Trigonometric Functions of Any...

LAST REVISED December, 2008

Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved.

Trigonometry Module T12

Trigonometric Functions of Any Angle

Module T12 − Trigonometric Functions of Any Angle

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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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Connection Activity Consider the following diagram. Given values for x and y you 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.

u 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.

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α α

P(x,y)

X

Y

θ

Note: θ is measured in standard position

y

x

r


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.

opposite

adjacent

hypotenuse

θ

sin θ = hypotenuse

opposite csc θ =

oppositehypotenuse

cos θ = hypotenuseadjacent

sec θ = adjacent

hypotenuse

tan θ = adjacentopposite

cot θ = oppositeadjacent

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 = ry

csc θ = yr

cos θ = rx

sec θ = xr

tan θ = xy

cot θ =

Y

x

y

X

P(x,y)

θ

r

3

yx

• Where r represents the radius of the circle

and is always a positive value.


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)).

Quadrant 1: x is positive (x > 0) y is positive (y > 0) r is positive (r is always positive) Therefore, all trigonometric functions of θ in quadrant 1 will be positive.

P(x,y)

X

Y

θ

y

x

r

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.

sin θ and csc θ will have positive values.

Example:

Determine the algebraic sign of cos 100º.

cos 100º =

4

xr

= negative evalupositive lueva

= negative value

cos 100º = negative value

P(x,y)

X

Y

θ

y

x

r

Note: θ is measured in standard position


P(x,y)

X

Y

θ

y

x

r

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. tan θ and cot θ will have positive values.

Example: Determine the algebraic sign of sin 210º.

sin 210º = yr

= negative valuepositive value

= negative

sin 210º = negative value

Quadrant 4: x is positive (x > 0) y is negative (y < 0) r is positive 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.

Example: Determine the algebraic sign of sec 315º.

sec 315º = xr = positive value

positive value = positive

sec 315º = positive value

P(x,y)

Y

θ

y

x

r

X

5


CAST SYSTEM The CAST system can be used for quick recall of algebraic signs of primary trigonometric functions.

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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.

X

Y

Quadrant I Quadrant II

Quadrant IV Quadrant III

T C

S A sine is positive all functions positive

tangent is positive cosine is positive

CAST C is cosine A is all S is sine T is tangent


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) ___________________

c) 5tan4

π ___________________

d) csc 310º ___________________

e) 7sec4

π− ___________________ Show Me.

f) cot 170º ___________________

g) 2tan3

π− ___________________

h) sin 34

π ___________________

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 c) positive d) negative

e) positive f) negative g) positive h) positive 2. a) 1, 2 b) 1,4 c) 3,4 d) 2,4 e) 2,3

f) 2 g) 1 h) 2 i) 4 j) 1

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OBJECTIVE TWO When you complete this objective you will be able to… Determine the reference angle of a given angle.


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 α = 180º − θ α = 180º − 150º α = 30º Therefore, 30º is the reference angle of 150º

Example 2: Determine the reference angle of 260º. The angle between P(x,y) and the x axis is 80º α = θ − 180º α = 260º − 180º α = 80º Therefore, 80º is the reference angle of 260º

P(x,y)

X

Y

θ = 150º α =

P(x,y)

X

Y

θ = 260º

α =

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.

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Experiential Activity Two 1. Determine the reference angle of the following:

Angle Reference Angle

a) 114º _____________

b) 34π _____________

c) 473º _____________

d) –212º _____________

e) 54π− _____________ Show Me.

f) 87º _____________

g) 320º _____________

h) –315º _____________

i) 87π− _____________

j) 73π− _____________

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 = _____

X

Y

X

Y

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c) θ = 305º α = _____

d) θ = −285º α = _____

X

Y

X

Y

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Experiential Activity Two Answers 1. a) 66º b) π/4 c) 67º d) 32º e) π/4 f) 87º

g) 40º h) 45º i) π/7 j) π/3 2.

a) θ = 165º α = 15º

b) θ = −125º α = 55º

c) θ = 305º α = 55º

d) θ = −285º α = 75º

X

Y

α θ

X

Y

α

θ

X

Y

α

θ

X

Y

α

θ

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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.


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.

Example: P(−3,4)

X

Y

θ

P(−3,4)

α

Step 2: Use the point to label the sides.

Example:

−3

X

YP(−3,4)

4

α

θ

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Step 3: Determine the length of the hypotenuse using the Pythagorean theorem.

Example: 222 bac +=

( ) 222 43 +−=c 252 =c

5=c

X

YP(-3,4)

−3

4

α

θ 5

Hypotenuse

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.

Example:

− 3

X

Y P( − 3,4)

4

α

θ 5

Hypotenuse

Adjacent

Opposite

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

hypopp

=θsin

54sin =θ

hypadj

=θcos

53cos −

=θ

adjopp

=θtan

4tan3

θ =−

opphyp

=θcsc

45csc =θ

adjhyp

=θsec

35sec

−=θ

oppadj

=θcot

43cot −

=θ

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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

Calculated ratio in exact form

Signs of calculated trig. ratios of θ in

Quadrant 2

sin Positive 4sin5

θ = Positive

cos Negative 3cos5

θ = − Negative

tan Negative 4tan3

θ = − Negative

csc Positive 5csc4

θ = Positive

sec Negative 5sec3

θ = − Negative

cot Negative 3cot4

θ = − Negative

The signs of the trigonometric ratios we calculated for θ in quadrant 2 are consistent with the CAST rule.

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Experiential Activity Three 1. Find the six trigonometric ratios for the following angles given a point on the

terminal side. a) (−5,8)

Leave your answers in exact form.

b) (−7.2, −4.1) Round your answers to 4 significant digits.

X

Y

X

Y

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 = 898 , cos θ =

895− , tan θ =

58−

csc θ = 889 , sec θ =

589− , cot θ =

85−

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

2. a) sin θ = 526 , cos θ =

524 , tan θ =

23 , csc θ =

652 , sec θ =

452 , cot θ =

32

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

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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

Ensure your calculator is in degree mode Example 1:

Evaluate tan 35º 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

csc 62º = °62sin

1

csc 62º = 8829.01

csc 62º = 1.1326 Note: Recall from the CAST rule that all acute angles have positive trigonometric ratios.

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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

cot 265º = °265tan

1

cot 265º = 4301.111

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.

Y

θ = 150º

X α = 30º

Trig. ratios for α: 500.030sin =° 8660.030cos =° 5774.030tan =° Trig. ratios for θ 500.0150sin =° 8660.0150cos −=° 5774.0150tan −=° 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.

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Module T12 − Trigonometric Functions of Any Angle 18

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) Ref. ∠

(α) sin θ cos θ tan θ cot θ csc θ sec θ

a) 107.0º 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


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Experiential Activity Four Answers Angle

(θ) Ref. ∠

(α) sin θ cos θ tan θ cot θ csc θ sec θ

a) 107.0º 73.0º 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 –0.1944 1.0187 –5.2408

h) 539.3º 0.7º 0.0122 –0.9999 –0.0122 –81.8470 81.8531 –1.0001

i) 905.0º 5.0º –0.0872 –0.9962 0.0875 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 1.1626 –1.5335 –1.3190

l) –362.6º 2.6º –0.0454 0.9990 -0.0454 –22.0217 –22.0444 1.0010

m) 1260.2º 0.2º –0.0035 –1.0000 0.0035 286.4777

−286.4795 –1.0000

n) 1.4 1.4 0.9854 0.1700 5.7979 0.1725 1.0148 5.8835

o) 6.2 0.0832 −0.0831 0.9965 −0.0834 −11.9936 −12.0352 1.0035

p) 0.65 0.65 0.6052 0.7961 0.7602 1.3154 1.6524 1.2561

q) 1.6 1.5416 0.9996 −0.0292 −34.2325 −0.0292 1.0004 −34.2471

r) −1.4 1.4 −0.9854 0.1700 −5.7979 −0.1725 −1.0148 5.8835 s) −4.9 1.3832 0.9825 0.1865 5.2675 0.1898 1.0179 5.3616


OBJECTIVE FIVE When you complete this objective you will be able to… Evaluate inverse trigonometric functions.


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 or θ = sin−1 0.9962 θ = 85º 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. 85º is in quadrant 1 sin 85º = 0.9962

95º is in quadrant 2 sin 95º = 0.9962 The reference angle of 95º is 85º. The ratio is positive because sine is positive in quadrant 2

Y

θ = 85º

X Y

θ = 95º

X

α = 85º

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265º is in quadrant 3 sin 265º = −0.9962 The reference angle of 265º is 85º. 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.

275º is in quadrant 4 sin 275º = −0.9962 The reference angle of 275º is 85º. 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

Y

θ = 265º

α = 85º

X

Y

θ = 275º

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

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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.

X

Y

Quadrant I

All Positive

Quadrant II

Sine Positive

Quadrant III

Tangent Positive

Quadrant IV

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 or α = sin−1 0.5976 α = 36.7º 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.

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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. If we graph the reference angle in quadrant 2 we can determine angle θ. 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

Y

θ = 143.3º

X α = 36.7º

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.

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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.

X

Y

Quadrant I Quadrant II

Quadrant IV Quadrant III

T C

S A θ = 180º − α θ = α

θ = 180º + α θ = 360º − α

CAST C is cosine A is all S is sine T is tangent


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 Check: cos 6.0336 = 0.9691 (off slightly due to rounding) Thus θ1 = 0.2496 and θ2 = 6.0336

X

Y

θ

α = 0. 2496

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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.

cos α = θsec

1 = 2408.51

α = cos−1 ⎟⎠⎞

⎜⎝⎛

2408.51

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

Check: sec 79.0º = °0.79cos

1 = 5.2408

Answer in Quadrant 4: θ = 360º − α θ = 360º − 79.0º θ = 281.0º

Check: sec 281.0º = °0.281cos

1 = 5.2408

Thus θ1 = 79.0º and θ2 = 281.0º

X

Y

θ

α = 79.0º

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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.

Answer in Quadrant 3: θ = 180 + α θ = 180º + 21.2º θ = 201.2º Check: sin 201.2º = −0.3616

X

Y

θ

α = 21.2º

26


Answer in Quadrant 4: θ = 360º − α θ = 360º − 21.2º θ = 338.8º Check: sin 338.8º = −0.3616 Thus θ1 = 201.2º and θ2 = 338.8º

X

Y

θ

α = 21.2º

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 The trigonometric ratio of the reference angle α is

the absolute value of the trigonometric ratio of θ. convert to sine function

sin α = 7965.11

α = sin−1 ⎟⎠⎞

⎜⎝⎛

7965.11

α = 33.8º

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X

Y

θ

α = 33.8º


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 = °8.213sin

1 = −1.7965

Answer in Quadrant 4: θ = 360º − α θ = 360º − 33.8º θ = 326.2º

Check: csc 326.2º = °2.326sin

1 = −1.7965

Thus θ3 = 213.8º and θ4 = 326.2.2º

X

Y

θ

α = 33.8º

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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 23π−

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.

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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.

When θ = 90º & θ = 2π

x = 0 r = 1. Since y falls on the positive y-axis y = 1.

We can now use the six trig functions from the circle method and substitute in the values

for x, y, and r for 90º. Remember that 90º = 2π and these concepts apply to radians as

well as degrees.

sin θ = ry

sin 90º = 11

sin 90º = 1

csc θ = yr

csc 90º = 11

csc 90º = 1

cos θ = rx

cos 90º = 1

0

cos 90º = 0

sec θ = xr

sec 90º = 0

1

sec 90º = undefined

tan θ = xy

tan 90º = 01

tan 90º = undefined

cot θ = yx

cot 90º = 10

cot 90º = 0

sin 2π

= 11

sin 2π

= 1

csc 2π

= 11

csc 2π

= 1

cos 2π

= 1

0

cos 2π

= 0

sec 2π

= 0

1

sec 2π

=

undefined

tan 2π

= 01

tan 2π

=

undefined

cot 2π

= 10

cot 2π

= 0

X

Y

y

x

r

θ

X

Yy = 1

x = 0

r = 1

θ

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TRIGONOMETRIC FUNCTIONS OF OTHER QUADRANTAL ANGLES.

θ = 180º & θ = π y = 0 r = 1. Since x falls on the negative x-axis x = −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 = ry

sin 180º = 10

sin 180º = 0

csc θ = yr

csc 180º = 01

csc 180º = undefined

cos θ = rx

cos π = 11−

cos π = −1

sec θ = xr

sec 180º = 1

1−

sec 180º = −1

tan θ = xy

tan 180º = 1

0−

tan 180º = 0

cot θ = yx

cot π = 01−

cot π = undefined

θ = 270º & θ = 2

3π

x = 0 r = 1. Since y falls on the negative y-axis y = −1.


sin θ = ry

sin 2

3π =

11−

sin 2

3π = −1

csc θ = yr

csc 2

3π =

11

−

X

Y

y = 0

x = −1 r = 1

θ

X

Y

y = −1 x = 0

r = 1

θ

= −1

cos θ = rx

cos 270º = 10

cos 270º = 0

sec θ = xr

sec 2

3π =

01

sec 2

3π =

undefined

tan θ = xy

tan 270º = 01−

tan 270º = undefined

cot θ = yx

cot 270º = 1

0−

cot 270º = 0

csc 2

3π

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QUADRANTAL ANGLES CONTINUED

θ = 0º or 360º & θ = 2π y = 0 r = 1. Since x falls on the positive x-axis x = 1.


sin θ = ry

sin 360º = 10

sin 360º = 0

csc θ = yr

csc 360º = 01

csc 360º = undefined

cos θ = rx

cos 2π = 11

cos 2π = 1

sec θ = xr

sec 360º = 11

sec 360º = 1

tan θ = xy

tan 2π = 10

tan 2π = 0

cot θ = yx

cot 2π = 01

cot 2π = undefined

X

Y

y = 0

x = 1

r = 1

θ

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 = yr =

01 = undefined

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 = xy =

10 = 0

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


SUMMARY OF QUADRANTAL ANGLES The following labeled diagram should be of assistance to you when you are solving inverse function problems of quadrantal angles

33

X

Y

90º sin 90º = 1 csc 90º = 1 cos 90º = 0 sec 90º = undefined tan 90º = undefined cot 90º = 0

0º and 360º sin 0º = 0 csc 0º = undefined cos 0º = 1 sec 0º = 1 tan 0º = 0 cot 0º = undefined

270º sin 270º = −1 csc 270º = −1 cos 270º = 0 sec 270º = undefined tan 270º = undefined cot 270º = 0

180º sin 180º = 0 csc 180º = undefined cos 180º = −1 sec 180º = −1 tan 180º = 0 cot 180º = 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.

cot 270º = °270tan

1

Using the calculator °270tan

1 evaluates as undefined

1 = undefined.

Since tan 270º = undefined, we cannot use °270tan

1 to evaluate cot 270º.


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 b) cos θ = 0.8200 g) sin θ = 1.0000 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 b) cos θ = −0.4555 g) sin θ = 0.7804 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.

θ sin θ cos θ tan θ

150.0º

233.5º

345.0º

-0.460

-0.951

0.680

164.3º

0.968

0.166

−2.32

360.0º


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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


Experiential Activity Five Answers 1. a) 65.5º, 114.5º b) 34.9º, 325.1º c) 22.3º, 202.3º d) 205.5º, 334.5º

e) 50.2º, 230.2º f) 45.0º, 315.0º g) 90.0º h) 14.0º, 194.0º i) 180.0º

2. a) 0.6213, 2.5203 b) 2.0437, 4.2395 c) 1.8483, 4.9899 d) 3.5290, 5.8958 e) 2.3196, 5.4612 f) 1.5987, 4.6845 g) 0.8953, 2.2463 h) 2.9234, 3.3597

3. θ sin θ cos θ tan θ

150.0º 0.500 −0.866 −0.577 233.5º −0.804 –0.595 1.35 345.0º −0.259 0.966 –0.268 207.4º – 0.460 –0.888 0.518 332.6º 0.888 –0.518 162.0º 0.309 – 0.951 –0.325 198.0º –0.309 0.325 34.2º 0.562 0.827 0.680 214.2º –0.562 –0.827

164.3º 0.271 –0.963 –0.281 75.5º 0.968 0.251 3.86

104.5º –0.251 –3.86 80.4º 0.986 0.166 5.91

279.6º –0.986 −5.91 113.3º 0.918 –0.396 –2.32 293.3º –0.918 0.396 360.0º 0 1 0

4. a) A is in quad 3 and 4, A=211.6º, 328.4º, b) A is in quad 1 and 3, A= 12.5º, 192.5º c) A is in quad 2 and 4, A= 139.2º, 319.2º d) A is in quad 1 and 4, A= 61.7º, 298.3º e) A is in quad 2 and 3, A= 144.1º, 215.9º f) A is in quad 2 and 3, A= 107.8º,

252.2º g) A is in quad 2 and 4, A= 167.5º, 347.5º h) A is in quad 1 and 2, A= 83.4º, 96.6º i) A is in quad 2 and 3, A= 139.9º, 220.1º j) A is in quad 3 and 4, A= 241.5º,

298.5º k) A is in quad 1 and 4, A= 62.9º, 297.1º 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


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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.

Trigonometric Functions of Any...

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