefined)
x
0
sin 270 
Mathematics C30
y
24
P(0, –3)
Lesson 1
Activity 1.2
Use a calculator to evaluate sin 30°, sin 150°, sin 210° and sin 330°. Sketch
each of these angles.
Angle
Evaluated
Sketch
y
30°
x
y
150°
x
y
210°
x
y
330°
x
What is the measure of the angle between the terminal arm and the nearest
x-axis?
Mathematics C30
25
Lesson 1
Exercise 1.2
1.
State the exact values of the sine and tangent functions of angle  in standard
position if the terminal side passes through the given point P. In each case sketch
the given angle. If a trigonometric function is undefined, state this.
a.
b.
c.
d.
e.
f.
2.
Find the exact values of the sine and cosine functions of angles whose measures are
given. See Example 2.
a.
b.
c.
d.
3.
P  4, 3
P 1,  3
P  3,  2
P 3, 0
P 0, 5
P 0,  5
120°
135°
150°
210°
Use the calculator to evaluate each function. Round the answer to 4 decimal places.
a.
b.
c.
d.
sin 10°, sin 170°
cos 10°, cos 350°, cos 10
tan 30°, tan 210°
tan 90°, tan 270°
1.3 Reference Angles
As shown in Activity 1.2, more than one angle has the same trigonometric
ratio. For example, cos 60° = 0.5, and cos 300° = 0.5. Conversely, if cos  = 0.5
is given, use of the inverse cos function on the calculator gives  = 60° without
any indication that  could also be 300°.
Mathematics C30
26
Lesson 1
The calculator gives only the smallest positive (or negative) angle when the inverse
function is used. Any other solutions to cos  = 0.5 must be found in other ways. The use
of reference angles solves this problem.
When an angle in standard position is drawn in any quadrant a right triangle is formed by
drawing a perpendicular from any point on the terminal arm to the x-axis. This is called
the reference triangle, and the acute angle between the terminal side and the x-axis is
called the reference angle.
In the diagrams the shaded triangle is the reference triangle and 0° <  < 90° is the
reference angle of angle A.
y
y
A
x


A
 = A – 180
x
= A
y
y
A

x
A

x
 = 180° – A
 = 360° – A
Example 1
Sketch each angle whose measure is given and determine the measure of the
reference angle.
a)
b)
c)
d)
Mathematics C30
A = 70
A = 150
A =  210
A = 800
27
Lesson 1
Solution:
a)
Make a sketch of the given angle.
y
x
–70°

Calculate the measure of the angle between
the terminal ray and the nearest x-axis.
b)
The reference angle has a
measure of 70°.
Make a sketch of the given angle.
y
150°

x
Calculate the measure of the angle between
the terminal ray and the nearest x-axis.
angle has a
Mathematics C30
180  150  30
The reference
measure of 30°.
28
Lesson 1
c)
Make a sketch of the given angle.
y

x
–210°
Calculate the measure of the angle between
the terminal ray and the nearest x-axis.
210  180  30
The reference
angle has a
d)
measure of 30°.
Make a sketch of the given angle.
y

800°
x
Calculate the measure of the angle between
the terminal ray and the nearest x-axis.
angle has a
800   2  360  80
The reference
measure of 80°.
All coterminal angles have the same reference angles.
The Sign of a Trigonometric Ratio
The radius vector in any quadrant is always considered to have positive length r. Since the
coordinates x and y vary in sign from quadrant to quadrant, the trigonometric functions
vary in sign from quadrant to quadrant.
Mathematics C30
29
Lesson 1
For example, all six trigonometric functions have positive values for angles in the first
quadrant because x, y, and r are each positive. In the second quadrant sine is positive but
cosine and tangent have negative values. For example, in the second quadrant,
y  ve
tan A  
 ve .
x - ve
The ASTC or CAST rule.
A:
S:
T:
C:
all functions are positive in Q1
sine is positive in QII
tangent is positive in QIII
cosine is positive in QIV
Activity 1.3
Complete the following table. A is an angle in standard position in the given
quadrant. Use the definitions to fill in each space with + or – to indicate whether
the values of the given function are positive or negative in the various quadrants.
I
II
sin A
+
cos A
–
tan A
–
III
IV
y
II
(–, +)
I
(+, +)
x
(–, –)
III
csc A
sec A
(+, –)
IV
cot A
Mathematics C30
30
Lesson 1
Example 2
If A is an angle whose sine is positive, sketch two possible angles in the correct
quadrants.
y
Solution:
Sine is positive in the first and second quadrants.
The measure of angle A is not known.
The main concern is to draw a typical angle in the correct
quadrant.
x
Example 3
Sketch an angle A in the correct quadrant which satisfies simultaneously the
conditions that csc A is negative and cot A is positive.
Solution:
y
y
x
For csc A < 0,
angle A is in
the 3rd or 4th
quadrant.
y
x
For cot A > 0,
angle A is in
the 1st or 3rd
quadrant.
x
For csc A < 0 and
cot A > 0
together, angle A
is in the 3rd
quadrant only.
Therefore, an angle in the third quadrant satisfies the conditions.
Mathematics C30
31
Lesson 1
Example 4
3
Evaluate the five other trigonometric functions of A if sin A   .
5
Solution:
y
Draw a diagram.

Since the quadrant is not specified, there are two
possibilities for A because sine is negative in the third
and fourth quadrants.
x
–3
x
x
–3
5
5
Q
P
Label the triangles.
sin A  
3
 3 (ordinate)

5 5 (radius vector)
Calculate the unknown sides and determine the coordinates of P and Q.
x 2  32  52
x 2  25  9
x 2  16

x  4
Therefore, P   4,  3, Q  4,  3 .
Evaluate the functions.
Third Quadrant
Fourth Quadrant
cos A  
4
5
cos A  
4
5
tan A  
3
4
tan A  
3
4
csc A  
5
3
csc A  
5
3
sec A  
5
4
sec A  
5
4
cot A  
4
3
cot A  
4
3
Mathematics C30
32
Lesson 1
Example 5
Find all the principal angles  for which tan   3 .
Solution:
Draw and label the reference triangles.
•
First, note that  could be in quadrant 1 or in quadrant 3 since tan  is a positive
value.
•
Notice that
3
3  3

.
1
1
y
y
P(1, 3)
r
3

1
–1
x

x
– 3
P(–1, – 3)

The two reference triangles are congruent since the corresponding sides are of equal
length, or congruent.

Therefore, the reference angles are equal in measure.

In the first quadrant   60  , and in the third quadrant   180   60   240  .
Check the solution with your calculator.
tan 240   tan 60   1 .7320508
Mathematics C30
33
Lesson 1
Example 6
Find all the principal angles  for which cos   0.6428 .
Solution:


Since the ratio is
positive, the angle 
could be in the first
quadrant or in the
fourth quadrant.
y
y
= 50°
x
 = 360° – 50°
x
50°
By calculator, we get
  50 .
Since the ratios are the same for both angles, the angles of the reference triangle must be
the same and, in particular, the reference angles must be the same.

In the first quadrant,  = 50°.

In the fourth quadrant,   360  50  310.
In summary, Examples 4, 5 and 6 show that the solutions to a trigonometric ratio have
the same reference angle.
Mathematics C30
34
Lesson 1
Example 7
Determine two angles,  , where tan   5 .
Solution:
tan   5
  78.7 .
Use a calculator to find one solution.
y
–78.7°
x
The reference angle is +78.7°.

The other solution must be in the second quadrant where tan is also negative, and it
must have the same reference angle.
y

78.7°
Use the reference angle to find the second solution.
x
  180  78.7
  101.3
The two angles are  78.7 and 101.3°.
Mathematics C30
35
Lesson 1
Evaluating Trigonometric Functions of any Angle
The trigonometric function of any angle can be expressed as a trigonometric function of
the reference angle with the appropriate + or – sign.
Example 8
Express cos 135° in terms of the reference angle and then evaluate.
Solution:
y
Sketch the 135° angle, and from any point P  a, b on the
terminal ray, draw a perpendicular to the x-axis.

P (–a, b)
r
b
The reference angle has a measure of 45°.
135°
x
45°
–a
Note: The sides of the triangle are labelled b and  a . The minus sign reminds us that
the abscissa is negative. In the reference triangle drawn without the coordinate system,
the side “a” has a positive length.
Determine the cos 135° using the definition of cosine.
a
r
a
  
r
cos 135 
 side adj. 45 angle 
  

hyp


 cos 45

Therefore, cos 135   cos 45  
Mathematics C30
1
2

r
45°
a
2
2
2
.
2
36
Lesson 1
Example 9
Express each of the following in terms of the reference angle and then, evaluate.
1.
2.
3.
sin 210
cos 210
tan 210
Solution:
y
–a
30°
–b
210°
x
r
P (–a, –b)


The reference angle is 30°.
Minus signs are used in labeling the sides to remind us that both coordinates are
negative.
1.
Determine sin 210°.
b
r
b
  
r
 side opp. 30 angle 
  

hyp


sin 210 
 sin 30

1
2
1
Therefore, sin 210   sin 30   .
2
Mathematics C30
37
Lesson 1
2.
Determine cos 210°.
a
r
a
  
r
cos 210 
 side adj. 30 angle 
 

hyp


 cos 30  
Therefore, cos 210   cos 30  
3.
3
2
3
2
Determine tan 210°.
b
a
b

a
 side opp. 30 angle 
 

 side adj. 30 angle 
tan 210 
 tan 30
Therefore, tan 210  tan 30 
1
3

3
.
3
The value of a trigonometric function of angle A is
equal to the value of the same trigonometric
function of the reference angle of A with the
appropriate (+ –) sign, depending on the quadrant of
the terminal side of angle A.
Mathematics C30
38
Lesson 1
Example 10 is similar to the two previous examples, except that the answer is obtained
quicker when this principle is applied.
Example 10
Express sin  225 as a function of the reference angle and evaluate. Do the same
for tan  225 .
Solution:

The angle is in the second quadrant and the reference angle is 45°.

Sine is positive in the second quadrant.
sin  225   sin 45


Tangent is negative in the second quadrant.
Mathematics C30
39
1
2

2
2
tan 225   tan 45
 1
Lesson 1
Exercise 1.3
1.
Determine the quadrant and the measure of the reference angle for each given angle
measure.
a.
b.
c.
d.
e.
2.
Determine the quadrant(s) that angle A lies in if the following conditions are
satisfied.
a.
b.
c.
d.
e.
f.
3.
4.
110°
110
480°
 70
 650
cos A > 0
cos A < 0, and tan A < 0
sec A > 0
cot A > 0, and csc A > 0
(sin A)(cos A) > 0
(tan A)(cos A) < 0
Evaluate the five other trigonometric functions of A for each of the possibilities for A.
3
5
a.
cos A  
b.
tan A  5
(give exact values)
5
5 


 give exact values,  5 

1
1 

Find all the principal angles  which satisfy the equation. Round to the nearest
degree.
a.
b.
c.
d.
e.
f.
g.
sin A = 0.9135
cos A = 0.5555
cos A =  0 .3
sin A =  0 .75
tan A = 10
sec A = 10
csc A =  17
Mathematics C30
40
Lesson 1
5.
Express each as a trigonometric function of the reference angle with the appropriate
sign (+, –) and give the exact value.
a.
b.
c.
d.
e.
f.
6.
cos  60 
sin  60 
tan 210 
cot  210 
sec 225°
csc 390°
Suppose that  is an acute angle. Sketch each angle and express each as a function
of  alone.
a.
b.
c.
d.
sin180  
sin180  
cos180  
tan180  
Mathematics C30
41
Lesson 1
Mathematics C30
42
Lesson 1
Answers to Exercises
Exercise
1.1
1.
The answers may vary. Those given here are possible answers.
a.
b.
c.
d.
375°,  345  , Principal angle: 15°
300°,  420  , Principal angle: 300°
330 ,  30  , Principal angle: 330°
165°,  195  , Principal angle: 165°
2.
a.
b.
c.
d.
30° + n360°, n  I
330° + n360°, n  I
150° + n360°, n  I
365° + n360°, n  I
3.
r  2 13
a.
b.
4.
2  2 13
13
13
cos A  3  3 13
13
13
tan A  2
3
csc A 
sin A 
13
2
sec A  13
3
cot A  3
2
3  3 13
13
13
cos B  2  2 13
13
13
tan B  3
2
sin B 
csc B 
13
3
sec B 
13
2
cot B  2
3
cos A 
15
4
sec A 
tan A 
1  15
15
15
cot A  15
4  4 15
15
15
csc A  4
Mathematics C30
43
Lesson 1
5.
sin A  1
2
csc A  2
tan A  1  3
3
3
cot A  3
sec A  2  2 3
3
3
3
2
1
cos  
2
6.
sin  
7.
a.
b.
c.
0.9998
1.0002
0.1763
d.
sec 47 .5  
a.
56°
1  10 .4226
sin 
1
sin  
 0 .0959
10 .4226
  5 .5 
84.5°
45°
25.1°
8.
b.
c.
d.
e.
9.
a.
b.
Mathematics C30
1
1

 1 .4802
cos 47 .5  0 .6756
tan 60   12 , b  12
 6 .928
b
tan 60 
sin 60   12 , c  12  13 .8564
sin 60 
c
tan 30   18 , y  18
 31 .1769
y
tan 30 
tan 30   30   x  18 , tan 60   x  18 , x  18  54 ,
y
31 .1769
44
x  36
Lesson 1
10.
a.
b.
1
3
1 
 2  
1 
3
3
2
1
 3 2  6
2
2
2
2
2
c.
 1 
 2 
5

  
 
3
 3
 3
d.
 1 
 1 

  
  1
 2
 2
e.
1  1 
f.
 3
 1  1  0
2

2
 2 
a.
sin  
2
 2
2
3 3
3
0
2
Exercise
1.2
1.
3
3
, tan   
5
4
y
(–4, 3)
5
3

–4
b.
c.
d.
e.
f.
Mathematics C30
sin  
3
10
 3 10
, tan   3
10
 2 13
2
, tan  
13
3
13
sin   0, tan   0
sin   1, tan   undefined
sin   1, tan   undefined
sin  
2

x

45
Lesson 1
2.
a.
sin  
3
2
cos   
1
2
y
60°
b.
3.
Exercise
1.3
1.
2.
sin  
1
2
1
2

2
2
cos   
2

x
2
2
3
2
c.
sin  
d.
sin   
a.
b.
c.
d.
0.1736
0.9848
0.5774
undefined
a.
b.
c.
d.
e.
2, 70°
3, 70°
2, 60°
4, 70°
1, 70°
a.
b.
1, 4
cos A < 0 in Quadrants 2 and 3
tan A < 0 in Quadrants 2 and 4
 A is in the 2nd quadrant only.
1, 4
1
For the product to be positive, the signs of sin A and cos A must be the
same.
Sin A and cos A are both positive in Quadrant 1, and are both negative
in Quadrant 3.
 1, 3
3, 4
c.
d.
e.
f.
Mathematics C30
cos   
1
120°
1
2
cos   
3
2
46
Lesson 1
3.
a.
In Quadrant 2
4
sin A 
5
4
tan A  
3
5
csc A 
4
5
sec A  
3
3
cot A  
4
y
5
4
–4
A
–3
x
5
In Quadrant 3
4
sin A  
5
4
tan A 
3
5
csc A  
4
5
sec A  
3
3
cot A 
4
b.
Mathematics C30
In Quadrant 2
5
5 26
sin A 

26
26
1
26
cos A  

26
26
26
csc A 
5
sec A   26
1
cot A  
5
y
5
1
x
–1
–5
47
r = 26
Lesson 1
In Quadrant 4
5
5 26
sin A  

26
26
1
26
cos A 

26
26
26
csc A  
5
sec A  26
1
cot A  
5
4.
a.
b.
c.
d.
e.
f.
g.
66°, 114°
56°, 304°
107°, 253°
311°, 229°
84°, 264°
84°, 276°
357°, 183°
5.
a.
cos  60   cos 60  
1
2
y
–60°
b.
sin  60    sin 60  
c.
d.
3
3
cot  210    cot 30    3
e.
f.
sec 225    sec 45    2
csc 390   csc 30   2
Mathematics C30
x
* cos is positive
in quadrant 4.
3
2
tan 210   tan 30  
48
Lesson 1
6.
a.
 sin 
y
180° + 
x

y
b.
sin 

c.
d.
Mathematics C30
180° – 
x
 cos 
 tan 
49
Lesson 1
Mathematics C30
50
Lesson 1
Mathematics C30
Module 1
Assignment 1
Mathematics C30
51
Lesson 1
Mathematics C30
52
Lesson 1
Optional insert: Assignment #1 frontal sheet here.
Mathematics C30
53
Lesson 1
Mathematics C30
54
Lesson 1
Assignment 1
Values
(40)
A.
Multiple Choice: Select the best answer for each of the following and place a
check () beside it.
1.
The one angle that is in standard position is ***.
____
____
____
____
GOH
EOF
BOA
COD
a.
b.
c.
d.
y
y
y
A
B
y
F
x
O
C
O
x
O
E
H
x
G
x
O
D
2.
The terminal arm of an angle in standard position whose measure is
 250 lies in quadrant ***.
____
____
____
____
3.
a.
b.
c.
d.
135°
45°
225°
 405 
The principal angle of a 430  angle is one whose measure is ***.
____
____
____
____
Mathematics C30
1
2
3
4
An angle in standard position coterminal with an angle whose measure
is 135 has a measure of ***.
____
____
____
____
4.
a.
b.
c.
d.
a.
b.
c.
d.
70°
290°
 290 
110°
55
Lesson 1
5.
In the given diagram sin A is the same as ***.
____
____
____
____
6.
7.
8.
B
cos A
cos B
tan B
csc B
A
C
In the given diagram, sec 20° is the same as ***.
____
a.
____
b.
____
____
c.
d.
cos 20 
1
sin 20 
sin 70°
csc 70°
B
20°
C
A
The one false equation is ***.
____
a.
____
b.
____
c.
____
d.
1
csc 
1
cos  
sec 
sin 
tan  
cos 
cos 
tan  
sin 
sin  
The expression for cos  is ***.
____
a.
____
b.
____
____
Mathematics C30
a.
b.
c.
d.
c.
d.
a
b
B
b
a
b  a2
2
A
b
a

b
C
b2  a 2
b
56
Lesson 1
9.
10.
The expression for tan is ***.
____
a.
____
b.
____
c.
____
d.
4
4
5
4
3
3
4

5
4
The terminal arm of an angle  in standard position passes through
P(2, 5). The expression for the value of sec  is ***.
____
a.
____
b.
____
c.
____
d.
29
2
21
2
2
29
5
21
11.
For the following triangle, given the length of the sides in cm, the
value of  is ***.
____
____
____
____
12.
5 .0 1 8 9
58.3°
31.7°
27.7°
62.3°

9 .5 6 1 2
The value of cos 79.3° to four decimal places is ***.
____
____
____
____
Mathematics C30
a.
b.
c.
d.
a.
b.
c.
d.
0.1857
0.9826
5.2924
0.1736
57
Lesson 1
13.
The value of csc 75° to four decimal places is ***.
____
____
____
____
14.
a.
b.
c.
d.
46.7°
20°
70°
170°
a.
b.
c.
d.
0.9962
0.0875
0.0872
0.2791
The expression sec 0  is equal to ***.
____
____
____
____
Mathematics C30
47.3°
42.7°
36.3°
53.7°
If tan  = 11.4301, then cos  is equal to ***.
____
____
____
____
17.
a.
b.
c.
d.
One solution to sec  = 1.0642 is ***.
____
____
____
____
16.
0.9659
0.2588
3.8637
1.0353
One solution to sin  = 0.7350 is ***.
____
____
____
____
15.
a.
b.
c.
d.
a.
b.
c.
d.
0
1
2
undefined
58
Lesson 1
18.
The expression cot 90° is equal to ***.
____
____
____
____
19.
90°
x
a.
b.
c.
d.
10°, 170°, 190°, 350°
10 ,  10 ,  170 , 170 
10 , 80 , 100 , 170 
10 , 20 , 30 , 40 
The one true statement about two solutions to sin  
____
____
____
____
Mathematics C30
(0, 1) = (x, y)
0
1
2
undefined
The measures of four positive angles, each with a reference angle of
10°, are ***.
____
____
____
____
20.
a.
b.
c.
d.
y
a.
b.
c.
d.
1
is ***.
2
their principal angles are equal
the angles are in the first and fourth quadrants
the angles are in the same quadrant
the reference angles have the same measure
59
Lesson 1
Mathematics C30
60
Lesson 1
Answer Part B and Part C in the space provided. Evaluation of your solution
to each problem will be based on the following criteria.
B.
(8)
•
A correct mathematical method for solving the problem is shown.
•
The final answer is accurate, and a check of the answer is shown
where asked for by the question.
•
The solution is written in a style that is clear, logical, well-organized,
uses proper terms, and states a conclusion.
Answer in the space provided. Show the calculations and the final solution in
a well organized form. Partial marks will be given for correct methods.
1.
Mathematics C30
Evaluate by giving the value in simplest exact form.
a.
5 sec 30  tan 60   sin 90 
b.
1  tan 2 45   sec 2 45 
c.
sin 2 30   sin 2 60 
d.
cos 0   sin 270   sec 180 
61
Lesson 1
(8)
2.
Find the exact values of the six trigonometric functions of the angle A
whose terminal ray passes through 6,  8 . Include a labeled diagram
of the angle with the reference triangle.
(8)
3.
Determine the quadrant(s) that angle A lies in for each of the
conditions. Include the method of arriving at your answer.
Mathematics C30
a.
sin A < 0 and cos A < 0
b.
cos A > 0 and tan A < 0
c.
cot A csc A   0
d.
tan A sec A   0
62
Lesson 1
(8)
4.
4
, give the exact
5
values of each of the other trigonometric ratios. A diagram of the
angles and
For each principal angle that satisfies sin A  
labeled reference triangles is to be included.
(5)
5.
Mathematics C30
Determine the coordinates of a point on the terminal ray of an angle
 given
a.
cos 180  1
b.
sin 60  
3
2
63
Lesson 1
(8)
6.
Mathematics C30
Express each as a trigonometric function of the reference angle and
evaluate. Give the exact value.
a.
sin  225 
b.
cos 780 
c.
tan 150 
d.
csc 300 
64
Lesson 1
(5)
C.
1.
In this lesson you learned that a trigonometric function of any angle is
equal to the same trigonometric function of the reference angle with
the correct + or – sign, depending on the quadrant of A.
Reference angles are defined to be positive and are between 0° and 90°
in measure.
In fact it is true that a trigonometric function of any non-quadrantal
angle is equal to some other trigonometric function of an angle
between 0° and 45° with the appropriate + or – signs.
Example
In the triangle to the right,
b
. What other trig
c
b
function has the ratio ? Using
c
60  , c is the hypotenuse and b is the adjacent side;
b
adj
b
cos 60  
 . Therefore sin 30    cos 60  .
c
hyp
c
sin 30  
(Since 30° and 60° are in quadrant 1, both sine and cosine are positive.)
Test this statement by expressing
i)
sin 70°
ii)
sin 130°
iii)
cos  250 
iv)
tan 120 
as some trigonometric function of an angle between 0° and 45°.
Make use of the information given in this triangle.
B
9
0
°–
A
Mathematics C30

C
65
Lesson 1
(10)
2.
Complete this question on a separate page apart from the main
assignment.
a.
Write a point form summary of this lesson. In this summary
include a list of definitions, formulas, techniques, and examples
which should help you review for exams.
b.
Write a brief paragraph introducing yourself to your instructor.
100 
Mathematics C30
66
Lesson 1
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