Download x - Hawker Maths 2017

Inverse Trig Functions
Unit Circle
Angles
and
Coordinates
 1 3
 2, 2 



(–1, 0)

3 1
 2 , 2 



2
 2 ,

90°
2
3 3 120°
5 4 135°
6 150°

2 2
 2 , 2 



3,1

 2 2


y (0, 1)
180°

2

60° 3
1 3
2, 2 


 2 2
 2 , 2 



4
45°

30° 6
 3 1
 2 ,2


x
0° 0
360° 2 (1, 0)
330°
11
315°
 3 1
7 210°
6
6  2 , 2 
225°


7
5
240° 300°
2  4 4
5 4  2 ,  2 

2 
2 
 2
3 270° 32 3
 1
3

,

 2
2 

(0, –1)
1
3
,

2
2 

Horizontal Line Test
• A test use to determine if a function is one-to-one. If a horizontal
line intersects a function's graph more than once, then the
function is not one-to-one.
• Note: The function y = f(x) is a function if it passes the vertical
line test. It is a one-to-one function if it passes both the vertical
line test and the horizontal line test.
This function is
not one-to-one.
Let us begin with a simple question:
What is the first pair of inverse functions that pop
into YOUR mind?
f ( x)  x 2
f 1 ( x)  x
This may not be your pair but
this is a famous pair. But
something is not quite right
with this pair. Do you know
what is wrong?
Congratulations if you guessed that the top function
does not really have an inverse because it is not 1to1
and therefore, the graph will not pass the horizontal
line test.
2
y

x
.
Consider the graph of
y
Note the two points
on the graph and
also on the line y=4.


f(2) = 4 and f(-2) = 4
so what is an inverse
function supposed
to do with 4?



f 1 (4)  2 or f 1 (4)  2 ?


By definition, a function cannot generate two different
outputs for the same input, so the sad truth is that this
function, as is, does not have an inverse.
x
So how is it that we arrange for this function to have an
inverse?
2
yx
y=x
We consider only one half
of the graph: x > 0.
4
The graph now passes
the horizontal line test
and we do have an
inverse:
f ( x)  x for x  0
2
y x
2




1
f ( x)  x
Note how each graph reflects across the line y = x onto
its inverse.
x
A similar restriction on the domain is necessary to
create an inverse function for each trig function.
Consider the sine function.
You can see right
away that the sine
function does not
pass the horizontal
line test.
But we can come up
with a valid inverse
function if we restrict
the domain as we did
with the previous
function.
y
y = sin(x)

y = 1/2





How would YOU restrict the domain?




x

Take a look at the piece of the graph in the red frame.
We are going to build
the inverse function
from this section of
the sine curve
because:
This section picks up
all the outputs of the
sine from –1 to 1.
This section includes
the origin. Quadrant I
angles generate the
positive ratios and
negative angles in
Quadrant IV generate
the negative ratios.
y









x


Lets zoom in and look at some
key points in this section.
I have plotted the special angles on the curve and the
table.
y
y = sin(x)
x





2

3

4

6
0

6

4

3

2
f ( x)

1
3
2
2

2
1

2
0
1
2
2
2
3
2

1






x

The new table generates the graph of the inverse.
The domain
1
x
sin ( x )
x sin( x)
of the chosen


1

To get a good

1
section of
2
2
 
look at the
the sineis
3

, 

3




 2 2
graph
of
the
2
3
3
2
So the range
2

inverse

2




of the arcsin
2
4
function, we
4
2
  
1

is

1


will “turn the
 2 , 2 


2
6
6
2
tables”
on
0
0
0
0
The range of
1


1
the sine
2
6
the chosen
6
2
function.
2


2
section of the
2
4
4
2
sine is [-1 ,1]
3


3
so the domain
2
3
3
2
of the arcsin is


1
1
[-1, 1].
2
2
Note how each point on the original graph gets “reflected”
onto the graph of the inverse.
    
 ,1 to 1, 
2   2
y = arcsin(x)
y
y = sin(x)
 3   3  
 ,
 

 3 2  to  2 , 3 

 


 2   2  
 ,
 

 4 2  to  2 , 4 

 



etc.
You will see the
inverse listed
as both:
arcsin( x) and sin 1 ( x)

x
In the tradition of inverse functions then we have:


 
sin    1  arcsin( 1)  or sin 1 (1) 
2
2
2
Unless you are
instructed to
 3 
 3 
use degrees,
3
 
1




sin   
 arcsin 

or
sin

 2  3
you should
3 2
 2  3


assume that
inverse trig
functions will
generate
outputs of real
numbers (in
radians).
The thing to remember is that for the trig function the
input is the angle and the output is the ratio, but for the
inverse trig function the input is the ratio and the output
is the angle.
The other inverse trig functions are generated by using
similar restrictions on the domain of the trig function.
Consider the cosine function:
What do you
think would be
a good domain
restriction for
the cosine?
Congratulations if
you realized that
the restriction we
used on the sine
is not going to
work on the
cosine.
y = cos(x)


y








x

The chosen section for the cosine is in the red frame. This
section includes all outputs from –1 to 1 and all inputs in
the first and second quadrants.
Since the domain and range for the section are 0,   and 1,1,
the domain and range for the inverse cosine are 1,1 and 0 ,  .
y = cos(x)
y
y
y = arccos(x)













x






x
The other trig functions require similar restrictions on
their domains in order to generate an inverse.
Like the sine function, the domain of the section of the
  
tangent that generates the arctan is  ,  .
 2 2
y
y
y=arctan(x)


y=tan(x)








x









  
D    ,  and R   ,  
 2 2
  
D   ,   and R    , 
 2 2

x

The table below will summarize the parameters we have
so far. Remember, the angle is the input for a trig function
and the ratio is the output. For the inverse trig functions
the ratio is the input and the angle is the output.
arcsin(x)
Domain
Range
arccos(x) arctan(x)
1  x  1 1  x  1    x  


2
x

2
0 x

2


2
x

2
The graphs give you the big picture concerning the
behavior of the inverse trig functions. Calculators are
helpful with calculations (later for that). But special
triangles can be very helpful with respect to the basics.
60
45
2
2
1
1
30
45

1
Use the special triangles above to answer the following.
Try to figure it out yourself before you click.
3
 3

arccos

2


csc 1 ( 2) 
3
 

30  or  because cos 30 
2
 6

 
 
30  or  because csc 30  2 / 1  2
 6

 
OK, lets try a few more. Try them before you peek.
60

45
2
1
1
2
30
45
3
2
 1 

1


arcsin 
  45 (or ) because sin 45 
4
2
 2
tan 1 ( 3 ) 

3
60 (or ) because tan 60 
 3
3
1
 1 
arcsin 
   45 (or   ) because sin  45   1
4
2
 2




Negative inputs for the arccos can be a little tricky.
y
60

2
2
3
1
30

-1
3
  180  60  120

 1 
arccos    
 2 

60

x

x 1
to check : cos120   
r
2

From the triangle you can see that arccos(1/2) = 60 degrees.
But negative inputs for the arccos generate angles in
Quadrant II so we have to use 60 degrees as a reference
angle in the second quadrant.

Example:

Find the exact value of tan arccos 2 .
3
adj 2
2
Let u = arccos , then cos u 
 .
3y
hyp 3
3
32  22  5
u


x
2
opp
2
tan arccos  tan u 
 5
3
adj
2
Finally, we encounter the composition of trig functions
with inverse trig functions. The following are pretty
straightforward compositions. Try them yourself before
you click to the answer.
 1   3  
  ?
sin  sin 


2



so
 1   3  
 3


  sin  
sin sin 


2
2




First, what do we know about
 ?
We know that  is an angle whose sine is
 3
.
2
Did you suspect from the beginning that this was the
answer because that is the way inverse functions are
SUPPOSED to behave? If so, good instincts but….
Consider a slightly different setup:
 

arcsin sin 120 
 3
  60.
arcsin 

2


This is also the
composition of two
inverse functions but…
Did you suspect the answer was going to be 120
degrees? This problem behaved differently because
the first angle, 120 degrees, was outside the range of
the arcsin. So use some caution when evaluating the
composition of inverse trig functions.
The remainder of this presentation consists of
practice problems, their answers and a few complete
solutions.
Find the
exact value
of each
expression
without using
a calculator.
When your
answer is an
angle,
express it in
radians.
Work out the
answers
yourself
before you
click.
 1 
1. sin  
 2 
2. arccos  1
10. sec 1 2 
1
 1 
11. arccos

 2
3. tan 1  1
 1 
4. arctan 

 3
5. arcsin 0 
 1 
6. cos 

 2
1

7. arctan  3
8. sin 1  1

 3

9. cos 

2


1
   
12. arcsin  sin    
  2 
13. arcsin sin 270
 


 1  
14. tan  arccos  
 2 

  
15. arccos cos
  3

 

 1  1  
16. sin  cos    
 2 

On most calculators, you access the inverse trig functions
by using the 2nd function option on the corresponding trig
functions. The mode button allows you to choose whether
your work will be in degrees or in radians. You have to stay
on top of this because the answer is not in a format that
tells you which mode you are in.
Use a calculator. For 17-20,
round to the nearest tenth
of a degree.
Use a calculator. For 21-24,
express your answers in
radians rounded to the
nearest hundredth.
17. cos (.6666)
21. tan 1 3.585
18. arctan( 2.345)
22. arcsin(. 7878)
19. arcsin( .1234)
23. cos 1  .2345
20. arccos( .8787)
24. arctan(. 7878)
1
Use a calculator. When your answer is an angle, express
it in radians rounded to the hundredth’s place. When
your answer is a ratio, round it to four decimal places, but
don’t round off until the very end of the problem.
25. arcsin sin 3.58
26. arcsin cos1
27. arctan(sin  2.34 )
28. cosarccos  .5758


29. cos sin 1  .5758
30. tan arccos .2345
Answers appear in the following slides.
Answers for problems 1 – 9.
 1   
1. sin   
6
 2 
2. arccos  1  
1

3. tan  1 
4
 1  
4. arctan 

 3 6
1
5. arcsin 0   0
 1  
6. cos 

 2 4

7. arctan  3 
3

1
8. sin  1 
2
  3  5

9. cos 
 6
2


1
Negative ratios for arccos
generate angles in Quadrant II.
y
1
2
 3

x
1


The reference angle is


6
6  5
 
so the answer is     
6
6 6
6
10. sec 1 2   cos 1 1 / 2   

11. arccos

y
3
 1  3

2 4
2
3
60
   

12. arcsin  sin      
2
  2 
 

  
15. arccos cos
  3
x


1 
   arccos  

2 3
 1  1  
 2
16. sin  cos      sin 
 2 
 3


-1

13. arcsin sin 270  arcsin  1  90 
2

 1  
 2 
14. tan  arccos    tan 
 3
y
 2 
 3 


14.
3


 2
15.
1
x

2
 3
Answers for 17 – 30.
1
17. cos (.6666)  48.2
21. tan 1 3.585  1.30

18. arctan( 2.345)  66.9
22. arcsin(. 7878)  0.91
19. arcsin( .1234)  7.1
23. cos 1  .2345  1.81
20. arccos( .8787)  151.5
24. arctan(. 7878)  0.67

25. arcsin sin 3.58  arcsin  0.4245...  0.44
26. arcsin cos1  arcsin 0.5403...  0.57
27. arctan(sin  2.34)  arctan  0.7184...  0.62
28. cosarccos .5758  .5758


29. cos sin 1  .5758  cos 0.6136...  0.8175
30. tan arccos.2345  tan 1.3341...  4.1455