Chapter 5
TRIGONOMETRIC FUNCTIONS
Section 5.1
 Angles and Degree Measure
 Learn how to convert decimal degree measures to degrees,
minutes, and seconds.
 Find the number of degrees in a given number of rotations
 Identify angles that are coterminal with a given angle
5.1 Angles and Degree Measure
 Vertex
 Endpoint of an angle
 Initial Side
 The ray of the angle that is fixed
 Terminal Side
 The second ray that rotates to form the angle
 Standard Position
 An angle with its vertex at the origin and
its initial side along the positive x-axis
5.1 Angles and Degree Measure
5.1 Angles and Degree Measure
 Degree

Most common unit to measure an angle
 Minutes

A degree is subdivided into 60 equal parts called minutes (1’)
 Seconds

A minute is subdivided into 60 equal parts called seconds (1”)
 Example 1:

Change 15.735 degrees to degrees, minutes, and seconds.
15.735o=15o+(0.735*60)’
 =15o+44.1’
 =15o+44’+(0.1*60)”
 =15o+44’+6”

5.1 Angles and Degree Measure
 Example 2:
 Change 329.125 degrees to degrees, minutes, and seconds.

329o+7’+30”
 Example 3:
 Change 39o+5’+34” to degrees
39o+5’+34”=39o+5’*(1 o/60’) +34” (1 o/3600’)
 39.093o

 Example 4:
 Change 35o+12’+7” to degrees

35.202o
5.1 Angles and Degree Measure
 Quadrantal Angle
 If the terminal side of an angle that is in standard position
coincides with one of the axes
 How many quadrantal angles are there?


4
What are their measures?
90 degrees
 180 degrees
 270 degrees
 360 degrees

5.1 Angles and Degree Measure
 Give the angle measure represented by the rotation about
the axis

5.5 Rotations clockwise
Which way do we rotate to go clockwise?
 Negative, clockwise rotations ALWAYS have negative measures
 5.5 * -360 = -1980 degrees


3.3 Rotations counterclockwise


9.5 Rotations clockwise


Which way do we rotate to go counterclockwise?
 Positive, counterclockwise rotations ALWAYS have positive
measures
 3.3*360 = 1188 degrees
-3420 degrees
6.75 Rotations counterclockwise

2430 degrees
5.1 Angles and Degree Measure
 Coterminal Angles
 If α is the degree measure of an angle, then all angles
measuring α+ 360k degrees, where k is an integer; are
coterminal with α.
 Two angles in standard position that have the same initial side
 Identify all angles that are coterminal with a 45 degree angle.
Find one positive and one negative angle that are also
coterminal.
All angles having a measure of 45o+ 360ko
 Positive Angle
 45o+ 360*(1)o=405o
 Negative Angle
 45o+ 360*(-2)o=-675o

5.1 Angles and Degree Measure

Identify all angles that are coterminal with a 294 degree angle.
Find one positive and one negative angle that are also
coterminal.
All angles having a measure of 294o+ 360ko
 Positive Angle
 294o+ 360*(1)o=654o
 Negative Angle
 294o+ 360*(-1)o=-66o

5.1 Angles and Degree Measure

If an angle with 775 degrees is in standard position, determine
a coterminal angle that is between 0 and 360 degrees. State
the quadrant in which the terminal side lies.
Find the number of rotations about the axis by dividing 775 by 360
= 2.15278
 Since we need an angle between 0 and 360 degrees, what rotation
number should we use?
 Subtract 2 and use .15278
 α= .15278*360
 Could also take the number and continue to subtract 360 until
you get a number between 0 and 360.
 55o
 What quadrant does the terminal side fall in?
 Quadrant 1

5.1 Angles and Degree Measure

If an angle with -777 degrees is in standard position, determine
a coterminal angle that is between 0 and 360 degrees. State
the quadrant in which the terminal side lies.
303o
 What quadrant does the terminal side fall in?
 Quadrant 4

5.1 Angles and Degree Measure
 Reference Angle
 The acute angle formed by the terminal side of the given angle
and the x-axis
 Reference Angle Rule
 For any angle α, 0o< α <360o, its reference angle α’ is defined
by:
α, when the terminal side is in Quadrant I.
 180o- α, when the terminal side is in Quadrant II.
 α – 180o, when the terminal side is in Quadrant III.
 360o- α, when the terminal side is in Quadrant IV.

5.1 Angles and Degree Measure
 Find the measure of the reference angle for an angle with a
measurement of 120 degrees

What Quadrant does this angle’s terminal side fall in?





Between 90o and 180o
Quadrant II
We use which formula?
180o- α
180o-120o=60o
 Find the measure of the reference angle for an angle with a
measurement of -135degrees

First we have to find a positive coterminal angle






360o-135o = 225o
What quadrant does this angle’s terminal side fall in?
Between 180o and 270o
Quadrant III
α – 180o
225o-180o=45o
5.1 Angles and Degree Measure
 Find the measure of the reference angle for an angle
with a measurement of 312 degrees

48o
 Find the measure of the reference angle for an angle
with a measurement of -195 degrees

15o
Section 5.2
 Trigonometric Ratios in Right Triangles
 Learn how to find the values of trigonometric ratios for actue
angles of right triangles
5.2 Trigonometric Ratios in Right Triangles
 What is a Right Triangle?
 A triangle with a 90 degree angle in it.
 How can we classify the other two angles in the right triangle?
Must be acute
 Are also complementary


What are the parts of a Right Triangle?
< 1 + < 2 = 90o
1
Side C is the hypotenuse
Sides A and B are the legs
2
5.2 Trigonometric Ratios in Right Triangles
 When looking at one specific acute angle in a
triangle, we can classify the legs by:



Adjacent Side: the leg that is a side of the acute angle
Opposite Side: the leg that is the side opposite the angle
Looking at Triangle ABC, what are the adjacent and opposite
sides for < B?
B
Hypotenuse
Adjacent
Opposite
5.2 Trigonometric Ratios in Right Triangles
 Trigonometric Ratios

The ratios of the sides of right triangles based on a specific acute
angle within the right triangle
 Easy way to remember Sin, Cosine, and Tangent Ratios

SOHCAHTOA
 Sine

Ratio of the side opposite Θ and the hypotenuse
 Cosine

Ratio of the side adjacent Θ and the hypotenuse
 Tangent

Ratio of the side opposite Θ and the side adjacent to Θ
5.2 Trigonometric Ratios in Right Triangles
5.2 Trigonometric Ratios in Right Triangles
 Find the values of sine, cosine, and tangent for <B.

What do we need to do first?
Find the third side’s length
 How do we do that?
 Pythagorean Theorem, A2 + B2 = C2
 C=3*(85)1/2



Which angle is our Θ?
Sin Θ
18 m
C
3*(85)1/2 m
33 m
Opposite/hypotenuse
 18/33


Cos Θ
Adjacent/hypotenuse
 3*(85)1/2/33


Tan Θ
Opposite/Adjacent
 18/3*(85)1/2

Θ
B
A
5.2 Trigonometric Ratios in Right Triangles
 Find the values of sine, cosine, and tangent for <A.
 Find the third side’s length



Which angle is our Θ?
Sin Θ


8/17
Cos Θ


C=17 m
15/17
Tan Θ

C
17 m
8m
8/15
B
Θ
15 m
A
5.2 Trigonometric Ratios in Right Triangles
 Cosecant
 Opposite of sine, cscΘ = 1/sin Θ
 Hypotenuse/side opposite
 Secant
 Opposite of cos, secΘ = 1/cosΘ
 Hypotenuse/side adjacent
 Cotangent
 Opposite of tan, cotΘ = 1/tanΘ
 Side Adjacent/ Side opposite
5.2 Trigonometric Ratios in Right Triangles
 If cos Θ = ¾, what is sec Θ?



sec Θ = 1/cos Θ
=1/(3/4)
=4/3
 If sin Θ=0.8, what is csc Θ?

=1.25
 If csc Θ = 1.345 find sin Θ?



sin Θ = 1/csc Θ
=1/1.345
=.7435
 If cot Θ = 6/5, what is tan Θ?

=5/6
5.2 Trigonometric Ratios in Right Triangles
 Find the values of the six trigonometric ratios for <E.
 First find third side using pythagorean theorem
 (58)1/2 m
 Cos E = 3* (58)1/2 /58
 Sin E= 7* (58)1/2 /58
 Tan E=7/3
D
1/2
 Sec E= (58)
/3
(58)1/2 m
 Csc E= (58)1/2 /7
7m
 Cot E=3/7
F
Θ
3m
E
5.2 Continued
 Special Triangles
 30O-60O-90O
 45O-45O-90O
 What are the special relationships we know about
these triangles?

30O-60O-90O
5.2 Continued
 What are the special relationships we know about
these triangles?

45O-45O-90O
5.2 Continued
 Trigonometric Ratios for 30o, 60o, 90o
5.2 Continued
 Cofunctions
 Trigonometric functions that are equal when their arguments
are complementary angles, such as sine and cosine, tangent
and cotangent, and secant and cosecant.
 Sin Θ = Cos (90O- Θ)
 Cos Θ = Sin (90O- Θ)
 Tan Θ=Cot (90O- Θ)
 Cot Θ=Tan (90O- Θ)
 Sec Θ=Csc (90O- Θ)
 Csc Θ=Sec (90O- Θ)
Section 5.3
Trigonometric Functions on the Unit Circle
 Find the values of the six trigonometric functions
using the unit circle
 Find the values of the six trigonometric functions of
an angle in standard position given a point on its
terminal side
Trigonometric Functions on the Unit Circle
 Unit Circle
 A circle with a radius of 1
 Usually with the center on the
origin on the coordinate system
 Symmetric with respect to the
x-axis, y-axis, and the origin
Trigonometric Functions on the Unit Circle
 Unit Circle
 Consider an angle between 0O and 90O in standard position
 Let P(x,y) be where the angle intersects with the unit circle
 Draw a perpendicular segment from intersection point back
down to the positive x axis
 Creates a right triangle
 Find the sin Θ and cos Θ

Sin Θ = y; Cos Θ = x
Trigonometric Functions on the Unit Circle
 Sine and Cosine on the Unit Circle
 If the terminal side of an angle Θ in standard position
intersects the unit circle at P(x,y), then cos Θ = x and sin Θ=y.
Trigonometric Functions on the Unit Circle
Trigonometric Functions on the Unit Circle
 Circular Functions
 Functions defined using the unit circle
 Ie Sin and Cosine
 How can we define the other cosine functions on the unit
circle?
Tan Θ = y/x
 Csc Θ =1/y
 Sec Θ =1/x
 Cot Θ = x/y

Trigonometric Functions on the Unit Circle
 Use the unit circle to find each value
 Cos (-180O)



Which way do you go for negative angles?
 Clockwise
What is the ordered pair of the intersection of this angle on the unit circle?
 (-1,0)
 Cos Θ = x-axis
 Cos (-180O) = -1
Sec(90O)


Where is the angle located ?
 Terminal Side is on positive y axis
Where is the intersection on the unit circle?
 Intersection at (0,1)
 Sec Θ = 1/x
 Sec(90O)=1/0 = undefined
Trigonometric Functions on the Unit Circle


Use the unit circle to find each value
Sin(-90O)


-1
Cot(270O)

0
Trigonometric Functions on the Unit Circle
 Use the unit circle to find the values of all six
trigonometric functions for a 210 degree angle.

What is the intersection with the unit circle?







(-√(3) /2, -1/2)
Sin Θ = -1/2
Cos Θ =- √(3) /2
Tan Θ = √(3) /3
Csc Θ =-2
Sec Θ =-2 √(3) /3
Cot Θ = √(3)
Unit Circle Quiz
 Use the unit circle to find each value
 Tan 3600
 Cos 450
 Sin(-600)
 Csc(-2100 )
 Sec(2250 )
 Name the six trigonometry functions for the angles
below


1500
4200
Trigonometric Functions on the Unit Circle
 What if the angle doesn’t fall within the unit circle?
 What if the length is greater or less than 1?
 Use length R instead of 1 in unit circle and R = (x2+y2)1/2
Trigonometric Functions on the Unit Circle
 What are the six trigonometric values using length r?
 Sin Θ =y/r
 Cos Θ =x/r
 Tan Θ =y/x
 Csc Θ =r/y
 Sec Θ =r/x
 Cot Θ =x/y
Trigonometric Functions on the Unit Circle
 Find the six trigonometric functions for angle Θ in standard position if a
point with the coordinates (-15, 20) lies on its terminal side.
 Draw Figure on xy axis
 What are the side measures?
 Leg = -3
 Leg = 4
 Hypotenuse = 5
 Where is the theta located?
 At the point they gave us
 What are the six functions?






Sin Θ =4/5
Cos Θ =-3/5
Tan Θ=-4/3
Csc Θ =5/4
Sec Θ =-5/3
Cot Θ =-3/4
Review Quiz 5.3
 Suppose Θ is an angle in standard position whose
terminal side lies in Quadrant III. If sin Θ = -4/5, find
the values of the remaining five trigonometric functions
of Θ?

What do we need to do first?


Draw figure
Next?
Find Missing Side
 +/-3, use negative 3 because Quadrant III


Find 5 remaining trig functions.
Cos Θ = -3/5
 Tan Θ =4/3
 Csc Θ =-5/4
 Sec Θ =-5/3
 Cot Θ =3/4

5.4 Applying Trigonometric Functions
 Use trigonometry to find the measures of the sides of
right triangles
 Make sure calculators are in degrees and not radians






Press Mode
Scroll down to third line and arrow left one
Hit enter
Quit
Quick Check
Cos(90) = 0
Applying Trigonometric Functions
 In triangle PRQ, P = 35o and r = 14. Find Q.
 First draw figure
 How is Q related to Θ?


Adjacent Side
What function should we use?
Cosine
 Cos P = q/r
 Cos 35=q/14
 14*cos(35)=Q
 Q is about 11.5

Applying Trigonometric Functions

Angles of Elevation


The angle between a horizontal line and the line of sight from an
observer to an object at a higher level
Angles of Depression

The angle between a horizontal line and thel ine of sight from the
observer to an object at a lower level
5.5 Solving Right Triangles
 Evaluate inverse trig functions.
 Find missing angle measurements.
 Solve right triangles.
Solving Right Triangles
 Inverse of a Trigonometric Function
 The arcsine, arccosine, and arctangent relations with their
corresponding trigonometric functions
 Arcsine
Sin x = √3/2
 Can be written as x = arcsine √3/2 or x = sin-1 √3/2
 Read this as x is an angle whose sine is √3/2


Same for other two trig functions
Arccosine
 Arctangent

Solving Right Triangles
 Sin x = √3/2
 X is an angle with sine √3/2
 X = arcsine √3/2 or x = sin-1 √3/2
 60 o, 120 o, or any coterminal angles with these
 Tan x = 1
 45 o,225o
 Sin x = -1/2
 210 o, 330 o
Solving Right Triangles
 Evaluate and assume angles are in Quadrant I
 Cos(arcsin 2/3)
Let B = arcsin 2/3
 Sin B = 2/3
 Draw figure in Quadrant I
 Find X
 = √5/3
 Try in calculator


Tan(cos-1 4/5)

3/4
Solving Right Triangles
 Solve the Triangle with A = 33o and B = 5.8


First fill in known amounts
Find Angle B
Two acute angles in a right triangle are
complementary
 B =90-33o = 57o degrees


B
Find side A
Tan A = a/b
 Tan 33o = a / 5.8
 5.8 Tan 33o = a
 a = 3.767


Find side C
Cos A = b/c
 Cos 33o = 5.8/c
 C=5.8/cos 33o
 c=6.916

A
C
Solving Right Triangles
 Solve the Triangle with K = 40o and k = 26
 < L = 50o
 l = 31.0
K
 j = 40.4
J
L
Section 5.6
The Law of Sines
 Solve triangles by using the Law of Sines if the
measures of two angles and one side are given
 Find the area of a triangle if the measures of sides
and the included angle or the measures of two angles
and a side are given.
The Law of Sines
 Let Triangle ABC by any triangle with a, b, and c
representing the measures of the sides opposite the
angles with measures A, B, and C, respectively. Then
the following is true:
a = b = c
SinA SinB SinC
The Law of Sines
 Solve triangle ABC if A = 33o, B = 105o, and b = 37.9
 Find < C by subtracting the other two angle measures from 180
 C= 42o
a
= 37.9
Sin33o Sin105o
A = 21.37o
c
= 37.9
Sin42o Sin105o
C=26.25o
The Law of Sines
 Area of a Triangle
 Let triangle ABC be any triangle with a, b, and c representing
the measures of the sides opposite the angles with
measurements A, B, C, respectively. Then the area K can be
determined using:
K = ½ * b* c* sinA
 K = ½ * a* c* sinB
 K = ½ * b* a* sinC

The Law of Sines
 Find the area of triangle ABC if a = 4.7, c = 12.4, and
B = 47o 20’





First convert B to decimal degrees
B = 47.33o
K = ½ * a* c* sinB
K = ½ * 4.7*12.4*sin47.33o
K = 21.4 units2
 Find the area of triangle ABC if b = 21.2, c = 16.5,
and A = 25o
 K = ½ * b* c* sinA
 K = 73.9 units2
The Law of Sines
 Area of a Triangle
 Let triangle ABC be any triangle with a, b, and c representing
the measures of the sides opposite the angles with
measurements A, B, C, respectively. Then the area K can be
determined using:
K = ½*a2*sinB*sinC
SinA
 K = ½*b2*sinA*sinC
SinB
 K = ½*c2*sinB*sinA
SinC

The Law of Sines
 Find the area of triangle DEF if d = 13.9, D = 34.4o,
and E = 14.8o


Find the measure of <F first, 180-34.4-14.8
<F = 130.8o
K = ½*d2*sinE*sinF
SinD
 K = ½*13.92*sin 14.8o *sin 130.8o
SinD 34.4o
K =33.065 units2

Section 5.8
The Law of Cosines
 Let Triangle ABC by any triangle with a, b, and c
representing the measures of the sides opposite the
angles with measures A, B, and C, respectively. Then
the following is true:



a2=b2 + c2 -2*b*c*cosA
b2=a2 + c2 -2*a*c*cosB
c2=b2 + a2 -2*b*a*cosC
The Law of Cosines
 Solve the triangle with the following information:
 A = 120o,b=9, c=5
 Solve for Side a
 a2=b2 + c2 -2*b*c*cosA
 a2=92 + 52 – 2*9*5*cos(120o)
 a2=426.97
 a=12.3
 Now use Law of Sines to find <B
 12.3
= 9
Sin120o sinB
B=arcsin (9*sin120o)
12.3
B= 39.3o
C= 180o – 120o-39.3o
C=20.7o
The Law of Cosines
 Try a triangle with A = 39.4o,b = 12, c = 14
 B=58.2o
 C = 82.4o
 a = 9.0
 Try a triangle with a =19, b = 24.3, c = 21.8
 A = 48.3o
 B = 72.7o
 C = 59o
The Law of Cosines
 Hero’s Formula
 If the measures of the sides of a triangle are a, b, and
c, then the area, K, of the triangle is found as follows:

K = √s*(s-a)*(s-b)*(s-c)

Where S = ½ (a+b+c)
The Law of Cosines
 Find the area of the triangle with the following
information:
 a=72 cm
 b=83 cm
 c=95 cm

K = √s*(s-a)*(s-b)*(s-c)
Where S = ½ (a+b+c)
 S=125
 K=2889.2 cm2
