Unit 8
Trigonometric Functions
•
•
•
•
•
•
•
•
Radian and degree measure
Unit Circle
Right Triangles
Trigonometric functions
Graphs of sine and cosine
Graphs of other Trigonometric functions
Inverse Trigonometric functions
Applications and Models
13.1 Right Triangle Trigonometry
• Evaluate trigonometric functions of acute angles
• Use the fundamental trigonometric identities
• Use trigonometric functions to model applied
problems
Remember the trigonometric
functions used to solve right
triangles?
opposite side
sine 
hypotenuse
hypotenuse

Adjacent
adjacent side
cosine 
hypotenuse
opposite side
tangent 
adjacent side
Opp.

Soh  Cah  Toa
In addition to the trigonometric functions we have already
learned, we have three more. Notice they are merely the
reciprocals of the three we already have.
opposite side
sine 
hypotenuse
hypotenuse
cosecant 
opposite side
adjacent side
hypotenuse
cosine 
secant 
hypotenuse
adjacent side
opposite side
adjacent side
tangent 
cotangent 
adjacent side
opposite side
Find the six trig functions of angle A:
4
Sin A 
5
5
Csc A 
4
3
Cos A 
5
5
Sec A 
3
4
Tan A 
3
3
Cot A 
4
B
5
A
4
3
C
It is also helpful to remember the special right
triangle relationships
30
2
3
45
2
1
60
1
45
1
Now with all of this information, we are able to
find trigonometric values of many angles on the
unit circle.
Inverse trigonometric functions help us to find the
angle when we are given the triangle ratios
Inverse Cosine Function
Inverse Sine Function
1
y  sin
1
x
y  cos x
• sin-1 x or arcsin x
• cos-1 x or arcos x
Inverse Tangent Function
1
y  tan x
•
1
arccos
2
tan-1 x or arctan x
What this question is asking is, “what
is the angle that has a cosine of 1 ?”
2
  60
arctan( 8.45)  1.453001
Solve for x.
5
cos 25 
x
x
25˚
5
x
cos 25
x  5.5169
5
Sketch a triangle that has an acute angle , and
then find .
8
8
csc  
5
5
1  5 
 sin  
 sin    
8
8

 38.68  
5


  angle of elevation
  angle of depression
If you look up from the middle of the football field to the top
of the gym, you see the angle of elevation is 41 degrees. If
you are 250 feet from the base of the gym, how tall is the
gym?
1) Draw picture first!
2) Set up trig ratio
x
3) Solve
tan 41 
x
250
41
x
(.8693) 
250
250(.8693)  x
217  x
250 feet
An airplane is flying at an elevation of 5150 ft, directly
above a straight highway. Two motorists are driving
cars on the highway on opposite sides of the plane,
and the angle of depression to one car is 35 and to the
other is 52. How far apart are the cars?
5150
tan 35 
y
11378.58 feet
52 ˚
5150
tan 52 
x
52 ˚ x
35 ˚
5150
y
35 ˚
13.2 Radian and Degree Measure
•
•
•
•
Describe angles
Use radian measure
Use degree measure
Use angles to model
applied problems
Terminal side
of angle
Positive
angle
Negative
angle
This is known as standard position
Initial side
of angle
Coterminal Angles: 2 angles who have
the same terminal sides.
 (alpha)
 (beta )
• To find positive or
negative coterminal
angles: Add or
Subtract 360 or 2.
Alpha and Beta have the
same initial and terminal
sides. Angles like this
are called coterminal.
A radian is the measure of the angle that
intercepts an arc whose length is equal to the
radius of the circle.
1 radian  57 degrees
Since the circumference of a circle is 2 and there
is a 360 rotation in circle; 2 radians = 360
360 180
 1 radian 

2

A quick conversion guide:
1 radian 
180

1 

180
Reminder- supplementary angles are two
angles that add up to 180 degrees or 
radians. Complimentary angles are two angles
that add up to 90 degrees or  radians
2
Find the radian measure of the angle: 150  5
6
Find the degree measure of the angle:

360
7

a) 
b) 2 
 114.59
 630
2

Find 1 positive & 1 negative angle that are
coterminal with the angle:
11
5
π
 3
23π

 and

and
a)
b)
6
2
2
2
6
6
32
11
Are the angles coterminal?
&
3
3
no
Arc Length
Arc length  s Central angle   r  radius
1) s  r In radians
r

s
Find s if the angle is 2
radians and the radius is 5cm
s = (5)(2) = 10 cm
Find the length of an arc that subtends a
central angle of 45 in a circle of radius 10 m.
5



   s     7.852 meters
s  r  s  10 
 2 
4
Area of a Circular Sector:
A r 
1
2
r
•
•
•

2
A = sector area
r = radius
 = angle measure (rad)
A sector of a circle of radius 24 mi has an area of 288 mi2.
Find the central angle of the sector.
A  r   288  24 
1
2
2
1
2
2
 576  576
   1 radian  57
13.6 Trigonometric functions: The
Unit Circle
• Identify a unit circle and
its relationship to real
numbers
• Evaluate trigonometric
functions using the unit
circle
• Use the domain and
period to evaluate sine
and cosine functions
• Use a calculator to
evaluate trigonometric
functions
(0,1)
(-1,0)
o
OR = x
p (x, y)

R
(1,0)
(0,-1)
What is the sin of  ?
What is the cos of  ?
What is the tan of  ?
OP = 1
RP = y
mR  90
x  y 1
y
sin    y
1
x
cos    x
1
y sin 
tan   
x cos 
2
2
This means the point P at (x, y) can be represented
by the ordered pair (cos x, sin x). It is easy for us to
figure out the trigonometric functions if the terminal
side of an angle falls on an axis.
(0,1)
(-1,0)
O
(0,-1)
(1,0)
sin 90  1
cos 90  0
tan 90  undefined
sin180  0
cos180  1
tan180  0
It is also helpful to remember the special right
triangle relationships
30
2
3
45
2
1
60
1
45
1
Now with all of this information, we are able to
find trigonometric values of many angles on the
unit circle.
A reference angle is the acute angle formed
between the terminal side and the x-axis
2 150
1

 3
1
sin150 
2
 3
cosine 150 
2
What is the
sine, cosine,
tangent of
150
degrees?
 is the reference angle
  30
1
3
tan150 

3
 3
The sine function repeats itself every 2 radians
(or 360). The smallest number in which the
function repeats itself is called the period. So, the
sine function has a period of 2
5

This means that the sine of
is the same as the sine of
2
2
Cosine also has a period of 2
Tangent has a period of

13.3 Trigonometric functions of any
angle
• Evaluate trigonometric functions of any
angle
• Use reference angles to evaluate
trigonometric functions
• Evaluate trigonometric functions of real
numbers
Let us re-visit the circle of radius r. We can still
find the length of x and y for any angle using trig.
p (x, y)
OP = r
o

y
x
What is the sin of  ?
What is the length of y?
y
sin    r sin   y
r
x
cos    r cos  x
r
What is the cos of  ?
What is the length of x?
This information allows us to find the sine and
cosine for any angle.
Use the reference angle formed between the
terminal side and the x-axis
The positive trig.
functions are listed.
ALL
sine
II
150
I

Students-sine
o
IV
III
tangent
All-all
cosine
 is the reference angle
Take-tangent
Calculus-cosine
  30
Find the reference angle for the given angle:
8

7

7

Evaluating Trig Functions:
1.
Draw a picture
2.
Find  (ref angle)
3.
Trig () = ±Trig ()
4.
Decide if it’s POS or NEG
Find the value of the other 5 trig functions of  given the
following: sec   5
sin   0
We know it is in 4th quadrant
1
5
 24  2 6
What is the value of y  sin  when :

3
  0?   ?    ?    ?
2
2
(0, 1)
(-1, 0)
1
(1, 0)
(0, -1)
  2 ?

2  3  
2
2
-1

2

3 2
2
y  sin 
The sine curve repeats every 2 radians, therefore
It has a period of 2 .
13.7/13.8 Graphs of trig functions
• Sketch the graphs of
basic sine and cosine
functions
• Use amplitude and
period to help sketch
the graphs of sine and
cosine functions
• Sketch the translations
of the graphs of sine
and cosine functions
• Use the sine and cosine
functions to model
applied problems
What is the value of y  sin  when :

3
  0?   ?    ?    ?
2
2
(0, 1)
(-1, 0)
1
(1, 0)
(0, -1)
  2 ?

2  3  
2
2
-1

2

3 2
2
y  sin 
The sine curve repeats every 2 radians, therefore
It has a period of 2 .
What is the value of y  cos  when :

3
  0?   ?    ?    ?
2
2
(0, 1)
(-1, 0)
1
(1, 0)
(0, -1)
  2 ?

2  3  
2
2
-1

2

3 2
2
y  cos 
The cosine curve repeats every 2 radians, therefore
It has a period of 2 .
If y  a sin  or y  a cos  . a represents the amplitude. The
amplitude is half the distance between the max and min
values of the function
1
y  sin 
2
y  sin 
1

2  3  
2
2
-1

2

3 2
2
If the sine or cosine function is multiplied by a negative
the entire curve is reflected across the x - axis.
y   cos 
y  cos 
1

2  3  
2
2
-1

2

3 2
2
When the angle is multiplied, the period is changed.
1

2  3  
2
2
-1

2

3 2
2
y  sin 
y  sin 2
This means there are two cycles in the period now rather
than just one. So the new period is  instead of 2 .
2
The period of y  sin kx or y  cos kx is
k
1

2  3  
2
2
-1

2

3 2
2
y  sin 


y  sin    
2

When the angle is added to or subtracted from a
number, the graph is shifted to the left or right. This
is called a phase shift
1

2  3  
2
2
-1
y  sin 

2

3 2
2
y   sin    1
When the entire functions is added to or subtracted
from a number, the graph is shifted up or down.
This is called a vertical translation
Review for Sine & Cosine Functions
period = 2 for f(x) = sin x & f(x) = cos x
* Domain = (–, ) Range = [–1, 1]
• y = asin(k(x – b))+ c OR y = acos(k(x – b))+ c
– Amplitude (half the distance between the max
and the minimum of the function. Or the
distance above or below the midline) = |a|
2
– Period = k
– Phase Shift (Lt or Rt) = b
– Graphing Interval
2 

b
,
b

=
k 

Range: c  a , c  a 
– Vertical Shift (Up or Dn)= c
Factoring may be necessary to get it in this form!!!
Sketch the graph. State the amplitude & period.
y  3 sin 2( x  2 )
Amplitude = 3
Period  
•Set graphing interval first
  
R   3,3
 2 , 2 
•Divide period into four parts
and plot the points accordingly

Each section is
So : 

 3
4
00

4
1
4
3

2
0


2
-1

2
Sketch the graph. State the amplitude & period.
y  cos(3x  2 )  1
y  cos(3( x  6 ))  1
Amplitude = 1
2
Period 
3
•Set graphing interval first
  3     
 6 , 6    6 , 2 
R  0,2
•Divide period into four parts
and plot the points accordingly
Each section is
So : 

2
6
0 1

6

1
6
0

3

1

2
2

6
-1

6

2
Tangent & Cotangent Functions
• y = a tan k (x – b)
• y = a cot k (x – b)

– Period = k
– Phase Shift (Lt or Rt) = b

– Graphing Interval = [b, b + ]
k
Tangent has a zero at the beginning, zero at the
end, and an asymptote in the middle.
Cotangent has an asymptote at the beginning and
end, and a zero in the middle.
y  tan 

2  3  
2
2

2

3 2
2
Tangent has a zero at the beginning, zero at the
end, and an asymptote in the middle.
y  cot 

2  3  
2
2

2

3 2
2
Cotangent has an asymptote at the beginning and
end, and a zero in the middle.
Secant & Cosecant Functions
• y = a sec k (x – b) + c
• y = a csc k (x – b) + c
– Amplitude (vertical distance between U’s and
midline) = a
2
– Period = k
– Phase Shift (Lt or Rt) = b
2
– Graphing Interval = [b, b + k ]
– C is the midline
y  sin 
y  csc 
1

2  3  
2
2
-1

2

3 2
2
Since it is the reciprocal, where the sine function
was equal to zero there is now an asymptote for
cosecant function at 0,180,360 degrees.
y  cos 
y  sec 
1

2  3  
2
2
-1

2

3 2
2
Similarly, where the cosine function was equal to
zero there is now an asymptote for the secant
function at 90,270 degrees.
Key Chapter points:
• Radian and degree
measure
• Unit Circle
• Right Triangles
• Trigonometric functions
• Graphs of sine and
cosine
• Graphs of other
Trigonometric functions
• Inverse Trigonometric
functions