Download Trigonometric Functions of Any Angle

Trigonometry
Module T12
Trigonometric
Functions of Any
Angle
Copyright
This publication © The Northern
Alberta Institute of Technology
2002. All Rights Reserved.
LAST REVISED December, 2008
Trigonometric Functions of Any Angle
Statement of Prerequisite Skills
Complete all previous TLM modules before completing this module.
Required Supporting Materials
Access to the World Wide Web.
Internet Explorer 5.5 or greater.
Macromedia Flash Player.
Rationale
Why is it important for you to learn this material?
When exploring introductory trigonometry concepts the learner is generally exposed to
angles that are less than 90º. This keeps things simple and allows the learner to
concentrate on the topic. The learner will encounter many situations involving angles
greater than 90º in applied situations. This module will provide the guidance necessary to
apply the trigonometry skills that have been learned to angles greater than 90º.
Learning Outcome
When you complete this module you will be able to…
Evaluate trigonometric functions of any angle.
Learning Objectives
1. Determine whether the value of a given trigonometric function is positive or
negative.
2. Determine the reference angle of a given angle.
3. Determine the six trigonometric function values for any angle in standard position
when the coordinates of a point on the terminal side are given.
4. Evaluate trigonometric functions of any angle.
5. Evaluate inverse trigonometric functions.
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Module T12 − Trigonometric Functions of Any Angle
Connection Activity
Consider the following diagram.
Given values for x and y youu are able to figure out the trigonometric ratios of α. What is
your estimate of the angle represented by θ? Is it 135º? Could it be 495º? Could it be
−205º? Without some indicator as to how many times r rotated around the origin we do
not know the measure of the angle. What we do know is that the trigonometric ratios
have not changed no matter which measure θ turns out to be. We can see that the
trigonometric ratios for θ and α will be true for many different angles. This module will
help you understand this concept and apply it to several situations.
Y
P(x,y)
y
r
x
α
Note: θ is measured
in standard position
θ
X
2
Module T12 − Trigonometric Functions of Any Angle
OBJECTIVE ONE
When you complete this objective you will be able to…
Determine whether the value of a given trigonometric function is positive or negative.
Exploration Activity
REVIEW
The study of earlier modules has introduced one set of definitions of trigonometric
functions:
METHOD 1: Triangle Method
The following are the six basic trigonometric functions derived using the triangle
method.
sin θ =
opposite
hypotenuse
csc θ =
hypotenuse
opposite
hypotenuse
adjacent
cos θ =
hypotenuse
hypotenuse
sec θ =
adjacent
opposite
tan θ =
adjacent
adjacent
cot θ =
opposite
opposite
θ
adjacent
METHOD 2: Circle Method
Another way of defining trigonometric functions is the circle method.
For the circle method remember that:
a) Positive angles rotate counterclockwise.
b) Negative angles rotate clockwise.
The following are the six basic trigonometric functions derived using the circle method.
sin θ =
csc θ =
r
y
x
cos θ =
r
r
sec θ =
x
y
x
x
cot θ =
y
tan θ =
•
y
r
Y
P(x,y)
r
θ
y
X
x
Where r represents the radius of the circle
and is always a positive value.
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Module T12 − Trigonometric Functions of Any Angle
USING THE CIRCLE METHOD IN ALL QUADRANTS
When using the circle method of defining trigonometric functions, the algebraic sign of a
trigonometric function may be determined by noting the quadrant which contains the
terminal side of angle θ (or the point P (x,y)).
Y
Quadrant 1:
x is positive (x > 0)
y is positive (y > 0)
r is positive (r is always positive)
r
θ
Therefore, all trigonometric functions of θ in
quadrant 1 will be positive.
Quadrant 2:
x is negative (x < 0)
y is positive (y > 0)
r is positive
Therefore, cos θ, sec θ, tan θ and cot θ will
have negative values for quadrant 2 angles.
P(x,y)
y
X
x
Y
P(x,y)
y
r
Note: θ is measured
in standard position
θ
x
sin θ and csc θ will have positive values.
Example:
Determine the algebraic sign of cos 100º.
x
r
negative value
=
= negative value
positive value
cos 100º =
cos 100º = negative value
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Module T12 − Trigonometric Functions of Any Angle
X
Y
Quadrant 3:
x is negative (x < 0)
y is negative (y < 0)
r is positive
Therefore, sin θ, csc θ, cos θ, and sec θ will
have negative values for quadrant 3 angles.
x
θ
X
r
tan θ and cot θ will have positive values.
y
P(x,y)
Example:
Determine the algebraic sign of sin 210º.
sin 210º =
y
negative value
=
= negative
r
positive value
sin 210º = negative value
Quadrant 4:
x is positive (x > 0)
y is negative (y < 0)
r is positive
Y
θ
x
Therefore, sin θ, csc θ, tan θ, and cot θ will have
negative values for quadrant 4 angles.
cos θ and sec θ will have positive values.
X
r
y
P(x,y)
Example:
Determine the algebraic sign of sec 315º.
sec 315º =
r
positive value
=
= positive
x
positive value
sec 315º = positive value
5
Module T12 − Trigonometric Functions of Any Angle
CAST SYSTEM
The CAST system can be used for quick recall of algebraic signs of primary
trigonometric functions.
Y
CAST
C is cosine
A is all
S is sine
T is tangent
Quadrant II
Quadrant I
S
A
sine is positive
all functions positive
X
Quadrant III
T
tangent is positive
Quadrant IV
C
cosine is positive
Note: Knowing the CAST rule will be very useful when determining inverse trig
functions (of any angle) later in this module.
Note: The cosecant, secant, and cotangent functions have the same algebraic signs as
their reciprocals.
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Module T12 − Trigonometric Functions of Any Angle
Experiential Activity One
Enter positive, negative, undefined, or 0 for the following questions:
1. Determine the algebraic sign of the following expressions:
a) sin 160º
___________________
b) cos( –120º)
___________________
5
tan π
4
___________________
c)
d) csc 310º
___________________
7
e) sec − π
4
___________________ Show Me.
f) cot 170º
___________________
2
g) tan − π
3
___________________
h) sin
3
π
4
___________________
2. Identify the quadrant(s) in which θ is located for each of the following
conditions:
a) sin θ is positive
__________________
b) cos θ is positive
__________________
c) sin θ is negative
__________________
d) tan θ is negative
__________________
e) cos θ is negative
__________________
f) sin θ is positive, cos θ is negative __________________
g) tan θ and sin θ both positive
__________________
h) cot θ negative, cos θ negative
__________________ Show Me.
i)
tan θ negative, cos θ positive
j)
all trigonometric functions of θ are positive
__________________
__________________
Experiential Activity One Answers
1. a) positive
b) negative
e) positive
f) negative
b) 1,4
c) 3,4
2. a) 1, 2
f) 2
g) 1
h) 2
c) positive
g) positive
d) 2,4
e) 2,3
i) 4
j) 1
d) negative
h) positive
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Module T12 − Trigonometric Functions of Any Angle
OBJECTIVE TWO
When you complete this objective you will be able to…
Determine the reference angle of a given angle.
Exploration Activity
DEFINITION OF REFERENCE ANGLE
The reference angle of a given angle is the positive acute angle formed by the terminal
side of the given angle and the x-axis.
In this book we will use the greek letter alpha (α) to represent the reference angle and the
greek letter theta (θ) to represent angles measured in standard position.
Example 1:
Determine the reference angle of 150º.
The angle between P(x,y) and the x axis is 30º
Y
P(x,y)
θ = 150º
α=
α = 180º − θ
α = 180º − 150º
α = 30º
X
Therefore, 30º is the reference angle of 150º
Y
Example 2:
Determine the reference angle of 260º.
θ = 260º
The angle between P(x,y) and the x axis is 80º
α = θ − 180º
α = 260º − 180º
α = 80º
X
α=
Therefore, 80º is the reference angle of 260º
P(x,y)
NOTE:
1. The reference angle is always positive.
2. The reference angle is always acute.
3. The reference angle is always measured between the terminal arm of the angle and
the nearest x-axis.
4. The trigonometric ratio of α is the absolute value of the trigonometric ratio of θ.
8
Module T12 − Trigonometric Functions of Any Angle
Experiential Activity Two
1. Determine the reference angle of the following:
Angle
Reference Angle
a) 114º
b)
_____________
3π
4
_____________
c) 473º
_____________
d) –212º
_____________
e)
−5π
4
_____________
f) 87º
_____________
g) 320º
_____________
h) –315º
_____________
i)
−8π
7
_____________
j)
−7π
3
_____________
Show Me.
2. Determine the reference angles (α) of the following angles measured in standard
position (θ) and then draw and label both α and θ on the diagram provided.
a) θ = 165º
α = _____
b) θ = −125º
α = _____
Y
Y
X
X
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Module T12 − Trigonometric Functions of Any Angle
c) θ = 305º
α = _____
d) θ = −285º
α = _____
Y
Y
X
10
Module T12 − Trigonometric Functions of Any Angle
X
Experiential Activity Two Answers
1. a) 66º
b) π/4
g) 40º
h) 45º
2.
a) θ = 165º
α = 15º
c) 67º
i) π/7
d) 32º
j) π/3
e) π/4
f) 87º
b) θ = −125º
α = 55º
Y
Y
θ
α
X
X
α
c) θ = 305º
α = 55º
θ
d) θ = −285º
α = 75º
Y
Y
θ
α
X
α
X
θ
11
Module T12 − Trigonometric Functions of Any Angle
OBJECTIVE THREE
When you complete this objective you will be able to…
Determine the six trigonometric function values for any angle in standard position when
the coordinates of a point on the terminal side are given.
Exploration Activity
FOUR STEPS
There are four steps needed to determine the trigonometric function values (ratios) when
the coordinates of a point on the terminal side of the angle are given.
Step 1:
Draw a right triangle using the terminal arm and the nearest x-axis.
Y
P(−3,4)
Example:
P(−3,4)
θ
α
X
Step 2:
Use the point to label the sides.
P(−3,4)
Y
Example:
4
θ
α
−3
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Module T12 − Trigonometric Functions of Any Angle
X
Step 3:
Determine the length of the hypotenuse using the Pythagorean theorem.
P(-3,4)
Example:
c2 = a2 + b2
2
c 2 = (− 3) + 4 2
Y
Hypotenuse
5
4
θ
c 2 = 25
c=5
α
X
−3
Step 4:
Identify the hypotenuse, the side opposite angle α and the side adjacent to angle α. Use
the sides to determine the six trigonometric ratios.
P(−3,4)
Example:
5
4
Opposite
Y
Hypotenuse
θ
α
Adjacent
X
−3
Since the reference angle (α) and the angle measured in standard position (θ) have the
same trigonometric ratios, we can now determine the six trigonometric ratios of θ.
Six Trig Ratios in exact form
opp
adj
opp
sin θ =
cos θ =
tan θ =
hyp
hyp
adj
4
−3
4
sin θ =
cos θ =
tan θ =
5
5
−3
hyp
opp
5
csc θ =
4
csc θ =
hyp
adj
5
sec θ =
−3
sec θ =
adj
opp
−3
cot θ =
4
cot θ =
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Module T12 − Trigonometric Functions of Any Angle
We can now compare the CAST rule to trigonometric ratios of θ.
Trig.
Functions
Signs of trig. ratios of θ
in Quadrant 2 according
to the CAST Rule
sin
Positive
cos
Negative
tan
Negative
csc
Positive
sec
Negative
cot
Negative
Calculated ratio in
exact form
Signs of calculated
trig. ratios of θ in
Quadrant 2
4
5
3
cos θ = −
5
4
tan θ = −
3
5
cscθ =
4
5
secθ = −
3
3
cot θ = −
4
sin θ =
Positive
Negative
Negative
Positive
Negative
Negative
The signs of the trigonometric ratios we calculated for θ in quadrant 2 are consistent with
the CAST rule.
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Module T12 − Trigonometric Functions of Any Angle
Experiential Activity Three
1. Find the six trigonometric ratios for the following angles given a point on the
terminal side.
b) (−7.2, −4.1)
a) (−5,8)
Leave your answers in exact form.
Round your answers to 4 significant
digits.
Y
Y
X
X
2. Find all six trigonometric ratios for the angles that have terminal arms passing
through the points given below.
a) (4,6) (Leave your answers in exact form.)
b) (0.8,−6.7) (Round your answers to 4 decimal places.)
3. Give the value of sin A and cos A for the angles, which have a terminal arm that
passes through the points. (Round your answers to 4 decimal places.)
a) (3.0,−4.2)
b) (−2.45,−7.32) Show Me.
c) (−4.82,1.83)
Experiential Activity Three Answers
1. a) sin θ =
8
89
, cos θ =
−5
89
, tan θ =
−8
5
−5
89
− 89
, sec θ =
, cot θ =
8
5
8
b) sin θ = −0.4948, cos θ = −0.8690, tan θ = 0.5694
csc θ = −2.0209, sec θ = −1.15081, cot θ =1.7561
6
4
3
52
2. a) sin θ =
, cos θ =
, tan θ = , csc θ =
, sec θ =
2
6
52
52
b) sin θ = −0.9929, cos θ = 0.1186, tan θ = −8.3750
csc θ = −1.0071, sec θ = 8.4345, cot θ = −0.1194
3. a) sin A = −0.8137, cos A = 0.5812
b) sin A = −0.9483, cos A = −0.3174
c) sin A = 0.3549, cos A = −0.9349
csc θ =
2
52
, cot θ =
4
3
15
Module T12 − Trigonometric Functions of Any Angle
OBJECTIVE FOUR
When you complete this objective you will be able to…
Evaluate trigonometric functions of any angle.
Exploration Activity
Calculator Comment
In the pre-calculator era, evaluating trigonometric functions of angles greater than 90°
required the use of reference angles. Using calculators, the 'reference angle' step is
usually left out, however, it is still necessary to use the reference angle when evaluating
inverse trigonometric functions (the next objective).
DETERMINING THE TRIGONOMETRIC RATIOS OF ANY ANGLE.
Acute Angles
In module 7 we evaluated trigonometric ratios of acute angles
Example 1:
Evaluate tan 35º
Ensure your calculator
is in degree mode
To evaluate tan 35º simply enter tan 35º into your calculator and press the equal key.
tan 35º = 0.7002
Example 2:
Evaluate csc 62º
1
csc 62º =
sin 62°
1
csc 62º =
0.8829
csc 62º = 1.1326
Note: Recall from the CAST rule that all acute angles have positive trigonometric
ratios.
16
Module T12 − Trigonometric Functions of Any Angle
Trig Ratios of any Angle
Evaluate trig ratios of any angle in the same manner you evaluate acute angles.
Example 1:
Evaluate cos 145º
To evaluate cos 145º simply enter cos 145º into your calculator and press the equal key.
cos 145º = -0.8192
Note: Notice that the trigonometric ratio for cos 145º is negative. 145º has a terminal
arm in quadrant 2, and the CAST rule tells us that the trigonometric ratio of cosine is
negative in quadrant 2.
Example 2:
Evaluate cot 265º
1
cot 265º =
tan 265°
1
cot 265º =
11.4301
cot 265º = 0.0875
Note: Notice that the trig ratio for cot 265º is positive. 265º has a terminal arm in
quadrant 3, and the CAST rule tells us that the trigonometric ratios for tangent and
cotangent are positive in quadrant 3.
Trig Ratios of References Angles
We know that references angles are always acute and therefore the trigonometric ratio of
any reference angle will always be positive.
How do the trigonometric ratios of reference angles relate to the trigonometric ratios of
angles measured in standard position?
In the diagram to the right θ = 150º and its reference
angle is 30º
Since both angles share the same terminal arm they
have the same trigonometric ratios. However, the
trigonometric ratio of the reference angle is always
positive because it is acute and we use the CAST
rule to determine the sign of the trigonometric ratio
of θ.
Trig. ratios for α:
Trig. ratios for θ
sin 30° = 0.500
sin 150° = 0.500
Y
θ = 150º
α = 30º
cos 30° = 0.8660
cos 150° = −0.8660
X
tan 30° = 0.5774
tan 150° = −0.5774
The terminal arm of 150º is in quadrant 2. The CAST rule tells us that the trigonometric
ratio of sine will be positive, tangent will be negative, and cosine will be negative.
17
Module T12 − Trigonometric Functions of Any Angle
Experiential Activity Four
1. Find the reference angle, and the sine, cosine, tangent, cotangent, cosecant
and secant of the following (If undefined, answer undefined):
Angle
(θ )
a) 107.0º
Ref. ∠
(α )
sin θ
cos θ
tan θ
cot θ
csc θ
sec θ
b) 147.5º
c) 183.0º
d) 180.9º
e) 208.0º
f) 349.9º
g) 461.0º
h) 539.3º
i) 905.0º
j) –17.1º
k) 940.7º
l) –362.6º
m) 1260.2º
Note the following angles are measured in radians. Ensure your calculator is in radian
mode to proceed.
n) 1.4
o) 6.2
p) 0.65
q) 1.6
r) –1.4
s) –4.9
18
Module T12 − Trigonometric Functions of Any Angle
Experiential Activity Four Answers
Angle
(θ )
a) 107.0º
Ref. ∠
(α )
73.0º
sin θ
cos θ
tan θ
cot θ
csc θ
sec θ
0.9563
–0.2924
–3.2709
–0.3057
1.0457
–3.4203
b) 147.5º
32.5º
0.5373
−0.8434
−0.6371
−1.5697
1.8612
−1.1857
c) 183.0º
3.0º
–0.0523
–0.9986
0.0524
19.0811
–19.1073
–1.0014
d) 180.9º
0.9º
–0.0157
–0.9999
0.0157
63.6567
–63.6646
–1.0001
e) 208.0º
28.0º
–0.4695
–0.8829
0.5317
1.8807
–2.1301
–1.1326
f) 349.9º
10.1º
–0.1754
0.9845
–0.1781
–5.6140
–5.7023
1.0157
g) 461.0º
79.0º
0.9816
–0.1908
–5.1446
1.0187
–5.2408
h) 539.3º
0.7º
0.0122
–0.9999
–0.0122
81.8531
–1.0001
i) 905.0º
5.0º
–0.0872
–0.9962
0.0875
–0.1944
–
81.8470
11.4301
–11.4737
–1.0038
j) –17.1º
17.1º
–0.2940
0.9558
–0.3076
–3.2506
–3.4009
1.0463
k) 940.7º
40.7º
–0.6521
–0.7581
0.8601
–1.5335
–1.3190
l) –362.6º
2.6º
–0.0454
0.9990
-0.0454
–22.0444
1.0010
m) 1260.2º
0.2º
–0.0035
–1.0000
0.0035
n) 1.4
1.4
0.9854
0.1700
5.7979
−286.479
5
1.0148
o) 6.2
0.0832
−0.0831
0.9965
−0.0834
p) 0.65
0.65
0.6052
0.7961
q) 1.6
1.5416
0.9996
−0.0292
r) −1.4
1.4
−0.9854
0.1700
0.7602
−34.232
5
−5.7979
1.1626
–
22.0217
286.477
7
0.1725
−11.993
6
1.3154
s) −4.9
1.3832
0.9825
0.1865
5.2675
–1.0000
5.8835
−12.0352
1.0035
1.6524
1.2561
−0.0292
1.0004
−34.2471
−0.1725
−1.0148
5.8835
0.1898
1.0179
5.3616
19
Module T12 − Trigonometric Functions of Any Angle
OBJECTIVE FIVE
When you complete this objective you will be able to…
Evaluate inverse trigonometric functions.
Exploration Activity
CONDITIONS FOR EVALUATING INVERSE TRIGONOMETRIC FUNCTIONS
We have seen that evaluating expressions like sin 85º results in one answer.
ie: sin 85º = 0.9962
However, if we wish to find the angle θ that has a sine equal to 0.9962 then we would
write:
sin θ = 0.9962
Solving for θ we would get:
θ = arc sin 0.9962
θ = 85º
or
θ = sin−1 0.9962
θ = 85º
This method works well when all of the angles for θ are acute. However, there are many
other values of θ that also have a sine equal to 0.9962.
Example:
According to the CAST rule, sine is positive in quadrant 2 as well. The following circle
diagram show that sin 85º and any angle with a reference angle of 85º in quadrant 2 will
have the same trigonometric ratio.
Y
85º is in quadrant 1
sin 85º = 0.9962
θ = 85º
X
Y
95º is in quadrant 2
sin 95º = 0.9962
The reference angle of 95º is 85º.
α = 85º
θ = 95º
The ratio is positive because sine is
positive in quadrant 2
20
Module T12 − Trigonometric Functions of Any Angle
X
265º is in quadrant 3
sin 265º = −0.9962
Y
The reference angle of 265º is 85º.
θ = 265º
The ratio is negative because sine is negative in
quadrant 3
Although 265º has a reference angle of 85º it is not a
solution because the sine of 265º is negative.
X
α = 85º
275º is in quadrant 4
sin 275º = −0.9962
The reference angle of 275º is 85º.
Y
θ = 275º
The ratio is negative because sine is negative in
quadrant 4
Although 275º has a reference angle of 85º it is not a
solution because the sine of 275º is negative.
X
α = 85º
Summary
Any angle that has a reference angle of 85º and is in quadrants 1 and 2 will have a sine of
positive 0.9962. In the example the solutions were 85º and 95º.
However, any angle coterminal with 85º and 95º will also be solutions:
445º and −275º are all coterminal with 85º
455º and −265º are all coterminal with 95º
Therefore, sin 445º = sin −275º = sin 455º = sin −265º
All of these evaluate to 0.9962.
To get around this problem of many possible answers, we will include a condition with
each question.
0º ≤ θ < 360º
This condition indicates that the answer for θ must be greater than or equal to 0º and less
than 360º.
Conclusion
Find θ for sin θ = 0.9962 given 0º ≤ θ < 360º
With this condition, the values of θ for the question sin θ = 0.9962 would be 85º and 95º
only.
Other angles coterminal with 85º and 95º would not satisfy the condition of
0º ≤ θ < 360º
21
Module T12 − Trigonometric Functions of Any Angle
EVALUATING INVERSE TRIGONOMETRIC FUNCTIONS
There are three steps needed to evaluate inverse trigonometric functions.
We will evaluate the following question to demonstrate the three steps.
Find the value θ for:
sin θ = 0.5976 given 0º ≤ θ < 360º
Step 1: Determine the quadrant(s) θ will be in given the sign of the trig ratio and the
condition.
The condition 0º ≤ θ < 360º indicates that θ must be greater than or equal to 0º and less
than 360º.
The trigonometric ratio of sine of θ is positive. The CAST rule shows that sine is
positive only in quadrants 1 and 2.
Y
Quadrant II
Quadrant I
Sine Positive
All Positive
X
Quadrant III
Quadrant IV
Tangent Positive
Cosine Positive
Therefore, we will have two solutions for θ. One solution in quadrant 1 and one solution
in quadrant 2.
Step 2: Determine the reference angle.
The trig ratio of the reference angle α is always the absolute value of the trig ratio of θ.
sin θ = 0.5976
sin α = 0.5976
solving for α
α = arc sin 0.5976
α = 36.7º
or
α = sin−1 0.5976
Step 3: Use the reference angle to determine θ.
From step 1 we know we will have answers in quadrants 1 and 2.
Answer in Quadrant 1:
The reference angle α and angle θ are the same in quadrant 1
Therefore, one solution is θ =36.7º
Check: Enter sin 36.7º into your calculator to check.
22
Module T12 − Trigonometric Functions of Any Angle
Answer in Quadrant 2:
We know that the reference angle α is 36.7º. We need to find the angle θ in quadrant 2
that has a reference angle of 36.7º.
Y
If we graph the reference angle in quadrant 2
we can determine angle θ.
θ = 143.3º
From the graph we can see that the sum of α
and θ in quadrant 2 =180º.
Therefore, θ = 180º − α.
θ = 180º − 36.7º
θ = 143.30º
α = 36.7º
X
Check: Enter sin 143.3º into your calculator to check.
Thus, θ1 = 36.7º and θ2 = 143.3º
It is sometimes desirable to find the angle in radians. Again your calculator will do this
operation as long as you change it to radian mode.
DETERMINING θ GIVEN THE REFERENCE ANGLE.
In step three above, once you have determined the reference angle you will need to
determine angle θ using the reference angle. The following diagram will help you when
determining angle θ from the reference angle α in any quadrant.
Y
CAST
C is cosine
A is all
S is sine
T is tangent
Quadrant II
Quadrant I
S
A
θ = 180º − α
θ=α
X
Quadrant III
Quadrant IV
T
C
θ = 180º + α
θ = 360º − α
Calculator comment
Most calculators will not give more than one answer for an inverse function. Be sure not
to just copy the calculator answer. There usually are more answers, so think it through.
23
Module T12 − Trigonometric Functions of Any Angle
MORE EXAMPLES
Example 1:
Find the value θ given 0 ≤ θ < 2π
cos θ = 0.9690
Step 1: Determine the quadrant(s) θ will be in.
The cosine of θ is positive. According to the CAST rule cosine is positive in quadrants
1 and 4.
The condition 0 ≤ θ < 2π indicates that θ must be greater than or equal to 0 radians and
less than 2π radians.
There will be two solutions for θ.
One solution in quadrant 1 and one solution in quadrant 4.
Step 2: Determine the reference angle α. (Make sure your calculator is in radian mode.)
cos θ = 0.9690
cos α = 0.9690
α = cos−1 0.9690
α = 0.2496
Step 3: Use the reference angle to determine θ.
From step 1 we know we need solutions in quadrants 1 and 4.
It will help to draw the angles to determine θ.
Answer in Quadrant 1:
θ=α
θ = 0.2496
Check: cos 0.2496 = 0.9690
Answer in Quadrant 4:
θ = 2π − α
θ = 2π − 0.2496
θ = 6.0336
Y
θ
Check: cos 6.0336 = 0.9691 (off slightly due to
rounding)
Thus θ1 = 0.2496 and θ2 = 6.0336
24
Module T12 − Trigonometric Functions of Any Angle
X
α = 0.2496
Example 2:
Find the value θ given 0º ≤ θ < 360º
sec θ = 5.2408
Step 1: Determine the quadrant(s) θ will be in.
The secant of θ is positive. According to the CAST rule secant is positive in quadrants 1
and 4 since it is the reciprocal of cosine.
The condition 0º ≤ θ < 360º indicates that θ must be greater than or equal to 0º and less
than 360º.
There will be two solutions for θ.
One solution in quadrant 1 and one solution in quadrant 4.
Step 2: Determine the reference angle α.
sec θ = 5.2408
sec α = 5.2408
Since calculators do not have sec functions, it is necessary to first convert this to the
cosine function.
1
1
=
cos α =
sec θ
5.2408
⎛ 1 ⎞
α = cos−1 ⎜
⎟
⎝ 5.2408 ⎠
α = 79.0º
Step 3: Use the reference angle to determine θ.
From step 1 we know we need solutions in quadrants 1 and 4.
It will help to draw the angles to determine θ.
Answer in Quadrant 1:
θ=α
θ = 79.0º
1
Check: sec 79.0º =
= 5.2408
cos 79.0°
Answer in Quadrant 4:
θ = 360º − α
θ = 360º − 79.0º
θ = 281.0º
Check: sec 281.0º =
Y
θ
1
= 5.2408
cos 281.0°
Thus θ1 = 79.0º and θ2 = 281.0º
X
α = 79.0º
25
Module T12 − Trigonometric Functions of Any Angle
Example 3:
Find the value θ given 0º ≤ θ < 360º
sin θ = −0.3616
Step 1: Determine the quadrant(s) θ will be in.
The sine of θ is negative. According to the CAST rule sine is negative in quadrants 3
and 4.
The condition 0º ≤ θ < 360º indicates that θ must be greater than or equal to 0º and less
than 360º.
There will be two solutions for θ.
One solution in quadrant 3 and one solution in quadrant 4.
Step 2: Determine the reference angle α.
sin θ = −0.3616
The trigonometric ratio of the reference angle α is
the absolute value of the trigonometric ratio of θ.
sin α = 0.3616
α = sin−1 0.3616
α = 21.2º
Step 3: Use the reference angle to determine θ.
Y
Answer in Quadrant 3:
θ = 180 + α
θ = 180º + 21.2º
θ = 201.2º
θ
X
Check: sin 201.2º = −0.3616
α = 21.2º
26
Module T12 − Trigonometric Functions of Any Angle
Y
Answer in Quadrant 4:
θ = 360º − α
θ = 360º − 21.2º
θ = 338.8º
θ
X
Check: sin 338.8º = −0.3616
α = 21.2º
Thus θ1 = 201.2º and θ2 = 338.8º
Example 4:
Find the value θ given 0º ≤ θ < 360º
csc θ = −1.7965
Step 1: Determine the quadrant(s) θ will be in.
The cosecant of θ is negative. According to the CAST rule cosecant is negative in
quadrants 3 and 4 since it is the reciprocal of sine.
The condition 0º ≤ θ < 360º indicates that θ must be greater than or equal to 0º and less
than 360º.
There will be two solutions for θ.
One solution in quadrant 3 and one solution in quadrant 4.
Step 2: Determine the reference angle α.
csc θ = −1.7965
csc α = 1.7965
convert to sine function
1
sin α =
1.7965
⎛ 1 ⎞
α = sin−1 ⎜
⎟
⎝ 1.7965 ⎠
α = 33.8º
The trigonometric ratio of the reference angle α is
the absolute value of the trigonometric ratio of θ.
Y
θ
X
α = 33.8º
27
Module T12 − Trigonometric Functions of Any Angle
Step 3: Use the reference angle to determine θ.
Answer in Quadrant 3:
θ = 180º + α
θ = 180º +33.8º
θ = 213.8º
Check: csc 213.8º =
1
= −1.7965
sin 213.8°
Answer in Quadrant 4:
θ = 360º − α
θ = 360º − 33.8º
θ = 326.2º
Y
θ
1
Check: csc 326.2º =
= −1.7965
sin 326.2°
Thus θ3 = 213.8º and θ4 = 326.2.2º
28
Module T12 − Trigonometric Functions of Any Angle
X
α = 33.8º
ALGEBRAIC SIGNS OF TRIGONOMETRIC FUNCTIONS OF QUADRANTAL
ANGLES
The last trigonometric functions that we’ll look at in this module are trig functions of
quadrantal angles.
Quadrantal Angle Definition
Angles with a terminal side on the x or y-axis are called quadrantal angles.
Example
90º, π, 270º, 2π, 450º, −90º, −π, and
−3π
2
Trigonometric Functions of Quadrantal Angles
The following demonstrates how to determine the trigonometric functions of quadrantal
angles less than 360º. The same rules apply for angles ≥ 360º. Your calculator will also
give the trigonometric functions of angles ≥ 360º by entering the angles directly.
29
Module T12 − Trigonometric Functions of Any Angle
Trigonometric functions of 90º
Let’s assume r = 1 remembering that r is always positive.
By using the circle method, we notice that as θ
approaches 90º, x approaches zero and y
approaches r.
Y
r
y
θ
x
Y
π
When θ = 90º & θ =
X
y=1
2
x=0
r = 1.
Since y falls on the positive y-axis y = 1.
r=1
θ
X
x=0
We can now use the six trig functions from the circle method and substitute in the values
π
and these concepts apply to radians as
for x, y, and r for 90º. Remember that 90º =
2
well as degrees.
sin θ =
y
r
sin 90º =
1
1
sin 90º = 1
sin
sin
π
2
π
2
=
r
csc θ =
1
1
cos θ =
y
csc 90º =
1
1
csc 90º = 1
csc
=1
csc
π
2
π
2
=
1
1
=1
x
sec θ =
r
cos 90º =
0
1
cos 90º = 0
cos
cos
π
2
π
2
=
0
1
=0
r
tan θ =
x
sec 90º =
1
0
sec 90º =
undefined
sec
sec
π
2
π
2
=
1
0
=
undefined
y
x
tan 90º =
cot θ =
1
0
tan 90º =
undefined
tan
tan
π
2
π
2
=
1
0
=
undefined
30
Module T12 − Trigonometric Functions of Any Angle
x
y
cot 90º =
0
1
cot 90º = 0
cot
cot
π
2
π
2
=
0
1
=0
TRIGONOMETRIC FUNCTIONS OF OTHER QUADRANTAL ANGLES.
θ = 180º & θ = π
Y
y=0
r = 1.
Since x falls on the negative xaxis x = −1.
θ
y=0
X
x = −1 r = 1
Using the six trig functions from the circle method and substituting in the values for x, y,
and r for 180º we get the following.
sin θ =
y
csc θ =
r
sin 180º =
0
1
sin 180º = 0
θ = 270º & θ =
r
cos θ =
y
csc 180º =
1
0
csc 180º =
undefined
cos π =
x
sec θ =
r
−1
r
sec 180º =
1
cos π = −1
tan θ =
x
1
−1
sec 180º = −1
y
x
tan 180º =
cot θ =
0
−1
tan 180º = 0
cot π =
x
y
−1
0
cot π =
undefined
Y
3π
2
θ
x=0
r = 1.
Since y falls on the negative
y-axis y = −1.
X
r=1
x=0
y = −1
Using the six trig functions from the circle method and substituting in the values for x, y,
and r for 270º we get the following.
sin θ =
y
r
csc θ =
r
y
−1
3π
=
sin
2
1
csc
1
3π
=
2
−1
3π
= −1
2
csc
3π
= −1
2
sin
cos θ =
x
sec θ =
r
cos 270º =
0
1
cos 270º = 0
r
x
1
3π
=
sec
2
0
sec
3π
=
2
tan θ =
y
x
tan 270º =
tan 270º =
undefined
cot θ =
−1
0
x
y
cot 270º =
0
−1
cot 270º = 0
undefined
31
Module T12 − Trigonometric Functions of Any Angle
QUADRANTAL ANGLES CONTINUED
θ = 0º or 360º & θ = 2π
Y
y=0
r = 1.
Since x falls on the positive xaxis x = 1.
θ
x=1
X
r=1
y=0
Using the six trig functions from the circle method and substituting in the values for x,
y, and r for 360º we get the following.
sin θ =
y
csc θ =
r
sin 360º =
0
1
sin 360º = 0
r
cos θ =
y
csc 360º =
csc 360º =
undefined
1
0
cos 2π =
x
r
1
1
cos 2π = 1
sec θ =
r
tan θ =
x
sec 360º =
1
1
sec 360º = 1
tan 2π =
y
x
0
1
tan 2π = 0
cot θ =
cot 2π =
x
y
1
0
cot 2π =
undefined
Trigonometric functions that are undefined
Any trigonometric definition that results in a denominator containing a zero (0) value will
produce an undefined trigonometric function at that particular angle.
Example:
csc 360º =
1
r
= undefined
=
y
0
Trigonometric functions that result in a zero (0) value
Any trigonometric definition that results in a zero numerator (denominator not = 0) will
produce a zero value.
Example:
tan 2π =
y
0
=
=0
x
1
Note: In cases where questions may ask you to indicate the algebraic sign of a
trigonometric function, if the answer is 0, enter 0, not positive or negative.
32
Module T12 − Trigonometric Functions of Any Angle
SUMMARY OF QUADRANTAL ANGLES
The following labeled diagram should be of assistance to you when you are solving
inverse function problems of quadrantal angles
Y
180º
90º
sin 180º = 0
csc 180º = undefined
cos 180º = −1 sec 180º = −1
tan 180º = 0
cot 180º = undefined
sin 90º = 1
cos 90º = 0
tan 90º = undefined
csc 90º = 1
sec 90º = undefined
cot 90º = 0
X
270º
sin 270º = −1
cos 270º = 0
tan 270º = undefined
csc 270º = −1
sec 270º = undefined
cot 270º = 0
0º and 360º
sin 0º = 0
cos 0º = 1
tan 0º = 0
csc 0º = undefined
sec 0º = 1
cot 0º = undefined
You could memorize the above relationships. They are not as difficult as they may
appear, i.e., note the pattern for sine as you go from 0º to 90º to 180º to 270º: sine goes
0, 1, 0, –1. Similar patterns exist for cosine and tangent.
CALCULATOR USAGE AND QUADRANTAL ANGLES
Use your calculator to evaluate the following and then compare your answers to the chart.
tan (−90º)
csc π
cot 270º
You should get an “ERROR” answer. Does this mean that every “ERROR” answer is
equivalent to an undefined trigonometric function?
No, it means that cot 270º and cot 90º cannot be evaluated using your calculator!!!
Why?
There isn’t a cot button on your calculator so we use the following relationship to
evaluate the cot 270º.
1
cot 270º =
tan 270°
1
1
evaluates as
= undefined.
Using the calculator
tan 270°
undefined
1
Since tan 270º = undefined, we cannot use
to evaluate cot 270º.
tan 270°
33
Module T12 − Trigonometric Functions of Any Angle
Experiential Activity Five
1. Find all the values of θ given 0º ≤ θ < 360º (nearest tenth of a degree)
a) sin θ = 0.9100
f) cos θ = 0.7070
g) sin θ = 1.0000
b) cos θ = 0.8200
c) tan θ = 0.4100
h) tan θ = 0.2500
d) sin θ = –0.4300
i) cos θ = –1.0000
e) tan θ = 1.2000
2. Find all the values of θ given 0 ≤ θ < 2π (answer to 4 decimals)
a) sin θ = 0.5821
f) cos θ = −0.0279
g) sin θ = 0.7804
b) cos θ = −0.4555
c) tan θ = −3.5105
h) cos θ = –0.9763
d) sin θ = –0.3778
e) tan θ = −1.0761 Show Me.
3. Complete the table for 0º ≤ θ < 360º.
Note: Unless θ is given, you will have two angles for θ and two values for any
trigonometric functions not given.
Find θ to the nearest tenth of a degree and other answers to 3 significant digits
Trigonometric ratios
to 3 significant digits.
cos θ
tan θ
sin θ
θ
150.0º
233.5º
345.0º
-0.460
-0.951
0.680
164.3º
0.968
0.166
−2.32
360.0º
34
Module T12 − Trigonometric Functions of Any Angle
4. Solve each of the following for A when 0º ≤ A < 360º.
Round answers to one decimal place.
(Each of these will have TWO answers; determine the quadrant in which the terminal
arm lies for those angles first.)
a) csc A = − 1.9084
Show Me.
b) cot A = 4.5067
c) tan A = − 0.8645
d) sec A = 2.1114
e) sec A = − 1.2352
f) sec A = −3.2633
g) cot A = − 4.5294
h) csc A = 1.0067
i) cos A = − 0.7654
Show Me.
j) sin A = −0.8789
k) cos A = 0.4563
l) tan A = 5.4000
5. Without using a calculator, give the trigonometric ratios of each of these
quadrantal angles.
Angle
sin A
cos A
tan A
csc A
sec A
cot A
0º
90º
180º
270º
35
Module T12 − Trigonometric Functions of Any Angle
Experiential Activity Five Answers
1. a) 65.5º, 114.5º
e) 50.2º, 230.2º
i) 180.0º
2. a) 0.6213, 2.5203
e) 2.3196, 5.4612
3.
θ
150.0º
233.5º
345.0º
207.4º
332.6º
162.0º
198.0º
34.2º
214.2º
164.3º
75.5º
104.5º
80.4º
279.6º
113.3º
293.3º
360.0º
b) 34.9º, 325.1º
f) 45.0º, 315.0º
c) 22.3º, 202.3º
g) 90.0º
d) 205.5º, 334.5º
h) 14.0º, 194.0º
b) 2.0437, 4.2395
f) 1.5987, 4.6845
c) 1.8483, 4.9899
g) 0.8953, 2.2463
d) 3.5290, 5.8958
h) 2.9234, 3.3597
sin θ
0.500
−0.804
−0.259
cos θ
−0.866
–0.595
0.966
–0.888
0.888
– 0.460
0.309
–0.309
0.562
–0.562
0.271
tan θ
−0.577
1.35
–0.268
0.518
–0.518
–0.325
0.325
– 0.951
0.827
–0.827
–0.963
0.251
–0.251
0.968
0.986
–0.986
0.918
–0.918
0
0.680
–0.281
3.86
–3.86
5.91
−5.91
0.166
–0.396
0.396
1
4.
a) A is in quad 3 and 4, A=211.6º, 328.4º,
c) A is in quad 2 and 4, A= 139.2º, 319.2º
e) A is in quad 2 and 3, A= 144.1º, 215.9º
g) A is in quad 2 and 4, A= 167.5º, 347.5º
i) A is in quad 2 and 3, A= 139.9º, 220.1º
k) A is in quad 1 and 4, A= 62.9º, 297.1º
–2.32
0
b) A is in quad 1 and 3, A= 12.5º, 192.5º
d) A is in quad 1 and 4, A= 61.7º, 298.3º
f) A is in quad 2 and 3, A= 107.8º,
252.2º
h) A is in quad 1 and 2, A= 83.4º, 96.6º
j) A is in quad 3 and 4, A= 241.5º,
298.5º
l) A is in quad 1 and 3, A= 79.5º, 259.5º
5.
Angle
sin A
cos A
tan A
csc A
sec A
cot A
0º
0
1
0
undefined
1
undefined
90º
1
0
undefined
1
undefined
0
180º
0
−1
0
undefined
−1
undefined
270º
−1
0
undefined
−1
undefined
0
36
Module T12 − Trigonometric Functions of Any Angle
Practical Application Activity
Complete the Trigonometric Functions of Any Angle assignment in TLM.
Summary
This module showed the student how to deal with the trigonometric functions of angles
that are greater than 90º.
37
Module T12 − Trigonometric Functions of Any Angle