Precalculus 1330 – Hollyer
Chapter 5 Lecture A
F’12
(5.4 and 5.1 A)
Popper 08 today
Homework 17 and Homework 18 is on this material
Quiz 6 and Quiz 7 BOTH due Sunday
Note: 4 lectures and 6 homeworks – some are due on TUESDAYS
HELPFUL website on our Course homepage under Chapter 5!
5.4
Recall



Inverse trig functions: arcsine, arccosine, arctangent
1:1 functions have inverse functions
if ( a, b ) is on a 1 to 1 function’s graph, then ( b, a ) is on it’s inverse function’s graph
f 1 (f (x))  f (f 1 (x))  x
So let’s start with the sine function: f ( x)  sin( x)
Note that it’s NOT 1:1!
Domain: all real numbers
Range: [−1, 1]
To make it 1:1
we restrict it’s domain to [ 
 
, ]
2 2
and the range is still [−1, 1]
So we have sine restricted!
Note: Q4 written negatively and Q1
Only these quadrants are used.
If ( a, b ) is on a 1 to 1 function’s graph,
then ( b, a )
is on it’s inverse function’s graph
Let’s make a table of points for Q4 (expressed with negative rotations) and Q1 and then we’ll
swap them:
Sine Restricted
f ( x)  sin 1 ( x)  arcsin( x)
(
(
(

2

3
, 1)
,
3
)
2

1
, )
6 2
(0, 0)
 2
( ,
)
4 2

( ,1)
2
HOMEWORK heads up: The official way to figure these out:
 
Formally:
sin 1 ( x)  the number in the interval [  , ] whose sine value is x.
2 2
Sine restricted:
domain [ 
 
, ] and the range is still [−1, 1]
2 2
sin 1 ( x)  the number in the interval [ 
 
, ] whose sine value is x.
2 2
Let’s say this while we work the following:
Homework!
arcsin(1) 
Calculating:
sin 1 ( 
2
)
2
sin 1 (0)
The graph of arcsine looks like:
Domain:
[−1, 1]
Range:
[
 
, ]
2 2
Taking the inverse function of a function value:
sin(arcsin(1)) 
f 1 (f (x))  f (f 1 (x))  x

sin 1 (sin(  ))
3
Be careful of expressions like:
Domain:
[−1, 1]
Range:
sin 1 (
3
)
4
[
Popper 08, Question 1
Hint: sin 1 ( x)  the number in the interval [ 
this is undefined! HW
 
, ]
2 2
 
, ] whose sine value is x.
2 2
Now on to the cosine function:
f ( x)  cos( x)
Domain: all real numbers
Range: [−1, 1]
NOT 1:1
We’ll switch to cosine
with a domain of [0,  ] , Q1
work with. Note that we’ll
domain and range, and all
that taking the inverse
cosine will return the x
Cosine restricted:
Domain:
[0,  ]
Range:
[−1, 1]
The graph of arccos(x):
Domain:
[−1,1]
Range:
[0,  ]
HOMEWORK heads up:
restricted:
and Q2 to
swap the
the points and
cosine of a
value to us.
Formally:
cos1 ( x)  the number in the interval [0,  ] whose sine value is x
Example:
cos 1 (
3
)
2
Let’s make a table of graph points for arccos(x) by starting with some points from cosine
restricted and swapping the values:
f 1 ( x)  cos 1 ( x)  arccos( x)
Cosine of x
(0,1)
 3
( , )
6 2
 2
( ,
)
4 2
( , 0)
(
2 1
, )
3
2
(
5
3
, )
6
2
( , 1)
Note: cos1 ( x)  the number in the interval [0,  ] whose sine value is x
Homework/Quiz
7
cos 1 ( )
6
Calculating:
arccos( 
cos 1 (0)
2
)
2
Popper 08, Question 2
Hint: cos1 ( x)  the number in the interval [0,  ] whose sine value is x
Now:
Taking the inverse function of a function value:
Homework/Quiz:
cos 1 (cos(0)) 
cos(arccos(
cos(cos 1 (
f 1 (f (x))  f (f 1 (x))  x
2
)) 
2
7
)) =
4
hint: what’s the domain of arccosine?
HW!
And the last one we’ll work with:
The inverse function to tangent.
Let’s look at the graph of tangent: it’s not one-to-one! Chop it off!
Tangent:
Domain:
all real numbers except odd integer multiples of
Range:
all real numbers

2
To get it to be a 1:1 function (tangent restricted)
 
we will restrict the domain to (  , ) and the
2 2
range is still all real numbers.
So for arctan x = tan 1 ( x ) , the domain is now all real numbers and the range is ( 
the graph looks like:
Let’s check out some points on
arctan(x)
On tan(x), we’ve got:
on arctan(x):
 3
( , )
6 3

( ,1)
4
(
2
,  3)
3
Note the lack of a domain restriction!
Note that THE sentence to learn is:
 
, )
2 2
tan 1 ( x) is the number in the interval ( 
 
, ) whose tangent is x.
2 2
Let’s say the sentence while we work these:
tan 1 (0)
Evaluate:
arctan( 1)
tan 1 (  3)
Popper 08 Question 3
Hint:
tan 1 ( x) is the number in the interval ( 
Popper 08, Question 4
 
, ) whose tangent is x.
2 2
Popper 08, Question 5
tan 1 ( x) is the number in the interval ( 
 
, ) whose tangent is x.
2 2
Switching gears back to sine and cosine, we will now work with negative rotations and
coterminal angles. Basically, if you have an unfamiliar angle, it has exactly the same trig
function values as it’s Quadrant 1 reference angles within a plus or a minus sign. These are 30°,
45°, and 60° for the famous angles.
For negative rotations, we need to review even and odd:


1
( cos( )  cos( ) 
3
3
2
cosine functions are “even”
Symmetric about the y-axis.
BUT

sine functions are “odd” (symmetric about the origin)

sin( )  sin( )
6
6
they have the opposite signs!


sin( )   sin( )
6
6
Here’s a typical problem:


2
cos( )  cos( ) 
4
4
2
Here’s a typical problem that combines negative rotations with a coterminal angle situation:
sin(
11
11

1
)   sin(
)    sin( ) 
6
6
6
2
Let’s review those steps again
This helps solve problems efficiently.
Let’s look first at cos(
5
)
2
Divide out the fraction!

You get cos(  2 ) and now we know that we can invoke coterminal angles!
2

cos( )  0
2
Let’s look at another one
sin(
17
)
3
sin(
17
17
)   sin(
)
3
3
Now what about that
because sine is odd.
17
?
3
Divide it out!
Well, when you divide that out you get 5 
2
which isn’t quite a co-terminal angle…I need
3
for the result to be an even multiple of pi!
So let’s get clever:
5 
Ah! The angle that the
2
2
3 2
5
 4   
 4 

 4 
3
3
3
3
3
17
5
is co-terminal with is
So I really want:
3
3
Putting it all together:
sin(
17
17
5
3
3
)   sin(
)   sin( )   

3
3
3
2
2
Let’s look at another one:
cos( 
19
)
6
From even/odd, with cosine being even, we know:
cos(
19
19
)  cos(
)
6
6
Now about that
19
…when you divide it out
6
You get 3 

BUT
6

19
isn’t co-terminal with
!
6
6
Let’s do the same clever shift on the extra pi:
3 

6
There!
cos(
 2   

6
 2 
6 
7
  2 
6 6
6
7
19
is co-terminal with
so know we know that
6
6
19
19
7
3
)  cos(
)  cos( )  
6
6
6
2
The work making the change to co-terminal angles to solve the problem is called using the
PERIODICITY of the functions or using the Coterminal Angle to solve the problem.
Let’s review:

13
sin(  4 )  sin(
)
3
3
Now use coterminal angles!
16
cos(
)
3
divide it out!
tan( 
9
)
4
sin(765) 
cos(120) 
Popper 08, Question 6
Popper 08, Question 7
Popper 08
Question 8
Popper 08, Question 9
Popper 08, Question 10