Download Session 15

Session 15
Agenda:
• Questions from 8.1-8.2?
• 8.3 – Trigonometric Identities
• 8.4 – Graphs of Trigonometric Functions
• 8.5 – Inverse Trigonometric Functions
• Things to do before our next meeting.
Questions?
8.3 – Trigonometric Identities
• Summary of Trig Identities Already Seen:
sin 2 x  cos 2 x  1
tan( x) 
sin( x)
cos( x)
1
cos( x)
1
csc( x) 
sin( x)
1
cos( x)
cot( x) 

tan( x) sin( x)
sec( x) 
Simplify the following and write as a single trig function.
tan( x)
sec( x)sin 2 ( x)
• When simplifying trig functions, other techniques such as
finding common denominators and factoring may still
need to be used.
• Simplify the following to a single trig function
cos3 ( x)  cos( x)sin 2 ( x)
csc( x)  csc( x) cos 2 ( x)
Simplify the following as much as possible. Express your
answer using only sines and cosines
sin 2 ( x)
cos( x)

sin( x) cos( x)  cos 2 ( x) sin( x)  cos( x)
• The following identities are obtained by dividing the
2
2
2
fundamental identity sin ( x)  cos ( x)  1 by either cos ( x)
2
or sin ( x) .
sin 2 ( x)  cos 2 ( x)  1
sin 2 ( x)  cos 2 ( x)  1
sin ( x) cos ( x)
1


cos 2 ( x) cos 2 ( x) cos 2 ( x)
sin 2 ( x) cos 2 ( x)
1


sin 2 ( x) sin 2 ( x) sin 2 ( x)
tan 2 ( x)  1  sec 2 ( x)
1  cot 2 ( x)  csc 2 ( x)
2
2
Simplify the following completely. (If necessary, assume x is
an angle in Quadrant I.) Express your answer using sines
and cosines.
4cot 2 x  4
sin x sec2 x  sin x
For any x, find the value of:
2sec2 x  2 tan 2 x
3cot 2 x  3csc2 x
Double Angle Identities
sin(2 x)  2sin( x) cos( x)
cos(2 x)  cos 2 ( x)  sin 2 ( x)
 2 cos 2 ( x)  1
 1  2sin 2 ( x)
• Use a double angle identity to simplify the following to a
an expression involving a single trig function.
cos( x)  cos3 ( x)
sin(2 x) cos( x)
• Given that csc( x) 
value of cos(2 x) .
7
and that x lies in Quadrant II, find the
6
• Given that θ is in Quadrant I and that cos( )  x , express
sin( ) in terms of x.
• Express sin(2 ) in terms of x.
8.4 – Graphs of Trigonometric Functions
f ( x)  sin( x)
• Domain: (-∞, ∞)
• Range: [-1, 1]
• Due to the nature of the unit
circle, we know that many
angles have the same sine
value. Thus, the graph of the
sine function repeats itself over
and over. We refer to such functions as
periodic. The period of the function is the
length of the smallest interval that is repeated over and
over to form the graph. For sine, the period is 2π.
f ( x)  cos( x)
• Domain: (-∞, ∞)
• Range: [-1, 1]
• Period: 2π
f ( x)  tan( x)
•
Vertical Asymptotes: x 
k
2
where k is any odd integer.
• Range: (-∞, ∞)
• Period: π
The graphs of the other three trig functions can be
remembered by using the fact that each is a reciprocal of
sine, cosine, or tangent.
• For example, we know that
csc( x) 
1
sin( x)
. The graph of
cosecant is shown below along with the graph of sine.
f ( x)  csc( x)
• Vertical Asymptotes: x  k
where k is any integer.
• Range: (,  1]  [1, )
• Period: 2π
• We know that sec( x) 
1
.
cos( x)
The graph of secant is shown
below along with the graph of cosine.
f ( x)  sec( x)
• Vertical Asymptotes:
x
k
2
where k is any odd integer.
• Range: (,  1]  [1, )
• Period: 2π
• We know that cot( x) 
1
.
tan( x)
f ( x)  cot( x)
• Vertical Asymptotes: x  k
where k is any integer.
• Range: (-∞, ∞)
• Period: π
Transformations of Trigonometric Graphs
• All the transformations we have applied previously still
hold for trigonometric functions.
• Examples:
f ( x)  3cos( x  3) -- Shift the graph of cosine to the left 3
and then vertically stretch by a factor of 3.
g ( x)   sin( x  2) -- Shift the graph of sine to the right 2
and then reflect the graph across the x-axis.
h( x)  tan( x)  4 -- Reflect the graph of tangent across
the y-axis and then shift the graph up 4 units.
Changing the Period
• We can also change the period of a trigonometric graph,
which essentially means stretching or shrinking the graph
horizontally.
• Consider the function f ( x)  sin(kx).
First note that this is NOT equal
to k sin( x) . The k inside the function changes
2
the period of the graph to
.
k
• The graph to the right is the graph
of f ( x)  sin(2 x) . The period of this
2
  . The graph has been
new graph is
2
“shrunk” horizontally.
• For a function of the form f ( x)  cos(kx) , the same is true.
• Consider the function
1 
f ( x)  2 cos  x  .
2 
• The period of the function is
2
1
2
 4
The graph was stretched horizontally.
• The final step is to stretch the graph
vertically by a factor of 2 so that its
range is now [-2, 2].
• Since the graph of tangent has a period of π, then a
function of the form f ( x)  tan(kx) has a period of  .
• Consider the function
 
 
f ( x)   tan  4  x    .
8 
 
• The first thing that is done is to

change the period to , essentially
4
shrinking the graph horizontally.
k
 
 
f ( x)   tan  4  x   
8 
 
The second thing done is to

shift the graph to the right 8 .
The last thing to be done
is to reflect the graph
across the x-axis.
• If a function is written in a form such as f ( x)  cos(3x   ) ,
you must first factor out the 3 to obtain
This would give you a period of
 
 
f ( x)  cos  3  x   
3 
 

.
2
and a shift left of .
3
3
2
3
Sketch a graph (or one period) of f ( x)  3sin  x 

.
3
4
6
Sketch a graph (or one period) of f ( x)  2 cos  x 
5
5

 1 .

8.5 – Inverse Trigonometric Functions
•
Although sine, cosine, and tangent are NOT one-to-one functions,
we can restrict the domains of each of them to obtain a one-to-one
function that still encompasses the whole range of each function.
The restricted domains are:
  
  2 , 2 
•
[0,  ]
  
 , 
 2 2
Now that we have restricted each to a one-to-one function, we
can define an inverse function for each.
Inverse Sine Function
•
The notation for the inverse sine function is
f ( x)  arcsin( x) OR sin 1 ( x)
•
The function is defined by:
y  sin 1 ( x)  sin( y)  x
•
1
Essentially, this means that sin ( x) is the ANGLE whose sine is x.
•
1
The graph of the f ( x)  sin ( x) is shown.
•
Domain: [-1, 1] (was range of sine)
•
Range:
•
IMPORTANT: Although there are many
angles which have the same sine value,
by definition, the angle chosen for the arcsine
function must be in the interval  2 , 2  , i.e.,


Quadrants I and IV on the unit circle.
  
  2 , 2 
(was restricted domain of sine)
Evaluate the following.
1
arcsin  
2

3
sin  

2


1
arcsin(3)
   
sin 1  sin    
  6 
  3  
arcsin  sin   
  4 
Inverse Cosine Function
•
The notation for the inverse cosine function is
f ( x)  arccos( x) OR cos1 ( x)
•
The function is defined by:
y  cos1 ( x)  cos( y)  x
•
1
Essentially, this means that cos ( x) is the ANGLE whose cosine is x.
•
1
The graph of the f ( x)  cos ( x) is shown.
•
Domain: [-1, 1]
•
Range: [0,  ]
•
IMPORTANT: Although there are many
angles which have the same cosine value,
by definition, the angle chosen for the arccosine
function must be in the interval 0,   , i.e.,
Quadrants I and II on the unit circle.
Evaluate the following.
 1 
arccos 

 2

3
cos 1  

2


arccos( )

  
cos 1  cos    
 3 


 7  
arccos  cos 

 6 

Inverse Tangent Function
•
The notation for the inverse tangent function is
f ( x)  arctan( x) OR tan 1 ( x)
•
The function is defined by:
y  tan 1 ( x)  tan( y)  x
•
1
Essentially, this means that tan ( x) is the ANGLE whose tangent is x.
•
1
The graph of the f ( x)  tan ( x) is shown.
•
Domain: (-∞, ∞)
•
Range:   2 , 2  ; Hor. Asymptotes:
•
  


y

2
IMPORTANT: Although there are many
angles which have the same tangent value,
by definition, the angle chosen for the
  
arctangent function must be in the interval   2 , 2 ,
i.e., Quadrants I and IV on the unit circle.
Evaluate the following.
arctan 1
 1 
tan 1  

3


 4
tan  tan 
 3

1




 5  
arctan  tan 

6



Inverse Cosecant and Secant Functions
• For arccsc(x) and arcsec(x), you can rewrite the function
in terms of arcsin(x) and arccos(x).
1
csc1 ( x)  sin 1  
 x
• Examples:
csc1 (2)  sin 1 (
 2 
1
sec1 
  cos (
 3
)
)
1
sec 1 ( x)  cos 1  
 x
Inverse Cotangent Function
•
The graph of cot(x) (with restricted domain in red) and
arccot(x) are shown below.
Domain: (0, π)
Domain: (-∞, ∞)
Range: (-∞, ∞)
Range: (0, π) [Quad. I and II
on unit circle.]
•

Evaluate cot 1  

1 
.
3
• Express the following as an algebraic expression involving
only x. [Hint: First draw a right triangle where an acute
angle has a sine of x.]
tan(arcsin( x))
• Are there any domain restrictions on x?
Things to Do Before Next Meeting
• Work on Sections 8.3-8.5 until you get all green
bars!!
• Write down any questions you have.
• Continue working on mastering 8.1-8.2.