chapter 5 Trigonometric Functions - Glendaleweb.gccaz.edu/~MICOW00001/HoltfrerichHaughnPreCalc… · Web viewWith trigonometric functions for each output ... Example 1: Finding

Chapter 5 Trigonometric Functions (picture here of something involving weather)

Trigonometric functions are cyclical functions. What we mean by this is that they repeat themselves

over and over again. They have a cycle. One everyday event that is cyclical is the weather,

specifically the temperature of the air. In fact you can view it as having two cycles. One is a yearly

cycle the other is a daily cycle. If you were to look at what happens, temperature wise, in St. Louis,

Missouri you would see that in the winter the average high temperature is 37.7° Fahrenheit and in the

summer the average high temperature is 89.3° Fahrenheit. With this little of information we can

construct a function that would give us the average high temperature for any month of the year for St.

Louis that would be fairly accurate. The formula we would use would be a trigonometric function

called sine. The formula would look like this: . Here is what the graph

looks like with the high temperatures in January and July for three years marked by dots.

This idea along with other is what weather men and women use to help them give you the weather

forecast each day.

Chapter 5 Trigonometric Functions 2

Section 5.1 Geometry Review

Objectives

Understanding some basic terms of geometry

Understanding basics about angles

Understanding triangles

As we begin to look at our next family of functions, trigonometric functions, we need to take some

time to review geometry. We will begin with some basics.

BASIC TERMS OF GEOMETRY

Discussion 1: Basic Terms

Let’s review the definitions of a few terms: point, line and plane.

Point A point is a dot in space that has no size. In other words, it is zero dimensional. It

has no width or height or length (we usually use capital letters to name them).

Line − A line is made up of an infinite number of points, is one dimensional and goes on

forever in both directions (we usually use the middle of the alphabet to name them).

Plane − A plane is two dimensional and is made up of an infinite number of points which

go on forever in all directions on a flat surface.

Here are illustrations of the three defined terms: point, line, plane.

Two points are called collinear if they lie on the same line. If two or more points or lines lie on the

same plane then we call them coplanar.

Question 1: Which of the following are collinear, which are coplanar and which are neither?

AB

C

m

l

n

k

Holtfrerich & Haughn 3 Section 5.1 Geometry Review

In geometry we also talk about line segments and rays. A line segment is a piece of a line that has a

starting point and an ending point. A ray is half of a line. It has a starting point but no ending point.

Here are examples of each.

When we talk about angles in geometry we are talking about the measurement between two rays that

have a common end point (vertex). We usually use small Greek letters to represent angle names.

Here are a couple of examples of angles.

In the first example θ, read “theta”, is the name of the angle between the two rays. In the second

example α, read “alpha”, is the name of the angle between its two rays. The point where the two rays

start is called the vertex. In trigonometry we will further define angles such that they can have

positive or negative measures of any amount. One unit of measure used for angles is degrees and the

symbol used is °. There is defined to be 360° in one full rotation around a circle. Therefore a quarter

of a circle, which creates a right angle, is 90°.

SOME BASICS ABOUT ANGLES

Let’s talk about the names we give angles in Geometry. Angles are either, acute,

obtuse, right or straight. These words have the following definitions.

Acute An acute angle is one whose measure is less than 90 degrees (θ above).

Obtuse − An obtuse angle is one whose measure is more than 90 but less than 180 degrees

(α above).

Right − A right angle is one whose measure is exactly 90 degrees.

Straight − A straight angle is one whose measure is 180 degrees.

Discussion 2: Special relationships of AnglesIf the measures of two angles add up to 90 degrees then we call the two angles complementary

angles. If the measures of two angles add up to 180 degrees then we call them supplementary angles.

Here are two examples. In example (a) α and θ are complementary. In example (b) they are

supplementary.

a) b)

Line segment Ray

θα

θ αα

θ


We also have relationships that make angles interior or exterior among other things. Let’s illustrate

some of these relationships when we have parallel lines l and m.

Interior angles 3, 4, 5, and 6

Exterior angles 1, 2, 7, and 8

Alternate interior angles 3 & 6 , 4 & 5

Corresponding angles (are equivalent) 1 & 5 , 2 & 6 , 3 & 7 , 4 & 8

Vertical angles (are equivalent) 1 & 4 , 2 & 3 , 5 & 8 , 6 & 7

In the above illustration line n is called a transversal because it intersects two coplanar lines. Also we

know from geometry that alternate interior and alternate exterior angles are supplementary.

Example 1: Finding Angle RelationshipsIf we know that angle 4 in the above illustration is a 120° angle and that lines l and m are parallel

then what do we know about all of the other seven angles?

Solution:

From Geometry we know that corresponding

angels will be equivalent if you have parallel

lines. (m 4 = 120°)

m 8 = m 4

m 8 = 120°

Vertical angles are always equivalent m 1 = m 4

m 1 = 120°

m 5 = m 8

m 5 = 120°

Angles 2 and 4 are supplementary (they add m 2 = 180° − m 4

Answer Q1:All points and lines are coplanar with each other except of line k which isn’t coplanar with any thing else in this example. Points A and B are collinear as are points B and C. Also points A and C are collinear.

l

n

m

1 2

3 4

5 6

7 8


up to 180°) so, m 2 = 180° − 120° = 60°

Angle 2 is a corresponding angle to angle 6 m 6 = m 2

m 6 = 60°

Vertical angles are equivalent m 3 = m 2

m 3 = 60°

m 7 = m 6

m 7 = 60°

To summarize m 1 = m 4 = m 5 = m 8 = 120°

m 2 = m 3 = m 6 = m 7 = 60°

Question 2: If in example 1 instead of the m 4 being 120°, we knew that the m 2 = 35°, what

would be the measures of the other seven angles?

TRIANGLES

Triangles are geometric shapes that have three sides and three angles. One fact about triangles is that

the sum of the three angles of any triangle must be 180 degrees.

Example 2: Triangle Angles Equal 180°

Given the following triangle what are the measures of all of its angles?

Solution:

First we know that the sum of the three

angles is 180°

40 + (θ + 10) + (2θ − 20) = 180°

We need to solve for θ. 30 + 3θ = 180°

3θ = 150°

θ = 50°

Now figure out what are the values of

(θ + 10) and (2θ − 20).

θ + 10 = 60°

2θ − 20 = 80°

The three angles are 40°, 60°, 80°

There are three special triangles that we need to talk about. They are: right,

isosceles, and equilateral.

Right A right triangle is one that has a 90 degree angle as one of its angles. It is made up

of two legs and a hypotenuse which is the side opposite the right angle.

Isosceles − An isosceles triangle is one where two of the legs of the triangle have the same

length. This makes two of its angles equal to each other.

Equilateral − An equilateral triangle is one where all three sides are the same length. This

40°

θ + 10

2θ − 20


makes all three of its angles equal to 60 degrees.

Here are illustrations of these three types of triangles.

Question 3: If one of the angles in a right triangle is a 90 degree angle, then what could we say about

the other two angles?

While we are discussing triangles we need to talk about similar triangles.

Question 4: What do you think it will take for two triangles to be similar?

Discussion 3: Similar Triangles

You should have answered the last question in one of two ways. Either you should

have thought that both triangles would need to have angles of equal measure or that

the corresponding sides would need to be proportional. Let’s illustrate these.

Figures a) and b) are similar

because they have

corresponding angles which

have equal measurements.

a) b)

Figures c) and d) are similar

because the corresponding

sides are proportional.

c) d)

Example 3: Similar TrianglesWhat values of x, and y will make these two triangles similar?

Solution:

7

5

2 x

y

6

Isosceles

θ θ

Equilateral

60°

60°60°

60°

75°

45° 60°

75°

45°

3 2

4

64

8

Right

α

β

HypotenuseLeg


Since we are looking for values which will make these two

triangles similar, we need to make sure that the

corresponding sides have the same ratio.

This leads to the following values for x, and y.

Discussion 4: Right Triangles

Lastly, we need to point out some special right triangles and a property of right triangles.

30-60-90 right triangle

This is an important triangle to have

memorized.

45-45-90 right triangle

This is also an important triangle to have

memorized.

If an altitude is drawn from the 90° angle to

the hypotenuse then it divides the right

triangle into two similar triangles that are

also similar to the original one.

Now because we know that similar triangles have sides that are proportional, we can find the sides of

any 30-60-90 or 45-45-90 right triangle from knowing these special triangles.

60°

30°

21

3

45°

45°

1

12

Altitude

Answer Q2:m 2=35° and,m 2 = m 3 = m 6 = m7 so, they all are 35° .m 2=35° and is supplementary to 4 which must then have a measure of 145°.m 4 = m 1 = m 5 = m8 so, they all are 145° .


Example 4: Finding Sides of a Right TriangleGiven the following triangles what are the lengths of all of their sides?

a) b)

Solution:

a) In the first triangle since we see that it is a right

triangle with a 30° angle we can use what we know

about a 30-60-90 triangle to find the other two

sides.

We can see that the ratio of the two hypotenuses is

= 5. So our triangle has sides that are 5 times

larger then the basic one. Therefore we know the

following:

Each side is 5 larger.

b) In the second triangle since we see that it is a right

triangle with a 45° angle we can use what we know

about a 45-45-90 triangle to find the other two

sides.

We can see that the ratio of two of the legs is =

3. So our triangle has sides that are 3 times larger

then the basic one. Therefore we know the

following:

Question 5: What are the other sides for the following triangle?

You should have noticed that the general form of the 30-60-90 and 45-45-90 triangles are as follows.

30°

10

45°

3

60°

30°

21

3

30°

10

5() = 5

5(1) = 5

45°

45°

1

12

45°

3

3(1) = 3

3 = 3

60°

1

Answer Q3:The other two angles must add up to 90°. This means that the other two angles are complementary angles.

Answer Q4:Two triangles will be similar if the have the same angles or if the sides are proportional.


Section Summary: Important Truths for the Study of TrigonometryIf you have two parallel lines intersected by a transversal then the follow are true:

Truth 1 Alternate interior angles have equal measure.

Truth 2 Corresponding angles have equal measure.

Truth 3 Alternate Exterior angles have equal measure.

With any intersecting lines the following is true:

Truth 4 Vertical angles have equal measure.

Two triangles will be similar when:

Truth 5 Corresponding sides are proportional.

Truth 6 They have the equivalent angles.

Angles

Truth 7 The sum of complementary angles is 90°.

Truth 8 The sum of supplementary angles is 180°.

Truth 9 The sum of three angles of a triangle is 180°.

Special Right Triangles

Truth 10 The ratio of the sides of a 30-60-90 right triangle are;


Truth 12 The acute angles of a right triangle are complementary.

45°

45°

x

x2 x

60°

30°

2xx

3 x

60°

30°

2xx

3 x

45°

45°

x

x2 x


SECTION 5.1 PRACTICE SET(1−6) Use the figure below:

1 2

1. m : m 2. m : m

3. m : m 4. m : m

5. and : m and m

6. and : m and m

(7−12) Use the figure below:

1 2

7. 8.

9. 10.

11.

12.

(13−20) Use the figure below: t 7 8 m line l is parallel to line m, with a transversal of t 5 6

4 3 l 2 1

13. Find the measure of each of the other angles.



16. Find the measure of all the angles.



Answer Q5:

60°

1 13

12


(21—28) Use the figure below:

C

A B

21. 22.

23.

24.

25.

26.

27.

28.

(29−34)

D B e f F E c a d A C b

Given the two triangles are similar; answer the following questions:

29. 30.

31. 32.

33. 34.


(35−42) 30o b c 90o 60o a

The given right triangle is a 30−60−90 right triangle.

35. 36.

37. 38.

39. 40.

41. 42.

(43−48) 45o

a c

900 45o

bThe given right triangle is a 45−45−90 right triangle.

43. 44.

45. 46.

47. 48.

(49−52) Given 1 and 2 are complementary angles:

49. m1 = 53o m 2 = 50. m2 = 48o m1 =

51. m1 = 3x+10 m2 = 2x+30 m1 = m2 =

52. m1 = 5x−12 m2 = 3x−18 m1 = m2 =

(53−56) Given 1 and 2 are supplementary angles:

53. m1 = 78o m2 = 54. m2 = 103o m2 =

55. m1 = 3x+50 m2 = 2x+30 m1 = m2 =

56. m1 = 5x−10 m2 = 3x−50 m1 = m 2 =


(57−58) b a d e

f c

57. a= 4; b = 8; d = 2; c = e = f =

58. d = 6; f = ; b = 12; a = c = e =


Section 5.2 Basics of Angles

Objectives

Understanding angles

Converting between degrees and radians

Understanding linear and angular speed

ANGLES

In trigonometry, angles are made up of three parts: a vertex, an initial side, a terminal side. The initial

side is the ray where the angle begins its rotation and the terminal side is the ray where it stops. The

vertex, as stated in section 5.1, is the point where the two rays start.

When the angle rotates in a counterclockwise manner the angle will have a positive measurement

(therefore θ in the above figure has a positive measure). When it rotates in a clockwise manner it will

have a negative measurement (therefore α in the above figure has a negative measure).

Discussion 1: Measuring Angles in Degrees

There are two ways used to measure angles. The first is with degrees. In one full revolution there are

defined to be 360 degrees (360°). Here are a few examples of angles measured in degrees.

When we talk about angles we often will want to talk about them in what is

called standard position.Standard Position An angle is said to be in standard position if the vertex is at the origin

of an x, y coordinate plane and the initial side is lying on the positive x-axis.

The coordinate plane is divided up into four regions called quadrants. Here is how we name each

quadrant.

θTerminal side

Initial sideα

30°−60°

405° −450°

x

y

Quadrant IQuadrant II

Quadrant III Quadrant VI


Example 1: QuadrantsIn which quadrant are the terminal sides of the following angles when placed in standard position?

a) 35° b) 230° c) −190°

Solution:

a) 35°

First quadrant

b) 230°

Third quadrant

c) −190°

Second quadrant

Question 1: With the angle, θ = 1045°, in standard position, in which quadrant does the terminal side

lie?

Angles may rotate through as many revolutions as they want. Therefore it is possible for many

different angles, an infinite number of possibilities, to have the same initial and terminal sides. When

this happens we call the angles conterminal angles.

Coterminal Angles Two angles are called coterminal if they both have the same initial

and terminating sides. This means that when you subtract the two angle measurements you

will get a multiple of 360°. In the picture below θ − α = 360°.

Example 2: Coterminal AnglesWhat are some coterminal angles to the following angles?

35°

230°

−190°

1045°

θTerminal side

Initial sideα


a) 80° b) −135° c) 340°

Solution:

a) Some coterminal angles to an 80° would

be the following since when they are

subtracted from 80° you get a multiple of

360.

−280° (80 − (−280) = 360)

440° (80 − (440) = −360)

800° (80 − (800) = −720)

b) Some coterminal angles to a −135°

would be the following since when they

are subtracted from −135° you get a

multiple of 360.

225° (−135 − (225) = −360)

−495° (−135 − (−495) = 360)

585° (−135 − (585) = −720)

c) Some coterminal angles to a 340° would

be the following.

−20° (340 − (−20) = 360)

−380° (340 − (−380) = 720)

700° (340 − (700) = −360)

Question 2: What is a coterminal angle to θ = −210° ? (hint: there are many answers)

In the degree system of measuring there are also minutes and seconds. They are defined as follows.

Minute One minute is of a degree. The symbol ' is used for minutes. Another way to

say this is that there are 60 minutes in one degree.

Second − One second is of a minute. The symbol ″ is used for seconds. Another way to

say this is that there are 60 seconds in one minute.

Example 3: Converting from degrees to degrees, minutes and seconds (dms)

Convert the following from degrees to degrees minutes and seconds or vise versa.

a) 328.32° b) −732.19° c) 15° 48' 12″

SOLUTION:a) Separate the whole number from the

decimal part of the number.

Now convert the 0.32° to minutes by

multiplying by 60 minutes.

Now do the same with the minutes that

328° 0.32°

0.32 60' = 19.2'

328° 19.2'

19' 0.2'

θ = −210°


we did with the degrees.

The final answer is

0.2 60″ = 12″

328° 19.2' 12″

b) Separate the whole number from the

decimal part of the number.

Now convert the 0.19° to minutes by

multiplying by 60 minutes.

Now do the same with the minutes that

we did with the degrees.

The final answer is

732° 0.19°

0.19 60' = 11.4'

−732° 11.4'

11' 0.4'

0.4 60″ = 24″

−732° 11.4' 24″

c) This time we are converting a degree

minute second measurement to just

degrees. You need to divide the minutes

by 60 and the seconds by 3600.

15° 48' 12″

15° + + = 15 + 0.8 + 0.0033333

15.80333333°

Let’s look at how the graphing calculator could have helped us with these last

three problems.a) 328.32° With the TI-83/84 simply type in

the 328.32 and then push the 2nd

MATRIX key (TI-83) or 2nd APPS

(TI-83+,84) for ANGLE and select

number 4 for DMS and push

ENTER twice.

With the TI-86 simply type in the

328.32 and then push the 2nd X key

(times key) for MATH then F3 for

ANGLE and F4 for DMS and then

ENTER.

b) −732.19° Do the same as for part a.

c) 15° 48'

12″

With the TI-83/84 type in 15 then

get the ° symbol from the ANGLE

With the TI-86 simply type in the

15 followed by the ‘ symbol found

Answer Q1:Fourth quadrant


menu. Then type in 48 and get the ‘

symbol from the ANGLE menu.

Lastly, type in 12 and then push

ALPHA + keys to get “. Push

ENTER to see the answer.

in the ANGLE menu. Then follow it

with 48 and the ‘ symbol and then

12 with the ‘ symbol. Now if you

have your calculator on degree

mode push ENTER to see your

answer. If you are in radian mode

go change it to degree mode first

then ENTER.

Discussion 2: Measuring Angles in Radians

This second way of measuring angles is the method used most often in calculus. It will be the

primary method used in this book. Radian measure of an angle is based on the length of the

circumference of a circle (arc of the circle) created by the angle being measured in multiples of the

radius of the circle. For example, the radian measure of an angle that is one complete revolution

would be 2π rad (radians). The circumference of a whole circle is 2πr. Thus, there are 2π radii

lengths in the circumference of a circle (the circumference of a circle is slightly longer that 6 radii,

2π 6.283). So the angle that creates a whole circle would have a radian measure of 2π rad.

There are a couple of terms we need to add at this point.

Central Angle − An angle with its vertex at the center of a circle.

Arc of a Circle − An arc of a circle is a piece of the circumference of a circle that is

intersected by a central angle.

Radian − One radian is the measure of a central angle (θ) that intercepts an arc (s) that is

equal in length to the radius of the circle.

Arc Length − The arc length (s) will equal the angle θ (in radians) times the radius (r) of

the circle (s = θr ).

Here are some examples of angles measured in radian.

Circumference of half a circle Circumference of a quarter of Circumference of a twelfth of

Half a circle2

One quarter of a circle

6

One 12th of a circle

Answer Q2:Some coterminal angles are 150°, −570°, 510°.


is = πr. So the angle that

rotates through half a circle

has a radian measure of π.

a circle is = . So the

angle that rotates through a

quarter of a circle has a radian

measure of .

a circle is = . So the

angle that rotates through one

twelfth of a circle has a radian

measure of .

Example 4: Arc Length and Sector AreaWhat is the arc length and sector area of the figure?

Solution:

To find the arc length (s) we use the

formula s = θr. (This gives us part of

the circumference of the whole circle

2πr.)

units

The outside arc, or the piece of the

circumference of a pizza, is 5.6π units in length.

To find the sector area (a sector is a

slice of pizza) the formula would be

. This comes from the fact that

the area of a circle is .

units squared

Another concept we need to take a look at is the concept of reference angle. A reference angle is an

acute angle (measurement between 0° and 90°) between the terminal side of an angle in standard

position and the x-axis. Let’s look at some examples.

Example 5: Reference Angles

Find the reference angle (r) for the following angles in standard position.

a) θ = 35° b) α = 230° c) β = −190° d) φ = 1045°

Solution:

a) Since θ is already an acute angle from

the terminal side to the x-axis, it is its

own reference angle.

r = 35°

θ =

r = 7

r35°


b) A 230° angle from standard position

means that the terminal side has gone

50° beyond the negative x-axis (180°).

r = 50°

c) A −190° angle has gone in a clockwise

rotation 10° past the negative x-axis.

r = 10°

d) A 1045° angle has made two complete

rotations and then another 325° which

means it is just 35° short of completing

its third revolution.

r = 35°

CONVERSION BETWEEN DEGREES AND RADIANS

Even though we will be focusing a lot of our attention on radian measure it is important that you are

comfortable with both degree and radian measurements. We also will need to be able to convert

measurements of one type into the other.

Discussion 3: Conversions from degrees to radians and vise versa

Since we know that there are 360° in a circle and 2π rad, we can create two ratios which will allow us

to convert from one form of measurement to the other. To find radian measure you will take the

degree measure times the ratio . To find degree measure you will take the radian

measure times the ratio .

For example let’s convert the following two angle measurements.

a) 240° b) rad

a) We can multiply the 240° by the fraction

b) We can multiply the rad by the fraction

Here is how we do these two on the calculator.

a) With the TI-83/84, type in 240 followed

by the ° symbol found in the ANGLE

menu and then push ENTER if you are

a) With the TI-86, type in 240 followed by

the ° symbol found in the ANGLE menu

and then push ENTER if you are in

230°

r

r

−190°

r1045°


in radian mode. If you are in degree

mode go change it to radian mode and

then push ENTER.

radian mode. If you are in degree mode

go change it to radian mode and then

push ENTER.

b) With the TI-83/84, type in with

parentheses followed by the r symbol

found in the ANGLE menu and then

push ENTER if you are in degree mode.

If you are in radian mode go change it to

degree mode and then push ENTER.

b) With the TI-86, type in with

parentheses followed by the r symbol

found in the ANGLE menu and then

push ENTER if you are in degree mode.

If you are in radian mode go change it to

degree mode and then push ENTER.

Question 3: What is the radian measure of the angle θ that is −135° ?

LINEAR AND ANGULAR SPEED

Discussion 4: Car SpeedVisualize a car and its tires. Now consider a nail that is in one of the tires. We can talk about two

ideas in this scenario. First, how fast is the car moving forward (linear speed). Second, how fast is the

nail going around (angular speed). Let’s consider how fast the car is moving first. The tire moving

around on its circumference is what is making our car go forward. So, as the tire traces through its

circumference the car moves forward that amount (2πr is the distance traveled by the car in one

revolution of the tires). Therefore if a car has a tire with a radius of 14 inches then with one rotation

of its tires the car will have traveled 87.96 inches (2 π 14 87.96).

Let’s say that we know that a car, with 14 inch radius tires with a nail in them, is

traveling such that the tires are making 8 revolutions per second, how fast is the car

moving forward?


The circumference of a tire is 2 π 14 = 87.96

inches. Multiply this by 8 since we get 8 revolutions

of the tire in one second. This will give us the

distance the car will travel in one second (in/sec).

We might prefer to convert this to mph in order to

better understand how fast this is or isn’t.= 39.98 40 mph

Discussion 5: Nail Speed

Now let’s take the same problem as discussion 4, but ask the question, how fast is the nail moving

around and around. Angles are a good way of measuring how we are going around and around. So, to

answer our question we need to think in terms of angles divided by time to get a rate value in terms

of radians per sec. In this example the nail is moving through 2π rad every one revolution so after 8

revolutions (one second of time) the nail will have traveled through 16π rad in one sec. Thus we

could say that the nail is traveling .

We have the following definitions:

Linear Speed The linear speed of an object connected to a wheel will be v = , where v

is the velocity of the object, s is the arc length and t is the time.

Angular Speed − The angular speed of an object on a wheel will be ω = , where ω read

“omega” is the angular speed, θ is the angle (in radians) and t is the time.

Question 4: What would be the linear speed (in feet/sec) of a car and the angular speed (in rad/sec) of

a nail in a tire on a car, if the tires are traveling at the rate of 10 revolutions per second with 1 foot

radii?

As you answered question 4 you should have noticed that when the radius is one the angular speed

and the linear speed have the same magnitude.

Section Summary: Vertex The common point where two rays begin that create an angle.

Initial side The ray where an angle begins.

Terminal side The ray where an angle ends.

Positive angles Angles that rotate in a counterclockwise manner.

Negative angles Angles that rotate in a clockwise manner.

Answer Q3:

−135°

= .


Minute one sixtieth of a degree.

Second One sixtieth of a minute.


Arc of a Circle − An arc of a circle is a piece of the circumference of a circle that is

intersected by a central angle.

Radian − One radian is the measure of a central angle (θ) that intercepts an arc (s) that is

equal in length to the radius of the circle.

Arc Length − The arc length (s) will equal the angle θ (in radians) times the radius (r) of the

circle (s = θr ).

A reference angle is an acute angle (measurement between 0° and 90°) between the terminal

side of an angle in standard position and the x-axis.

Conversion 180° is equivalent to π radians. When you need to convert from one form to

another use one of the following ratios: .

Linear Speed The linear speed of an object connected to a wheel will be v = , where v is

the velocity of the object, s is the arc length and t is the time.

Angular Speed − The angular speed of an object on a wheel will be ω = , where ω read

“omega” is the angular speed, θ is the angle (in radians) and t is the time. Answer Q3:


SECTION 5.2 PRACTICE SET(1–8) For each angle in standard position, the terminal side of the angle is in which quadrant?

1. 83o 2. 193o 3. –230o 4. –58o

5. radians 6. radians 7. radians 8. radians

(9–12) For each angle give all the coterminal angles from –360o to 360o:

9. 50o 10. 190o 11. –150o 12. –210o

(13–16) For each angles give all the coterminal angles from –2 to 2:


(17–20) Convert the decimal value of a degree measurement to degrees–minutes–seconds:

17. 58.32o 18. 233.24o 19. –583.45o 20. –78.54o

(21–24) Convert from degrees–minutes–seconds to a decimal degree answer:

21. 22. 23. 24.

(25–36) For each angle give the reference angle:

25. 158o 26. 321o 27. 238.5o 28. 538.4o

29. –273o 30. –162o 31. –323.8o 32. –58.7o


(37–44) Convert the following degree measurement of an angle to a radian measurement of that angle:

37. 60o 38. 240o 39. 165o 40. 324o

41. –120o 42. –300o 43. –230o 44. –55o

(44–52) Convert the following radian measurement of an angle to a degree measurement of that angle:



Answer Q4:Angular speed would be 10*2π since we have 10 revolutions per sec. Thus the answer is 20π rad/sec.Linear speed would be 10*2π(1) since we have 10 revolutions and the radius is one foot. Thus the answer is 20π feet/sec.


(53–54) For each of the following find: a. The linear speed b. The angular speed

53. A Ferris wheel has a radius of 30 feet and it takes 70 seconds for one revolution.

54. A bicycle wheel has a diameter of 26 inches and the wheel makes 425.5 revolutions every minute.

(55–60) A = The area of a sector (Area bounded by the interior of the angle and the circle) r = The radius of the circle S = The length of the arc intersected by the central angle = The measurement of the central angle of a circle in radians Use the formula A = ½ r2 and S = r r S

55. r = 8 meters = ½ radian Area of the sector = Length of arc =

56. r = 12 inches = 5 radians Area or the sector = Length of arc =

57. r = 10 centimeters = radians Area = Length of arc =

58. r = 3 feet = radians Area of the sector = Length of arc =

59. r = 10 meters = 60o Area of the sector = Length of arc =

60. r = 12 inches = 300o Areas of the sector = Length of arc =

61. A water sprinkler sprays water over a distance of 25 feet while rotating through an angle of 120o. What is the area of the lawn is watered?

62. A person that installs sprinkler systems is asked to design a water sprinkler that will cover a field of 50 square yards that is in the shape of a sector of a circle with radius of 20 yards. Through what angle should the sprinkler rotate?


Section 5.3 Sine, Cosine and the Unit Circle

Objectives

Understanding the definition of sine and cosine

Understanding the unit circle

Understanding the domain and range of sine and cosine

Understanding multiple inputs yielding same output

SINE AND COSINE

We are now ready to define two of the six new functions in the family called trigonometric functions.

Here is a picture of an angle θ in standard position. P is any point on the terminal side of angle θ. r is

the distance from the origin to the point P. Sine and Cosine are defined as follows.

Sine The trigonometric function sine is defined to be sin θ = .

Cosine − The trigonometric function cosine is defined to be cos θ = .

Notice that these two functions are defined to be a ratio of the sides of a right triangle formed

between the terminal side of an angle and the x-axis. In essence you will always be using the

reference angle to find function values for the trigonometric functions. We now have a way of

relating x and y with an angle.

We need to remind you here about the Pythagorean Theorem.

Pythagorean Theorem The Pythagorean Theorem states that the square of the two legs

of a right triangle added together will equal the square of the hypotenuse (a2 + b2 = c2).

Example 1: Finding Function Values

Given the following angles in standard position find the function values of sine and cosine?

θ

P(x, y)r

x

y

a

b

c


a) b)

Solution:

a) First we need to find the value of r.

Since we have a right triangle we can use

the Pythagorean theorem.

42 + (−3)2 = r2 r2 = 25 r = 5

(r is positive since it represents a distance)

sin θ = , cos θ =

b) First find r then sine and cosine 82 + (−15)2 = r2 r2 = 289 r = 17

sin θ = , cos θ =

Discussion 1: Any Point Will Do

It turns out that it doesn’t matter which point you pick on the terminal side of your angle the sine and

cosine values will always be the same. Let’s look at example 1a again.

You can see from the picture, that as you choose different points on the terminal side of the angle θ

you create similar triangles. And we know that similar triangles have proportional sides (Truth 5

from section 5.1). To further help you see this look at the point (−6, 8) we get sin θ = and cos

θ = . These are the same two answers that we got with the point (−3, 4). Therefore with this

angle the sin θ will always be and the cos θ will always be no matter which point you choose

on the terminal side of the angle.

Question 1: Since it doesn’t matter which point you pick when evaluating the sine and cosine of an

angle, what happens to the definitions of sine and cosine if you choose your point on the terminal

side of your angle such that r = 1?

θ

(−3, 4)

−3

4θ

(8, −15)

8

−15

θ

(−6, 8)

−3

4

8

−6

(x, y)

x

y


THE UNIT CIRCLE

The last question leads us to another way of defining sine and cosine. We can use what is called the

unit circle definition of trigonometric functions. A unit circle is a circle whose radius is one and has

this formula x2 + y2 = 1.

If we apply the earlier definitions for sine and cosine we would get sin θ = and cos θ = since

r = 1 in this picture. Therefore for a unit circle our original definition now becomes as follows.

When θ is in standard position as a central angle of a unit circle with point P on the

terminal side of the angle on the unit circle then,

The trigonometric function sine is defined to be sin θ = y.

The trigonometric function cosine is defined to be cos θ = x.

Notice, that since we are working with a unit circle θ = s, the arc length of a piece of a circle

intersected by the angle in standard position equals the angle measured in radians. So we really could

view this as a way of relating x and y to the arc length of an arc from a unit circle.

We now have two ways to view what sine and cosine do for us. One is that they relate an

angle to coordinates of a point on the terminal side of an angle in standard position. The second one

is that they relate the arc length of a unit circle with the coordinates of points on that unit circle. One

relates angles to real numbers the other relates real numbers to real numbers. This distinction is

important in the study of Calculus. Let’s take a look at some of the key points on a unit circle and

how we derive them.

From section 5.1 we know the following truths.

Truth 10 − The ratio of the sides of a 30-60-

90 right triangle are

Truth 11 − The ratio of the sides of a 45-45-

90 right triangle are

s

θ

P (x, y)

1

(1, 0)

60°

30°

2xx

3 x45°

45°

x

x2 x


Our radius is one so 2x = 1 x = and thus

the other sides are as follows.

Our radius is one so = 1 x = and

thus the other sides are as follows.

So, for a 30°, 45° and 60° central angle in a unit circle the points on the unit circle

would be as follows:

Many of the other angles of the unit circle have these three as reference angles, so the points are just

positive or negative versions of the above points. As for 0°, 90°, 180°, and 360°, their points are

obvious since the radius of the unit circle is one.

Example 2: Finding Sine and Cosine Outputs

What is the sine and cosine function outputs for the following inputs?

a) θ = b) θ = c) θ = d) θ = e) θ =

Solution:

60°

30°

1 12

32

45°

45°12

12

1

60°

30°

1 = y

= x

3 1,2 2

45°

45°= x

= y1

2 2,2 2

1 3,2 2

60°

30°1

= x

= y

3 1,2 2

1,0

0,1

2 2,2 2

1,0

0, 1

306

454

603

90

2

3 1,2 2

3 1,2 2

3 1,

2 2

2 2,2 2

2 2,

2 2

2 2,2 2

1 3,

2 2

1 3,

2 2

1 3,

2 2

1 3,

2 2

7315

4

5150

6

3135

4

2120

3

5225

4

4240

3

7210

6

180

11330

6

5300

3

3270

2

0 0 , 2 360


a) To find sin look at the y value of the

point on the unit circle that corresponds

to the angle . For the cos use the x

value.

sin = cos =

b) To find sin look at the y value of the

point on the unit circle that corresponds

to the angle . For the cos use the x

value.

sin = cos =

c) This time notice that we have a negative

angle so go clockwise around to −45°.

Now as before just look at the y and x

values of the point on the unit circle.

(−45° coterminal with 315°)

sin −45°= cos −45°=

d) As in part (c) go clockwise around to

. Now look at the y and x values of

the point on the unit circle.

( coterminal with )

sin = 1 cos = 0

e) Find the point at 600°.

(600° same terminal side as 240°, they

are coterminal angles) sin 600° = cos 600°=

We need to take some time now to talk about some patterns in the unit circle and how to use them to

help us find values for sin θ and cos θ for common angles.

Discussion 2: Ways of Finding Trigonometric Values with the Unit Circle

You should have noticed that as you go from quadrant to quadrant that many of the values for x and y

are repeated but occasionally with different signs. For example at the point on the unit circle

is and at the point is and at the point is . These

three points have the same magnitudes for x and y because all of their angles have the same reference

angle . You should notice as well, that as you go around the circle the values for x goes

Answer Q1: If r = 1 then the denominators in the definitions of sine and cosine become 1. Thus, sin θ = y and cos θ = x.


through this pattern: , , , , , , , , … and

then back to 1. The y values do a similar pattern: , , , , , , ,

, , , , , … and then back to 0. Also notice how, with

sin θ defined to be equal to y on the unit circle, that sin θ will always be positive in the 1st and 2nd

quadrants (above the x-axis) and negative in the 3rd and 4th (below the x-axis), likewise cos θ = x will

be positive in the 1st and 4th (right of the y-axis) but negative in the 2nd and 3rd (left of the y-axis).

Take a look at the next chart; it may help you to see these patterns more easily.

Reference angles 0° 30° 45° 60° 90°

rad

Q1 x value

Q1 y value

rad

Q2 x value

Q2 y value

rad

Q3 x value

Q3 y value

rad

Q4 x value


Q4 y value

Ultimately, all you need to remember is the points on the unit circle in the first quadrant and in which

quadrants sine and cosine are positive and negative. With this information memorized you can find

the sin θ and the cos θ of any angle (θ) that is a multiple of the special angles: 30, 45, 60, 90.

Example 3: Finding Sine and Cosine Function Values

What is the sin θ and cos θ for the following angles?

a) −330° b) c) −

Solution:

a) We first need to find the angle’s

reference angle and then decide

whether the function value will

be positive or negative.

A −330° angle terminates in the first quadrant 30°

short of the x-axis. Therefore the reference angle is

30° and the function value for both sine and cosine

will be positive since the terminating side of the

angle is in the first quadrant. The answers are:

sin −330° = sin 30° =

cos −330° = cos 30° =

b) We first need to find the angle’s




A angle terminates in the second quadrant

short of the x-axis (angle is over one revolution).

Therefore the reference angle is and the function

value for sine will be positive and cosine will be

negative since the terminating side of the angle is in

the second quadrant. The answers are:

sin = sin =

cos = − cos =

c) We first need to find the angle’s




A − angle terminates in the third quadrant

short of the x-axis. Therefore the reference angle is

and the function value for sine will be negative

and cosine will be negative since the terminating

Point on unit circle for a 30° angle.

Point on unit circle

for a angle.


side of the angle is in the third quadrant. The

answers are:

sin = − sin =

cos = − cos =

Question 2: What is the sin θ and cos θ for the following angles using the knowledge of the unit

circle you have just learned?

a) 120° b) −π

Example 4: Finding Sine and Cosine of non-special Angles

Find the values of the following using your graphing calculator.

a) sin 50° b) cos c) sin (−3.56)

Solution:

a) First make sure that your

calculator is on degree mode and

then push the SIN key followed

by 50, then ENTER.

b) This time the angle is in radians,

so first make sure the calculator

is on radian mode then push the

COS key followed by 5π/13,

then ENTER

c) The last example is also in

radians and you should get the

following on your graphing

calculator.

All of this brings us to two very important ideas discussed in the chapter on functions: domain and

range. Let’s take a look at what the domain and range are for sine and cosine and their graphs.

DOMAIN AND RANGE FOR SINE AND COSINE

If you look closely at the unit circle points you will first notice that any angle could be used as an

input with the sine and cosine functions. You can have either positive or negative angles (in degrees

or radians) and you may go as many times around the circle as you wish, so the domain is all

Point on unit circle

for a angle.


possible angles if you are thinking in degrees or all real numbers if thinking in terms of radians. We

want to focus on the radian usage so:

Domain for both sine and cosine is all real numbers (−∞, ∞).

The range is also visible from the unit circle points. If you look back at the points on the unit circle

you will see that the largest x or y ever are is 1, and the smallest they ever are is −1. Another way to

think about this is that you are talking about a unit circle (x2 + y2 = 1). On a circle of radius one, x and

y can’t be larger then 1 or smaller than −1. Therefore:

Range for both sine and cosine is [−1, 1].

From observing the points on the unit circle we can get this table for both functions.

0

cos θ 1 0 −1 0 1

sin θ 0 1 0 −1 0

Now, we can sketch a graph of both sine and cosine. First a partial graph of y = sin θ.

This isn’t the whole graph but only the graph for inputs from 0 to 2π radians (angles from the first

through the fourth quadrants). Notice that even though the angles used as inputs were from quadrants

1 through 4, the graph of the sine function (matching the inputs and outputs) is in the first and fourth

quadrants only so far.

Of course you should have realized by now that for inputs after 2π and before 0 the outputs

just repeat each other. The sine function is an example of what is called a periodic function (we go

dc

b

a

h

g

f

e

i

j

k

1

4

2

3

6 7

4 2

−1

54 3

23

4

θ

yd

c

a

h

g

f

e

i

j

k

b30

6

454

603

90

2

7315

4

3135

4

5225

4

180

3270

2

0 0 , 2 360


around and around the same circle). This leads to a graph that is going to go on forever to the left and

the right of the origin.

Periodic Function − A function is say to be periodic if there exists a smallest number c

such that f(x) = f(x + nc) where n is any integer.

For the sine function the value for c is 2π. Therefore the following is true for the sine function:

sin θ = sin (θ + 2nπ) where n is any integer.

Let’s look at cosine now. It is also periodic on a period of 2π. Here is its graph.

To summarize here are the graphs of both sine and cosine.

MULTIPLE INPUTS

Question 3: What are the exact values of the following and what do you think is the point of asking

this question?

a) sin b) sin c) sin

4

1

2

3

6 7

4 2

−1

54 3

23

4

θ

y

2π−2π

y

θ

y = sin θ1

−1

2π−2π

y

θ

y = cos θ1

−1

Answer Q2:

a) sin 120°=

cos 120°=

b) sin −π = 0 cos −π = −1


With trigonometric functions for each output there are many inputs (trigonometric functions are not 1

to 1). To be exact there are an infinite number of inputs that yield each possible output. So if we were

to ask you, “how many angles could you find the sine of that would yield 1 as an answer?”, you

would answer many. The way we write such an answer is this way.

sin x = 1 when x = (n any whole number)

The 2πn part of the answer is telling everybody that all revolutions around the circle that end in the

same place as the angle will have the same answer, 1.

Example 5: Finding Angles That Yield Same OutputsWhat angles yield the following?

a) sin θ = b) cos θ = 0 c) cos θ =

Solution:

a) So what angles in the unit circle yield

an answer of ?

Answer: and

Therefore the solutions to the question are:

and

b) So what angles in the unit circle yield

an answer of 0?

Answer: and


and or as just one

answer (all angles that make

cosine equal 0 are half a circle apart)

c) So what angles in the unit circle yield

an answer of ?

Answer: and


and

Section Summary

Pythagorean Theorem states that the square of the two legs of a right triangle added

together will equal the square of the hypotenuse (a2 + b2 = c2).

Sine is defined as the y value of the point on a unit circle on the terminal side of angle θ

(sinθ = y).


Cosine is defined as the x value of the point on a unit circle on the terminal side of angle θ

(cosθ = x).

The first quadrant of the unit circle

The sign of the outputs of sine and cosine are:

The domain for both sine and cosine is (−∞, ∞)

The range for both sine and cosine is [−1, 1]

A function is periodic if for some smallest c, f(x) = f(x + cn) where n is any integer

Sine and cosine have a period of 2π.

3 1,2 2

1,0

0,1

2 2,2 2

306

454

603

1 3,

2 2

0 0

sin θ > 0cos θ > 0

sin θ > 0cos θ < 0

sin θ < 0cos θ < 0

sin θ < 0cos θ > 0

Answer Q3:

sin =

sin =

sin =

The point is that these are all coterminal angles and thus they all have the same answer.


SECTION 5.3 PRACTICE SET(1–8) Use the Pythagorean theorem to find the following values:

a b

c

1. a = 9 b = 16 c = 2. a = 10 b = 24 c = 3. a = 6 c = 10 b =

4. a = 15 c = 39 b = 5. b = 5 c = 9 a = 6. b = 7 c = 11 a =

7. a = 4 b = 8 c = 8. a = 6 c = 11 b =

(9–18) Find the value of sin and cos for the following: (Don’t use the calculator)

9. 10. 11.

12. 13. 14.

15. 16. 17.

18. 19. 20.

21. 22. 23.

24. 25. 26.


27. 28. 29.

30. 31. 32.

33. 34. 35.

36. 37. 38.

39. 40. 41.

42. 43. 44.

45. 46.

(47–54) For each angle in standard position find the value of sin and cos .

(x, y) r

47. 48. 49.

50. 51. Terminal side in the 2nd quadrant


52. Terminal side is in the 4th quadrant 53.

54.

(55–64) Give the angle values for between –2 and 2 that yield the following answer. (Give the answers radians)

55. 56. 57.

58. 59. 60.

61. 62. 63.

64.

(65–74) Give the angle values for between –360o and 360o that yield the following answer. (Give the answers degrees)

65. 66. 67.

68. 69. 70.

71. 72. 73.

74.

75. What is the domain for sin ?

76. What is the domain for cos ?

77. What is the range for cos ?

78. What is the range for sin ?

79. What is the period for cos and what does the period tell us?

80. What is the period for sin and what does the period tell us?

(81–92) Use the calculator to approximate the value of sin and cos for the given value of . (Approximate to 4 decimal places)


81. 82. 83.

84. 85. 86.

87. 88. 89.

90. 91. 92.

(93–94) (a,b) r y b 1 x a

93. a = 3 and b =4; a.) x = b.) y = c.) r = d.) Find the value of sin and cos by using both the rectangular and unit circle definition. e.) How do the values compare?

94. a = 5 and b =12; a.) x = b.) y = c.) r = d.) Find the value of sin and cos by using both the rectangular and unit circle definition. e.) How do the values compare?


Section 5.4 Tangent, Cotangent, Secant, Cosecant

Objectives

Understanding the definition and graphs of tangent and cotangent

Understanding the definition and graphs of secant and cosecant

TANGENT AND COTANGENT

We are now ready to define the last four trigonometric functions using the unit circle.

Given a central angle θ in standard position whose terminal side intersects the unit circle at

the point (x, y) the following is defined.

Tangent − Tangent is defined to be tan θ = = .

Cotangent − Cotangent is defined to be cot θ = = .

Secant − Secant is defined to be sec θ = .

Cosecant − Cosecant is defined to be csc θ = .

Let’s look at a table of values for all six of the trigonometric functions using just the first quadrant.

0

y = sin θ 0 1

x = cos θ 1 0

= tan θ = 0 = = 1 = = undefined

= cot θ = undefined = = 1 = 0

= sec θ = 1 = = = 2 = undefined

= csc θ = undefined = 2 = = = 1

1

x

θ

P (x, y)

(1, 0)

y


As you look at the above table notice how a function’s output value for an angle is equal to its co-

function’s output of the angle’s complement. A couple of examples are that sin = cos or that the

tan 0 = cot . To state this more mathematically:

Cofunction Property − For any angle θ the following is true.

sin θ = cos ( − θ), tan θ = cot ( − θ), sec θ = csc ( − θ)

Discussion 1: The Graph of Tangent

Let’s look at the graphs of these new functions. First, let’s look at the graph of the tangent function.

If you look back at the table you see that as the inputs go from 0 to , the outputs, tan θ, are

increasing in value 0, , 1, , undefined. The question is what is happening between θ =

and θ = ? Let’s turn to our graphing calculator and see.

It looks as though the function values, as we get close to , are getting very large. In fact as θ

approaches the tan θ approaches ∞.

Let’s go backwards now from θ = 0 to θ = .

0

y = sin θ 0 −1

x = cos θ 1 0

= tan θ = 0 = = −1 = − = undefined

Once again the question is what is happening between θ = and θ = ?

Question 1: After looking at the first quadrant what is your thoughts on what will happen here in the

fourth quadrant?


If we continued to follow this process and looked in both the positive and negative directions we’d

quickly discover that the tangent function is periodic with a period of π. Here is the graph of tangent

near the origin.

Because the tangent function is undefined at , and every multiple of π added to , we get many

vertical asymptotes . From the graph of tangent we see that the domain for

tangent will be (−∞, ∞) except for and any added π multiple to {… , ,

…}. The range is (−∞, ∞). You can also see from the graph of the tangent function that it

does not have a period of 2π, like sine and cosine, but has a period of π.

Discussion 2: The Graph of CotangentSince the cotangent function is just the reciprocal of the tangent function all we need to do to graph it

is to look at the reciprocals of the output values of tangent.

0

tan θ 0 1 undefined −1 0 1 undefined −1 0

cot θ undefined 1 0 −1 undefined 1 0 −1 undefined

Remember that undefined happened when we had so its reciprocal would be .

From this we can determine the graph of cotangent.

2

2 3

23

2


From the graph of cotangent we see that the domain is (−∞, ∞) except for 0 and every multiple of π

added to 0 , {…(−2π, −π), (−π, 0), (0, π), (π, 2π)…}. Also, you can see from the

graph of the cotangent function that it has a period of π just like the tangent function. That should be

no surprise, cotangent was just the reciprocal of tangent. We’d expect them to have a lot of

similarities.

Question 2: What is the range of cotangent?

SECANT AND COSECANT

Discussion 3: The Graph of SecantSecant is the reciprocal of cosine, so we just need to look at the cosine function and use the

reciprocals of its output values.

0

cos θ −1 0 1 0 −1

sec θ −1 undefined 1 undefined −1

As with the other functions, when you approach an input where the output is undefined your outputs

are approaching either ∞ or −∞. The graph of secant is,

2

2 3

23

2

2

2 3

23

2

1

−1

Answer Q1:The same but going to negative infinite.


Once again like the tangent function (which has cosine in the denominator ) this graph

has vertical asymptotes at and every multiple of π added to . We get many vertical asymptotes

at . The domain is (−∞, ∞) except for and every multiple of π added to ,

{… , , …}. The range is (−∞, −1) (1, ∞). But unlike the tangent

function you can see from the graph that it has a period of 2π, just like the cosine function.

Discussion 4: The Graph of CosecantCosecant is the reciprocal of sine, so we just need to look at the sine function and use the reciprocals

of its output values.

0

sin θ 0 −1 0 1 0

sec θ undefined −1 undefined 1 undefined

As with the other functions, when you approach an input where the output is undefined your outputs

are approaching either ∞ or −∞. The graph of cosecant is,

This time, like with the cotangent function (which has sine in the denominator ), the

graph has vertical asymptotes at 0 and every multiple of π added to 0 . The

domain is (−∞, ∞) except for 0 and every multiple of π added to 0 {…(−2π, −π), (−π, 0), (0, π), (π,

2π)…}. The range is (−∞, −1) (1, ∞). But unlike the cotangent function you can see from the

graph that it has a period of 2π, just like the sine function.

Example 1: Finding Trigonometric Function Values

What are the following values?

2

2

32

32

1

−1


a) sec b) csc c) cot d) tan

Solution:

a) To find the sec we first must realize that secant is the

reciprocal of cosine (cos θ = x). Second we must notice

that the reference angle is . The unit circle point

associated with that reference angle is . Lastly we

realize that the original angle is in the first quadrant where

every trigonometric function has a positive answer.

sec =

= 2

b) To find the csc we first must realize that cosecant is

the reciprocal of sine (sin θ = y). Second we notice that the

reference angle is . The unit circle point associated with

that reference angle is . Third, we know that sine

is negative in the third quadrant.

csc =

=

=

= −2

c) To find the cot we first must realize that cotangent is

the reciprocal of tangent (tan θ = ). Second we notice

that the reference angle is . The unit circle point

associated with that reference angle is . Third, we

know that tangent is negative in the second quadrant.

cot = =

=

= =

d) To find the tan we first must notice that the reference

angle is . The unit circle point associated with that

reference angle is . Second, we calculate that

is just short of two revolutions, so it is in the fourth

quadrant where tangent is negative.

tan = − tan

= −1

Answer Q2:(−∞, ∞)


Here is a little idea that may help you remember which functions are positive and negative in which

quadrants. The acronym to memorize is ASTC (all students take calculus). All, stands for all trig

functions positive in the first quadrant. Students, stands for only sine and its reciprocal (cosecant) are

positive in the second quadrant. Take, stands for only tangent and its reciprocal (cotangent) are

positive in the third quadrant. Calculus, stands for only cosine and its reciprocal (secant) are positive

in the fourth quadrant.

Let’s do one more example to remind you of the first definition once again.

Example 2: Finding Trigonometric ValuesWhat are the values of the following given the picture of θ? (Hint: you will want to apply the first

definition and not the unit circle definition.)

a) tan θ b) sin θ c) sec θ d) cos θ e) cot θ

Solution:

a) Tangent is defined to be . tan θ =

b) Sine is defined to be . We need to find r

(x2 + y2 = r2).

sin θ =

c) Secant is the reciprocal of cosine so secant

is defined to be .

sec θ =

d) Cosine is defined to be . cos θ =

e) Cotangent is the reciprocal of tangent so it

is defined to be .

cot θ =

θ

P (5, 6)

AllStudents

Take Calculus


Example 3: Using the Calculator

What are the output values of the following trig functions?

a) tan 17° b) cot 126° c) sec (78°) d) csc 341°

Solution:

A word of caution is needed here. When using the graphing calculator with trigonometric

functions it is very important as to whether the calculator is set in radian or degree mode.

The best rule of thumb is: if the angles are given in radians then the calculator should be in

radian mode and if the angles are given in degrees then the calculator should be in degree

mode.

a) To find the tan 17° simply push the TAN

key followed by 17, then ENTER. Be

sure that your calculator is in degree

mode.

b) To find the cot 126° use the fact that

cotangent is the reciprocal of tangent.

Push 1 then / then TAN followed by

126, then ENTER.

c) To find the sec (78°) use the fact that

secant is the reciprocal of cosine. Push 1

then / then COS followed by 78, then

ENTER.

d) To find the csc 341° use the fact that

cosecant is the reciprocal of sine. Push 1

then / then SIN followed by 341, then

ENTER.

The last thing we should mention here is that sine is an odd function and cosine is an even function.

You might remember from chapter 3 that an odd function is a function whose graph is symmetric to

the origin and an even function is a function whose graph is symmetric to the y-axis. It is true that for

even functions f(−x) = f(x) and for odd functions f(−x) = −f(x). The other four trigonometric

functions can be found from these two, thus the following is true for the six

trigonometric functions.

Odd Properties − For any angle θ the following is true.

sin (−θ) = −sin (θ), csc (−θ) = −csc (θ), tan (−θ) = −tan (θ), cot (−θ) = −cot (θ)


Even Properties − For any angle θ the following is true.

cos (−θ) = cos (θ), sec (−θ) = sec (θ)

Section Summary: The graphs of the six trig functions are:

Notice that in order to graph cosecant, secant, and cotangent with the graphing calculator

you need to use the fact that they are the reciprocals of sine, cosine and tangent.

Domain and range of the six trig functions and their period.

Domain Range Periodic

Sine Function (−∞, ∞) [−1, 1] 2π

Cosine Function (−∞, ∞) [−1, 1] 2π

Tangent Function {… , , …} (−∞, ∞) π

Cotangent Function {…(−2π, −π), (−π, 0), (0, π), (π, 2π)…} (−∞, ∞) π

Secant Function {… , , …} (−∞, −1) (1, ∞) 2π

Cosecant Function {…(−2π, −π), (−π, 0), (0, π), (π, 2π)…} (−∞, −1) (1, ∞) 2π

Reciprocal relationships cot θ = , sec θ = , csc θ = .

The first definitions of the six trigonometric functions sometimes refered to as the coordinate

definitions are: sin θ = , cos θ = , tan θ = , csc θ = , sec θ = , cot θ = .

The second definitions which are the unit circle definitions are:

sin θ = y, cos θ = x, tan θ = , csc θ = , sec θ = , cot θ = .

The even functions are cosine and secant [ f(−x) = f(x)]


The odd functions are sine, cosecant, tangent, and cotangent [ f(−x) = −f(x)]


SECTION 5.4 PRACTICE SET(116) For the given value of find the exact value for tan , cot , sec and csc .

1. 2. 3.

4. 5. 6.

7. radians 8. radians 9. radians


13. 14. 15. radians

16. radians

(17−26) For the given value of approximate to 4 decimal places the value of tan , cot , sec and csc .

17. 18. 19.

20. 21. 22.


26. radians

(27−32) Give all the other angles in degrees from –360o to 360o that will give the same value as each of the following:

27. tan 105o 28. cot 225o 29. sin 321o

30. cos 78o 31. sec 70o 32. csc 330o

(33−38) Give all the other angles in radians from −2 to 2 that will give the same value as each of the following:

33. 34. 35.

36. 37. 38.


(39−46) The angle is in standard position and (x, y) is a point on the terminal side of the angle. For the given value of (x, y) find the value for tan , cot , sec and csc :

39. (3, 4) 40. (5, 12) 41. (5, 3)

42. (4, 2) 43. (0, 3) 44. (0, 4)

45. (2, 0) 46. (5, 0)

(4752)

47. If then csc =?

48. If then sin = ?

49. If then cot = ?

50. If then tan = ?

51. If then sin = ?

52. If then cos = ?

(5360) Use the calculator to test the following relationships:

53. tan 38o

54. cot 123o

55. sin 238o

56. cos 332o

57.

58. cot 59o


59.

60.

(6164) (a,b) r b 1 y x a

Use the value of (a, b) that is given to find r and (x, y). Use (a, b) and r to find the value of tan ,cot , sec and csc using the coordinate system definition. Use (x, y) and the unit circle definition to find the value of the four functions. (Hint: to find the value of (x, y) use similar triangles)

61. (a, b) = (3, 4) (x, y) =? tan = ? cot = ? sec = ? csc = ?




65. What did you find out about the coordinate definition answers and the unit circle answers from the problems 61 to 64?

66. What is the domain of tan ?

67. What is the range of tan ?

68. What is the range of cot ?

69. What is the domain of cot ?

70. What is the domain of sec ?

71. What is the range of sec ?

72. What is the range of csc ?

73. What is the domain of csc ?


Section 5.5 Inverse Trigonometric Functions

Objectives

Understanding the inverse of sine and cosine

Understanding all the other inverse trigonometric functions

Understanding the composition of trigonometric functions

As we begin our look at the inverses of trigonometric functions we need to review what was

discussed in chapter 1. Here is a list of facts from chapter 1.

Functions are simply special relations where for each input there is only one output.

Notation is y = f(x) where x is the input, the independent variable, and f(x) is the output, the

dependent variable.

Domain is the set of all possible inputs, and the range is the set of all resulting outputs.

You can add, subtract, multiply, divide and compose functions.

An inverse “undoes” something by doing the opposite operations in the opposite order.

A relation is one-to-one if for each output there is only one input.

For a relation to have an inverse function it must be a function and it must be one-to-one.

There are two ways to find an inverse function: one is to solve for the independent variable

and then switch the two variables; the second, is to switch the two variables and then solve

for the dependent variable. In short, either solve for x then switch x and y or switch x and y

and then solve for y.

If the function g is the inverse of function f , then the domain of f(x) (Df) is the same as the

range of g(x) (Rg) and the range of f(x) (Rf) is the same as the domain of g(x) (Dg) .

Inverse function graphs are mirror images of each other around the line y = x .

INVERSE SINE AND INVERSE COSINE

Discussion 1: Finding Inverse SineLet’s take a look at the sine graph once again.

At first glance you should notice that the graph passes the vertical line test so we can visually see that

it is a function (no vertical lines will cross our graph more than once). But next, you should also

notice that it doesn’t pass the horizontal line test for determining if something is one-to-one (no

2π−2π

y

θ

y = sin θ1

−1


horizontal lines cross the graph more than once). Thus, to quote a famous statement “Houston we

have a problem”! We will get around this problem by adding a restriction to the domain of the sine

function. This is similar to what you saw us do with the function x2. By restricting how much of the

graph we are going to look at we can get something that will pass the horizontal line test.

Question 1: What part of the graph, and thus what domain, should we restrict ourselves too?

There are many possible answers you could have given to question 1 that would be correct. The one

most often used in the science/mathematics world is to restrict the domain to . Notice that

this keeps the most amount of the graph that one can and still maintain a one-to-one function and it is

near the origin. Thus as we begin talking about the inverse of sine we will use this graph for sine.

In order to find an inverse we can switch the two variables and solve.

Original function y = sin x

Switch the two variables x = sin y

Now solve for y Can’t solve, don’t know how. This happened

to us when we discussed the logarithm

function. There we made a definition, so

here we will do the same.

Inverse Sine − Given, y = sin x with domain and range [−1, 1], inverse sine is

defined to be:

y = sin -1 x, where the domain is [−1, 1] and the range is and x = sin y.

(Notice how the independent and dependent variables switch positions in inverse functions.)

The notation sin -1 is sometimes written arcsin. Thus, the two ways you will see the function inverse

sine written are: y = sin 1 x, or y = arcsin x.

1

2

4

−1

θ

y

2

4


We can make a graph of sin -1 x by using the fact that the graph of an inverse is the mirror image

of the original graph about the line y = x. Here is a sketch of sin -1 x, a TI calculator screen shot of

inverse sine, and a calculator screen shot of the sine and the inverse sine functions and the line y = x.

Notice the mirror image we have in the third picture.

Example 1: Finding Inverse Sine Values

What are the missing values in the following table for the function y = sin-1 x?

x −1 0 1

y

Solution:

First remember that the inputs of the inverse

are the outputs of the original function, so a

table of values for y = sin x should be

helpful to us.

x y x y

−1

1

2

4

−1x

y

2

4


1

0 0

All we need to do is take the inputs from y = sin x and use those for the outputs of

y = sin -1 x.

x −1 0 1

y 0

Example 2: Finding Inverse Sine Values from the Calculator

What are the values of the following?

a) sin -1 (0.2) b) sin -1 (−0.92) c) sin -1 (1.2)

Solution:

a) To find the inverse sine of a number that

doesn’t yield a standard angle we will use

the graphing calculator. One word of

caution, make sure the calculator is on

radian mode or degree mode depending

on the type of angle answer you are

wanting.

In this example the angle answers are in

radian measure. Here is what we see on

our calculators. SIN −1 is above the SIN

key.

b) Type this into the calculator to see the

answer.

Answer Q1:We would look at the first part of the graph near the origin that goes uphill. The domain would be

.


c) Do the same as for the first two

examples.

Notice that we get an error message about

the domain. Our domain for sin -1 x is [−1,1]

and so 1.2 can’t be used as an input value.

Question 2: With the function y = sin x, we are either substituting in real numbers, that are angle

measurements in radian measure, or angles in degree measure for the input values and we get real

number outputs. What are the kinds of inputs and outputs that correspond to the function y = sin -1 x?

In essence, we have one function with angle inputs yielding number answers and the inverse function

where we have number inputs yielding angle answers. Keeping this in mind could really save you

some mistakes down the road in trigonometry.

Please keep in mind also that we defined an angle in radian measure to be the arc length of a

piece of a unit circle. So the angle is a real number which we need it to be for many areas of

mathematics but yet these same real numbers are representing an angle.

Discussion 2: Finding Inverse CosineHere is a graph of the cosine function.

Notice that cosine isn’t one-to-one just like the sine function isn’t one-to-one, unless we restrict its

domain. With this function the standard definition for inverse cosine is:

Inverse Cosine − Given y = cos x with domain and range [−1, 1], inverse cosine is

defined to be:

y = cos -1 x, where the domain is [−1, 1] and the range is and x = cos y.

The notation cos -1 is sometimes written arccos. Thus, the two ways you will see the function inverse

cosine written are: y = cos 1 x, or y = arccos x.

We can make a graph of cos -1 x by using the fact that the graph of an inverse is the mirror image of

the original graph about the line y = x. Here is a sketch of cos -1 x, a TI calculator screen shot of

inverse cosine, and a calculator screen of cosine, inverse cosine and the line y = x.

2π−2π

y

θ

y = cos θ1

−1


Notice the mirror image we have in the third picture.

Example 3: Finding Inverse Cosine Values

What are the missing values in the following table for the function y = cos-1 x?

x −1 0 1

y

Solution:

First remember that the inputs of the inverse

are the outputs of the original function, so a

table of values for y = cos x should be

helpful to us.

x y x y

0 1

π −1

1

2

−1x

y

34

4

π


0

All we need to do is take the inputs from y = cos x and use those for the outputs of

y = cos -1 x.

x

−1 0 1

y π 0

Example 4: Finding Outputs for sin -1 x and cos -1 x


a) cos -1 b) cos -1 c) sin -1 d) cos -1 (1.5)

Solution:

a) To find the cos -1 we need to think

find x such that = cos x. Inputs in the

inverse function are outputs in the

original function.

So, what angle makes cos x = ?

Answer: rad (the angle does also

along with many others but they aren’t in the

range of inverse cosine [0, π])

Thus, cos -1 =

b) To find the cos -1 we need to think

find x such that = cos x.

The angle is . (the angle does also


range of inverse cosine [0, π])

Thus, cos -1 =

c) To find the sin -1 we need to think The angle is . (the angle does also


Answer Q2:The inputs are now real numbers and the outputs are angles either, in real number form as radians or angle form in degrees.


find x such that = sin x. range of inverse sine )

Thus, sin -1 =

d) The value of cos -1 (1.5) can’t be

determined.

1.5 is not in the domain of the inverse cosine

function. Another way to look at this is that

there isn’t any angle where cos x = 1.5.

Here is a good rule of thumb.

If you are asked to find an inverse trigonometric function value, try thinking in terms of

what angle would yield the input for the inverse function. This is very much like finding

the log 5 125. There we talked about thinking in terms of “What exponent should 5 be

raised to in order to get 125 as an answer”. With trigonometry if you are asked to find

sin -1 (0.5) you should be thinking what angle does one input into the sine function in order

to get 0.5 as an answer. In this example the angle would be a angle.

THE OTHER FOUR INVERSES

Discussion 3: Finding More Inverses

We now must look at the other four trigonometric functions. Here are their graphs.

y = tan x y = cot x

y = sec x y = csc x

Notice that like sine and cosine none of these are one-to-one functions. We will need to restrict the

domain on each of these functions in order to find their inverse functions.

For y = tan x let’s take the branch of the tangent function near the origin.


Therefore, the domain we will consider is . This means that the inverse will have a range of

and a domain of (−∞, ∞). Here is the graph of tan -1 x or if you will arctan x.

For y = cot x let’s take the branch of the cotangent function to the right of the y-axis.

Therefore, the domain we will consider is (0, π). This means that the inverse will have a range of (0,

π) and a domain of (−∞, ∞). Here is the graph of cot -1 x or if you will arccot x.

There isn’t a standard way of restricting the secant function (y = sec x).

The two common choices for restricting the domain are or . The first

makes an identity work, the second allows for both positive and negative sloped tangent lines to the

curve. In this book we are going to choose the later, . This means that the inverse


will have a range of and a domain of (−∞, −1] [1, ∞). Here is the graph of sec -1 x

or if you will arcsec x.

There isn’t a standard way of restricting the cosecant function (y = csc x).

The two common choices for restricting the domain are or . The

first makes an identity work, the second allows for both positive and negative sloped tangent lines to

the curve. In this book we are going to choose the later, . This means that the inverse

will have a range of and a domain of (−∞, −1] [1, ∞). Here is the graph of csc -1 x

or if you will arccsc x.

Here is a summary of the six inverses and their domains, ranges and graphs. It will

be very important to have this information memorized!

Function Domain Range Graph

y = sin -1 x [−1, 1]


y = cos -1 x [−1, 1] [0, π]

y = tan -1 x (−∞, ∞)

y = cot -1 x (−∞, ∞)

y = sec -1 x (−∞, −1] [1, ∞)

y = csc -1 x (−∞, −1] [1, ∞)

Example 5: Evaluating Inverse Trigonometric Functions

What are the values of the following:

a) sin -1 (0.5) b) cos -1 c) tan -1 d) sec -1 (−2) e) csc -1 (1)

Solution:

a) To find the sin -1 (0.5) it might be easiest

to think solve 0.5 = sin x.

Inverse sine range is .

The unit circle points with y coordinate of 0.5

are associated with the angles of and .

(Answers must come from 1st or 4th quadrants)

Therefore, sin -1 (0.5) = .


b) As with part (a), solving for x in the

equation = cos x might be best.

Inverse cosine range is [0, π].

The unit circle points with x coordinate of

are associated with the angles of and .

(Answers must come from 1st and 2nd quadrants)

Therefore, cos -1 = .

c) Like the previous two examples, solve

− = tan x.

Inverse tangent range is .

The unit circle points, where is , are the

points associated with the angles and .

(Answers must come from 1st and 4th quadrants)

Therefore, tan -1 = .

d) To find the answer solve −2 = sec x. It

might be easier if we think,

−2= = cos x.

Inverse secant range is .

The unit circle points, where x is , are the

points associated with the angles and .

(Answers must come from 1st and 3rd quadrants)

Therefore, sec -1 (−2) = .

e) To find the answer solve 1 = csc x. It

might be easier if we think,

1 = 1 = sin x.

Inverse cosecant range is .

The unit circle point, where y is 1, is the point

associated with the angle .

There is only one choice and it is in our range.

Therefore, csc -1 (1) = .

Discussion 4: Evaluating Inverse Trigonometric Functions with the CalculatorLet’s see how we could evaluate the following using the graphing calculator.

a) b) c) d) e)

a) Here we just simply push 2nd and the COS

key and then our number followed by

ENTER (answer in degrees since the

calculator was set to degree mode).


b) Here we just simply push 2nd and the TAN

key and then our number followed by

ENTER (answer in degrees since the

calculator was set to degree mode).

c) To graph, or evaluate for some

value of x, you must know that

.

Thus for our problem we will use the

second half of the formula since our input

value is negative.

Make sure that the calculator is in radian

mode if you use π, or degree mode if you use

180° in the formula.

d) To graph, or evaluate for some


.

Thus for our problem we will use the

second half of the formula since our input

value is negative.


mode if you use 2π, or degree mode if you

use 360° in the formula.

e) To graph, or evaluate for some


.

Thus for our problem we will use the first

half of the formula since our input value is

positive.


mode if you use π, or degree mode if you use

180° in the formula.

COMPOSITION OF TRIGONOMETRIC FUNCTIONS

Discussion 5: Composing Trigonometric FunctionsTo serve as a refresher, remember that if we had f(x) = x2 − 4 and g(x) = x + 2 then (f g)(3) = f (g(3))

= f (5) (since g(3) = 3 + 2 = 5) = 21 (since 52 − 4 = 21). Composition stated simply, is just

substitution. We will need to be able to do the same with trigonometric functions. Let’s look at and

evaluate the following problems.


a) sin (cos -1 0.5) b) tan -1 (sin π) c) sec (sec -1 (−1)) d) sin -1

a) We start with the inner most

parenthesis and find the cos -1 0.5cos -1 0.5 = (since cos = 0.5 and the angle

is in the range of inverse

cosine.)

Now find the sin of our answer.sin = .

Therefore, sin (cos -1 0.5) = .

b) Once again begin by finding sin π. sin π = 0 (since π puts the terminal side on

the negative x-axis which has the

unit circle point of (−1, 0))

Now evaluate tan -1 0. tan -1 0 = 0

Therefore, tan -1 (sin π) = 0.

c) Start with finding sec -1 (−1) sec -1 (−1) = π (since negative input values yield

outputs in the interval )

Now evaluate sec π. sec π = −1 (you should not be surprised by this

answer. Remember that inverses

cancel each other out and you get

just the input as the answer. See

discussion following question 3.)

Therefore, sec (sec -1 (−1)) = −1

d) We start with = (the reference angle is in the

fourth quadrant)

Now evaluate sin -1 . sin -1 =

Therefore, sin -1 =

Question 3: What is the value of cos -1 ?


As you answered the last question you may have thought the answer was . That shows that you

are understanding the concept of the composition of inverses, but it isn’t correct in this case because

the inverse cosine function only exists with a restricted domain on cosine. Therefore, can’t be

the correct answer because when we defined inverse cosine we had used the part of cosine from 0 to

π. In order to evaluate cos -1 we need to first change the angle to an equivalent one in

the proper domain. For the cosine function that would be . Both and have the same

reference angle and the cosine of both is negative. Therefore, cos -1 can be changed to

cos -1 which then can be evaluated and we get the answer . Our statement in discussion

5c, “…inverses cancel each other out and you just get the input as the answer”, is only true when you

are working with 1 to 1 functions or, the restricted domains of non 1 to 1 functions. If you are using

input values that are not in the restricted domains of non 1 to 1 functions, this rule will not hold true.

For problems similar to these, but where the values are not the standard ones, we sometimes

need to go back to our original definitions with the right triangles. Let’s do one last example.

Example 6: Finding Compositions Using Triangles


a) cos (sin -1 ) b) cot (csc -1 (−3)) c) tan (cos -1 ) d) sec (sec -1 x)

Solution:

a) First we could use the graphing

calculator, but we’ll get to that

later. What we can do in order to

get the answer by hand is sketch a

picture of an angle where the sine

of it would yield the answer .

Remember from section 5.3 sine

was defined to equal .

By the Pythagorean Theorem we can find that x=4.

We now have a picture of the answer to sin -1 =θ.

cos θ = = .

Therefore, cos (sin -1 ) =

b) We will use the same approach

here as we did for part (a). We

need to draw a picture of an angle = csc θ = = = (notice that we put

θy = 3

r = 5

x = ?


where csc θ = −3

Notice that we drew the picture of

the angle in the third quadrant

since that is the defined restricted

domain for cosecant with negative

outputs.

Definition section 5.3, cot θ =

the negative sign in the denominator since r is in

the numerator and it can’t be a negative value) so,

we get a picture that looks like this.

x2 = 32 − (−1)2 = 9 − 1 = 8, so x = (x in

our picture is in the negative direction)

cot θ = = (reciprocal of tangent)

Therefore, cot (csc -1 (−3)) =

c) We need to use the same approach

as we have and start off by drawing

a picture of the angle cos -1 .

We know that cos θ = and the

definition cos θ = .

y2 = 92 − 72 = 81 − 49 = 32, so y =

tan θ = =

Therefore, tan (cos -1 ) =

d) We draw a picture of the angle

sec -1 x. We know that sec θ =

and the definition

sec θ = = = .

The answer should not be a

surprise. Inverses undo each other,

as long as we are working within

the restricted domains for

trigonometric functions.

We will draw our picture with r = x and the x

coordinate equal to 1.

sec θ = = = x

Therefore, sec (sec -1 x) = x

Question 4: What is the value of cos (sin -1 a)?

θy = −1r = 3

x = ?

θy = ?

r = 9

x = 7

θy = ?

r = x

x = 1

Answer Q3:

is not in the

domain for cosine inverse. Thus we change to an angle which has the same value,

which is .

So now we evaluate

cos -1

= .


Section Summary: Here are the domains, ranges and graphs of the six trigonometric inverses.


y = sin -1 x [−1, 1]

y = cos -1 x [−1, 1] [0, π]

y = tan -1 x (−∞, ∞)

y = cot -1 x (−∞, ∞)

y = sec -1 x (−∞, −1] [1, ∞)

y = csc -1 x (−∞, −1] [1, ∞)

Remember, when working with a problem like trig -1 (trig θ), the trigonometric inverse of a

trigonometric function, that we must stay within the restricted defined domains and ranges.


SECTION 5.5 PRACTICE SET(1–24) Evaluate each expression without using the calculator. (Give your answer in radians)

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 21.

22. 23. 24.

(25–48) Evaluate each expression without using the calculator. (Give your answer in degrees)

25. 26. 27.

28. 29. 30.

31. 32. 33.

34. 35. 36.

37. 38. 39.

40. 41. 42.

43. 44. 45.

46. 47. 48. (49–80) Evaluate each expression without using a calculator.

Answer Q4:We make a triangle first which is a picture of the angle sin -1 a = θ.

θ y = a

r = 1

x = 21 a The cos θ =


49. 50. 51.

52. 53. 54.

55. 56. 57.

58. 59. 60.

61. 62. 63.

64. 65. 66.

67. 68. 69.

70. 71. 72.

73. 74. 75.

76. 77. 78.

79. 80.

(81–96) Evaluate each expression without using a calculator.

81. 82. 83.

84. 85. 86.

87. 88. 89.


90. 91. 92.

93. 94. 95.

96.

(97–102) Give the domain and range of each function.

97. sin–1 98. arccos 99. arctan

100. cot–1 101. sec–1 102. arccsc

(103–126) Use the calculator to evaluate each expression. (If possible)

103. 104. 105.

106. 107. 108.

109. 110. 111.

112. 113. 114.

115. 116. 117.

118. 119. 120.

121. 122. 123.

124. 125. 126.

127. 128.

(129–140) Arcsin , Arccos , Arctan and Arccot represent the inverse relations of each corresponding trigonometric function. For each of the following give an expression that will give all the values for the inverse relations.


129. 130. 131.

132. 133. 134.

135. 136. 137.

138. 139. 140.

141. For problems 123 thru 128 why did the calculator give you an error reading?


Section 5.6 More On Graphing Trigonometric Functions

Objectives

Understanding sine and cosine translations

In chapter 3 we looked at vertical and horizontal shifts in a graph and what causes them along with

rotations about the x-axis (flipping), rotations about the y-axis (rotating), stretching and compressing.

Let’s look at these same ideas but with trigonometric functions.

TRANSLATIONS OF THE SINE AND COSINE FUNCTIONS

Let’s see a summary of the chapter 3 material that relates to what we wish to learn in this section.

Vertical translation: f(x) + k, means the graph shifts up (k > 0) or down (k < 0) k units. Notice

that by adding something to a function you change the y values (vertical shift).

Horizontal translation: f(x + h), means the graph shifts left (h > 0) or right (h < 0) h units.

Notice that by adding something to the x variable in a function you change the x values

(horizontal shift).

Rotations about the x-axis (flipping): f(x), means the graph flips upside down. Notice that by

multiplying by a negative a function changes its y value signs (Flipped upside down).

Rotations about the y-axis (rotating): f(x), means the graph rotates around the y-axis. Notice

that by multiplying the x variable by a negative a function changes its x value signs (rotating

around the y-axis).

Stretching or compressing: af(x), means the graph is narrower if is larger than 1 and wider

if is less than 1. Notice that multiplying a function by a number changes its y value

magnitudes (stretching or compressing the graph).

These same properties will hold true with any function so let’s see how it looks in the world of

trigonometry.

Question 1: How do you think the graph of y = sin (x) + 4 will differ from y = sin (x)?

Discussion 1: Vertical and Horizontal TranslationsLet’s look at how the following functions differ from their parent functions.

a) y = cos (x) − 2 b) y = sin (x − ) c) y = sin (x + ) + 3

a) The parent function would be

y = cos (x). Since we see that this

function is subtracting 2 from

As you can see from the graphing calculator the

graph of our function is 2 units lower.


every output in the parent function

the graph should move down 2

units.

b) The parent function would be

y = sin (x). Since we see that this

function tells us to subtract

from every input in the parent

function the graph should move

right units.


graph of our function is units to the

right.

c) The parent function would be


function is adding to every

input and adding 3 to every output

as compared to the parent function

the graph should move left

units and up 3 units.


graph of our function is units to the left

and 3 units up.

We have now seen a few examples that fit right in with what was learned in chapter 3. We see here

as before that by adding to the dependent variable (y) the graph moves up or down and that by adding

to the independent variable (x) the graph moves left or right. Let’s try a couple of more examples.

Example 1: Vertical Stretching and Compressing (Amplitude)How do the graphs of the following different from their parent graphs?

a) y = 3sin (x) b) y = cos (x) c) y = −5sin (x)

Solution:



function is multiplying every

output in the parent function by 3

the graph should stretch by a

factor of 3.


graph of our function is 3 times taller.

b) The parent function would be

y = cos (x). Since we see that this




output in the parent function by

the graph should compress by a

factor of .

graph of our function is times shorter.




output in the parent function by

−5 the graph should stretch by a

factor of 5 and be flipped upside

down.


graph of our function is 5 times taller but flipped

upside down because of the negative sign.

The magnitude of the coefficient in front of a trigonometric function is called the amplitude. It

effects how much the graph is stretched or compressed vertically. The sign of the coefficient

determines if the graph is rotated about the x-axis (flipped upside down or not) by whether the

coefficient is positive or negative.

It is now time to address horizontal stretching and compressing. This is a bit more complicated and

affects two concepts: period and phase shift. Let’s do a few examples.

Discussion 2: Horizontal Stretching and Compressing (Period)Here we are going to look at examples of the form y = sin (bx). Let’s see how the following functions

differ from their parent functions.

a) y = sin (2x) b) y = cos c) y = cos (3x) d) y = sin (−x)


y = sin (x). Let’s see how the two compare.

Notice that the second graph is more

compact. The parent graph must go from 0

to 2π before covering a whole period (sine

is 2π periodic). But, the function

y = sin (2x), as we can see, has a period of

only π. The second function has been

compressed horizontally by a factor of two.

2π 6.28, 3.14

b) The parent function would be 4π 12.57

Answer Q1:Probably it will be four units higher (up four).


y = cos (x). Let’s see how the two compare.


spread out. The parent graph must go from

0 to 2π before covering a whole period

(cosine is 2π periodic). But, the function

y = cos , as we can see, has a period of

4π. The second function has been stretched

by a factor of two.


y = cos (x). Let’s see how the two compare.


compact. The parent graph must go from 0

to 2π before covering a whole period. But,

the function y = cos (3x), as we can see,

has a period of . The second function

has been compressed by a factor of three.

d) The parent function would be

y = sin (x). Let’s see how the two compare.

Your first thought might be that the second

graph is flipped upside down from the first.

But, that can’t be correct because that only

happens when you are multiplying the

whole function by a negative. If you

remember from chapter 3, when

multiplying x by a negative you rotate the

graph around the y-axis. That is what has

happened in this case.

2π 6.28

Let’s see if you understand the pattern yet after just a few examples.


Question 2: How do we figure out whether ae function has been compressed or stretched horizontally

and by what factor? (In essence this question is asking how do we find the period of a sine or cosine

function.)

The period of a sine or cosine function isn’t difficult to evaluate but you do need to take some care or

you will quickly find out how easy it is to make a mistake.

If given a function y = asin (bx − c) or y = acos (bx − c), where b is a positive number, then

the period is calculated as follows.

Period = .

The last new concept is phase shift. Let’s once again look at a few examples.

Discussion 3: A More Complete Look at Horizontal Shifts (Phase shift)Let’s find the phase shift for each of the following functions.

a) y = sin (x − π) b) y = sin (2x + ) c) y = cos (3x − )

a) This is a problem just like the one we looked at

in discussion 1(b). This time though let’s

approach the problem not from a chapter 3

approach but from the idea of figuring out how

far we have moved from the origin (this method

could be used in chapter 3 as well). What we

will do is set the input quantity equal to zero to

see what x-value will yield a zero input value.

y = sin (x − π) given

(x − π) = 0 set equal to zero

x = π solved for x

We now know that the graph of the

function y = sin (x − π) will be shifted

π units to the right from its parent

function y = sin (x).

b) As in part (a) above let’s set the input quantity

equal to zero and then solve for x. This will

help us to find the phase shift for this function.

y = sin (2x + ) given

(2x + ) = 0 set equal to zero


function y = sin (2x + ) will be

shifted units to the left from its

parent function y = sin (x).


2x = subtracted

x = multiplied by

You should also notice that this graph will have

a period of π since .

c) As in part (a) above let’s set the input quantity

equal to zero and then solve for x. This will

help us to find the phase shift for this function.

y = cos (3x − ) given

(3x − ) = 0 set equal to zero

3x = added

x = multiplied by

You should also notice that this graph will have

a period of since .


function y = cos (3x − ) will be

shifted units to the right from its

parent function y = cos (x).

The General Form

If given a function y = asin (bx − c) + d or y = acos (bx − c) + d then the following are

true.

is the amplitude of the function (how the graph has been stretched or

compressed vertically).

= period of the function (how long it takes for the function to repeat itself).

bx − c = 0 x = is the phase shift (how much the graph has been shifted

horizontally).

d is the amount the graph has been shifted vertically.

Answer Q2:It looks as though you just divide 2π by the number in front of x.


If , then the graph is rotated about the x-axis (flipped vertically).

If , then the graph is rotated horizontally.

Domain (∞, ∞)

Range , a stretches the range and d shifts it up or down.

Example 2: Sketching Graphs of Sine and Cosine

What are the graphs of the following functions?

a) y = 2cos (6x + 5π) − 3 b) y = −sin (2x − 3π) + 1

Solution:

a) From the function we see that a = 2, b = 6,

c = −5π, and d = −3. Therefore, we know that

the graph is shifted down 3, that it will have

an amplitude of 2 ( ) and a period of

. We find the phase shift by

solving (6x + 5π) = 0 for x.

(6x + 5π) = 0

6x = − 5π

x =

As we sketch the graph we center it on

the line y = −3 (shifted down 3), make

it go up and down 2 units from the

center line (amplitude = 2), repeat itself

every , and we start the graph to

the left of the origin.

b) From the function we see that a = −1, b = 2,

c = 3π, and d = 1. Therefore, we know that

the graph is flipped vertically (turned upside

down) since a is negative, shifted up 1, that it

will have an amplitude of 1 ( ) and

a period of . We find the phase

shift by solving (2x − 3π) = 0 for x.

(2x − 3π) = 0

2x = 3π

x =

As we sketch the graph we center it on

the line y = 1 (shifted up 1), make it go

up and down 1 units from the center

line (amplitude = 1), repeat itself every

π, and we start the graph to the

right of the origin. The negative a

means the graph is flipped.


Let’s take a look at how we could use this information to help us model a real life event.

Discussion 4: Weather Model

As you are aware the weather from day to day doesn’t change much. Or the weather

from year to year follows a pattern, hot in the summer and cooler in the winter.

Since weather tends to be cyclical, repeats itself, the sine or cosine functions can be

used to model temperature fairly well. In section 1.1 discussion 8 we looked at how

temperature was a function of time. Here is the table we used.

Month Jan.

0

Feb.

1

Mar.

2

Apr.

3

May

4

June

5

July

6

Aug.

7

Sept.

8

Oct.

9

Nov.

10

Dec.

11

AverageTemp. 54 58 62 70 79 88 94 92 86 75 62 54

Let’s use the information we have learned about translating sine and cosine to derive a function we

could use to model this yearly temperature change. Let’s be loose about the exactness of our model.

We need to find values for a, b, c, and d for the formula y = asin (bx − c) + d.

Question 3: What do you think the d in the formula might represent in a real life example?

First, we should find a number for d. d gives

the function its vertical translation. It would

represent the average between the high and

the low. The a, amplitude, would tell us how

far the high and low are from the average.

Given that the high is 94 and the low is 54

the average would be = 74. Since

d = 74, a would be 20, the amount up and

down to the high and low.

d = 74, a = 20

Next, we need to find a value for b.

Remember that b is involved with the

period.

The period in this example is 12, it takes 12

months before you begin to repeat.

Therefore,

Lastly, we need to find a value for c. c is The x then is either month 3 or 4. They are


involved with the phase shift. We need to set

to find phase shift. In this example

we have found b, which is and x would

need to be the month when it is the average

temperature and increasing because this is

the starting point for sine. The point at the

origin of the parent sine function.

near the average value and the months

following them have higher temps. Thus we

need to solve this for c.

Our model then is,

y = 20sin + 74

Here is a graph of the model along with the data points.

As you can see our model does fit the data fairly well.

Section Summary:If given a function y = asin (bx − c) + d or y = acos (bx − c) + d then the following are true.

is the amplitude of the function (how the graph has been stretched or compressed

vertically). It tells you how far it is to the high and low points on the graph from the average

height (d).

= period of the function (how long it takes for the function to repeat itself).

bx − c = 0 x = is the phase shift (how much the graph has been shifted horizontally).

d is the amount the graph has been shifted vertically (average height).




SECTION 5.6 PRACTICE SET(1–46) For each function give:

a.) The amplitudeb.) The periodc.) The phase shiftd.) The vertical changee.) Flipped vertically or horizontallyf.) The domain of the functiong.) The range of the functionh.) Graph sketched

1. 2. 3.

4. 5. 6.

7. 8. 9.

10. 11. 12.

13. 14. 15.

16. 17. 18.

19. 20. 21.

22. 23. 24.

25. 26. 27.

28. 29. 30.

31. 32. 33.

34. 35. 36.

37. 38. 39.

40. 41. 42.

Answer Q3:It is the amount the graph gets shifted up so this would be the average temperature over the entire year. It’s the middle amount.


43. 44. 45.

46.

(47–48) 47. The following is a table for average temperatures by month for a certain area of the country.

Month Jan.0

Feb.1

March2

April3

May4

June5

July6

Aug.7

Sept.8

Oct.9

Nov.10

Dec.

11AverageTemp. 49 51 55 62 69 87 93 91 87 80 68 58

a. Find an equation of y = asin(bx – c) + d that fits this data.

b. Check the equation and the data by graphing the data points and the equation on the same graph using the graphing calculator.

48. The following is a table for average temperatures by month for a certain area of the country.

Month Jan.0

Feb.1

March2

April3

May4

June5

July6

Aug.7

Sept.8

Oct.9

Nov.10

Dec.

11AverageTemp. 54 57 63 74 83 92 96 94 90 81 67 59

a. Find an equation of y = asin(bx – c) + d that fits this data.

b. Check the equation and the data by graphing the data points and the equation on the same graph using the graphing calculator.


Chapter 5 Review

Topic Section Key Points

Intersecting

lines

5.1 If you have two parallel lines intersected by a transversal then the follow are

true:

Truth 1 Alternate interior angles have equal measure.

Truth 2 Corresponding angles have equal measure.

Truth 3 Alternate Exterior angles have equal measure

With any intersecting lines the following is true:

Truth 4 Vertical angles have equal measure

Triangles 5.1 Two triangles will be similar when:

Truth 5 Corresponding sides are proportional.

Truth 6 They have the equivalent angles.

Angles 5.1 Truth 7 The sum of complementary angles is 90°.

Truth 8 The sum of supplementary angles is 180°.

Truth 9 The sum of three angles of a triangle is 180°.

Special

Right

Triangles

5.1 Truth 10 The ratio of the sides of a 30-60-90 right triangle are;


Angles 5.1 Truth 12 The acute angles of a right triangle are complementary.

Basics of

Trig.

5.2 Vertex The common point where two rays begin that create an

angle.

Initial side The ray where an angle begins.

Terminal side The ray where an angle ends.

Positive angles Angles that rotate in a counterclockwise manner.

Negative angles Angles that rotate in a clockwise manner.

Minute one sixtieth of a degree.

Second One sixtieth of a minute.


Arc of a Circle − An arc of a circle is a piece of the circumference of

a circle that is intersected by a central angle.

Radian − One radian is the measure of a central angle (θ) that

intercepts an arc (s) that is equal in length to the radius of the circle.

60°

30°

2xx

3 x45°

45°

x

x2 x


5.2 A reference angle is an acute angle (measurement between 0° and

90°) between the terminal side of an angle in standard position and

the x-axis.

Conversion 180° is equivalent to π radians. When you need to

convert from one form to another use one of the following ratios:

.

Linear Speed The linear speed of an object connected to a wheel

will be v = , where v is the velocity of the object, s is the arc length

and t is the time.

Angular Speed − The angular speed of an object on a wheel will be ω

= , where ω read “omega” is the angular speed, θ is the angle (in

radians) and t is the time.

Sine Cosine

and the

Unit Circle

5.3 Pythagorean Theorem states that the square of the two legs of a right

triangle added

together will equal the square of the hypotenuse (a2 + b2 = c2).

Sine is defined as the y value of the point on a unit circle on the

terminal side of angle θ (sinθ = y).

Cosine is defined as the x value of the point on a unit circle on the

terminal side of angle θ (cosθ = x).

The first quadrant of the unit circle and the sign of the outputs of sine

and cosine are:

The domain for both sine and cosine is (−∞, ∞)

The range for both sine and cosine is [−1, 1]

A function is periodic if for some smallest c, f(x) = f(x + cn) where n

is any integer

Sine and cosine have a period of 2π.

3 1,

2 2

1,0

0,1

2 2,2 2

306

454

603

90

2

1 3,2 2

0 0

sin θ > 0cos θ > 0

sin θ > 0cos θ < 0

sin θ < 0cos θ < 0

sin θ < 0cos θ > 0


All Trig.

Functions

5.4 The graphs of the six trig functions are:

Domain and range of the six trig functions and their period.

Function Domain Range Periodic

Sine (−∞, ∞) [−1, 1] 2π

Cosine (−∞, ∞) [−1, 1] 2π

Tangent {… …} (−∞, ∞) π

Cotangent {… (−π, 0) …} (−∞, ∞) π

Secant {… …} (−∞, −1) (1, ∞) 2π

Cosecant {… (−π, 0) …} (−∞, −1) (1, ∞) 2π

Reciprocal relationships cot θ = , sec θ = , csc θ = .

The first definitions of the six trigonometric functions sometimes refered to as

the coordinate definitions are: sin θ = , cos θ = , tan θ = , csc θ = ,

sec θ = , cot θ = .

The second definitions which are the unit circle definitions are:

sin θ = y, cos θ = x, tan θ = , csc θ = , sec θ = , cot θ = .

The even functions are cosine and secant [ f(−x) = f(x)]

The odd functions are sine, cosecant, tangent, and cotangent [ f(−x) = −f(x)]


Inverse

Trig.

Functions

5.5 Here are the domains, ranges and graphs of the six trigonometric inverses.


y = sin -1 x [−1, 1]

y = cos -1 x [−1, 1] [0, π]

y = tan -1 x (−∞, ∞)

y = cot -1 x (−∞, ∞)

y = sec -1 x (−∞, −1] [1, ∞)

y = csc -1 x (−∞, −1] [1, ∞)

Remember, when working with a problem like trig -1 (trig θ), the trigonometric inverse of a trigonometric function, that we must stay within the restricted defined domains and ranges.

Shifting 5.6 If given a function y = asin (bx − c) + d or y = acos (bx − c) + d then:

is the amplitude of the function (how the graph has been stretched

or compressed vertically). It tells you how far it is to the high and low

points on the graph from the average height (d).

= period of the function (how long it takes for the function to

repeat itself).

bx − c = 0 x = is the phase shift (how much the graph has been

shifted horizontally).

d is the amount the graph has been shifted vertically (average height).



CHAPTER 5 REVIEW HOMEWORK

Section 5.1

1 2

(1−2)1. m 1 = 53o; m 2 = 2. m 1 = 3x + 20; m 2 = 5x −10; m 1= m 2 =

1 2 (3−4)3. m 2 = 33o; m 1 = 4. m 1 = 3x − 10; m 2 = 7x −20; m 1= m2 =

t 7 8 m line l is parallel to line m, with a transversal of t 5 6

4 3 l 2 1

(5−6)5. m1 = 1100 find the measure of each of the other angles.

6. m5 = 2x + 10 and m1 = 3x + 20 find the measure of all of the angles.

C

A B (7−8)7. mA = 33o; mC = 62o mB = 8. mB = 50o; mA = 2x + 10 mC =

D B e f F E c a d A C b (9−10) Given triangle DEF is similar to triangle ABC answer the following questions:9. e = 6; d = 10; f = 4; c = 12 a = b =

10. mE = 38o; mF =62o; mD = mA = mB = mC =


30o b c 90o 60o a(11−13) Given the right triangle is a 30o−60o−90o right triangle answer the following questions:11. a = 10 b = c = 12. b= a = c =

13. c = 12 a = b =

45o

a c

900 45o

b(14−16) Given the right triangle is a 45o−45o−90o right triangle answer the following questions:14. a = 9 b = c = 15. b = 15 a = c =

16. c = a = b =

(1718) Given 1 and 2 are complementary:17. m1 = 50o m2 = 18. m1 = 3x 20; m2 = 5x 20 m1 = m2 =

(1920) Given 1 and 2 are supplementary:19. m1 = 133o m2 = 20. m1 = 5x + 10; m2 = 3x + 50 m1 = m2 =

Section 5.2

(2124) For each angle in standard position the terminal side is in which quadrant?

21. 78o 22. 23. 153o 24.

(2528) Convert the following degree measurement of an angle to a radian measurement of that angle.25. 30o 26. 120o 27. 60o 28. 150o

(2933) Convert the following radian measurement of an angle to a degree measurement of that angle.

29. 30. 31. 32.

33. A bicycle wheel has a diameter of 24 inches and the wheel makes 440.5 revolutions every minute. a. What is the linear speed of the bicycle wheel? b. What is the angular speed of the bicycle wheel?

Chapter 5 Trigonometric Functions

r(3435) S

34. r = 8 inches = 45o a. area of the sector = b. length of the arc =

35. r = 8 feet =

a. area of the sector = b. length of the arc =

36. A water sprinkler sprays water over a distance of 20 feet while rotating through an angle of 90o. What is the area of the lawn watered?

Section 5.3

(3740) a b Use the Pythagorean theorem to find the following values:

c37. a = 12; c = 16 b = 38. a = 25; b = 65 c =

39. c = 10; b = 12 a = 40. a = 8; c = 5 b =

(4148) Without using the calculator find the value of sin and cos for each of the following.

41. 42.

43. 44.

45. 46.

47. 48.


(4952) For each angle in standard position find the value of sin and cos .

(x,y) r

49. 50.

51.

52.

(5356) Give the angle values of in radians between 2 and 2 that yield the following answers.

53. 54. 55. 56.

(5760) Give the angle values of in degrees between 360o and 360o that yield the following answers.

57. 58. 59. 60.

(6164) Use the calculator to approximate to 3 decimal places the value of sin and cos for the given value of .

61. = 67o 62. = 163o 63. = 64. =

Section 5.4

(6570) Without using the calculator find the exact value of tan , cot , sec and csc for each of the following values of .

65. = 60o 66. = 150o 67. = radians 68. = radians

69. = 180o 70. =

(7176) Use the calculator to approximated to 3 decimal places the value of tan , cot , sec and csc for each of the following values of .

71. = 135o 72. = 59o 73. = 74. =

75. = 583o 76. = 733o


(7780) Give the angle values of in degrees between 360o and 360o that yield the following answers.77. tan 233o 78. cot 83o 79. sec 157o 80. csc 321o

(8184) Give the angle values of in radians between 2 and 2 that yield the following answers.

81. cot 82. tan 83. csc 84. sec

(8588) For each angle in standard position find the value of tan , cot , sec and csc .

(x,y) r

85. x = 6 and y = 8 86. x = 10 and y = 26 87. x = 5 and y = 4

88. x = 3 and y = 6

89. If sin = then csc = ?

90. If tan = then cot = ?

91. If sec = then cos = ?

92. If cot = then tan = ?

Section 5.5

(93104) Evaluate each expression exactly without using the calculator. (Give your answers in degrees and radians)

93. 94. 95.

96. 97. 98.

99. 100. 101.

102. 103. 104.

(105114) Evaluate each of the following without using a calculator.


105. 106. 107.

108. 109. 110.

111. 112. 113.

114.

(115118) Evaluate each expression without using a calculator.

115. 116. 117.

118.

(119128) Use the calculator to evaluate each expression if possible.

119. 120. 121.

122. 123. 124.

125. 126. 127.

128.

Section 5.6(129146) For each function give: a. The amplitude b. The period c. The phase shift d. The vertical change e. Flipped vertically or horizontally f. The domain of the function g. The range of the function h. Sketch a graph

129. 130. 131.

132. 133. 134.

135. 136. 137.


138. 139. 140.

141. 142. 143.

144. 145.

146.


CHAPTER 5 EXAM

(1−2) Give the reference angle for each angle.

1. 232o 2.

3. Convert 305 o to radians.

4. Convert radians to degrees.

5. A Ferris wheel has a radius of 40 feet and takes 90 seconds for one revolution. a. What is the linear speed of the Ferris wheel? b. What is the angular speed of the Ferris wheel?

6. A water sprinkler sprays water a distance of 20 feet while rotating through an angle of 150o. What area of the lawn is watered?

(7−10) Find the value of sin θ, cos θ, tan θ, cot θ, sec θ and csc θ for each of the following angles.

7. 8. 9.

10.

(11−14) For each angle in standard position find the value of sin θ, cos θ, tan θ, cot θ, csc θ and csc θ.

(x,y) r

11. x = 6 y = −8 12. x = −15 y = −36

13. x = 9 r = 15, terminal side in the first quadrant

14. y = 24 r = 26, terminal side in the second quadrant

(15−20) Give the angle values in radians for θ between and that yield the following answer.

15. 16. 17.

18. 19. 20.

21. For sin θ a. What is the domain?

b. What is the range?


22. For cos θa. What is the domain?b. What is the range?

23. For tan θa. What is the domain?b. What is the range?

24. For cot θa. What is the domain?b. What is the range?

25. For sec θa. What is the domain?b. What is the range?

26. For csc θa. What is the domain?b. What is the range?

(27−32) Evaluate without a calculator

27. 28. 29.

30. 31. 32.

(33−38) Evaluate to 3 decimal places with a calculator

33. 34. 35.

36. 37. 38.

(39−44) Evaluate in radians to 3 decimal places with a calculator39. 40. 41.

42. 43. 44.

(45−46) For each function give: a. The amplitude

b. The periodc. The phase shiftd. The vertical changee. Flipped vertically or horizontallyf. The domain of the functiong. The range of the functionh. Sketch a graph

45. 46.

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chapter 5 Trigonometric Functions - Glendaleweb.gccaz.edu/~MICOW00001/HoltfrerichHaughnPreCalc… · Web viewWith trigonometric functions for each output ... Example 1: Finding