Transcript Document

Lesson 4.7. Inverse Trigonometric Functions.
Previously you have learned
To find an inverse of a function, let every x be y and every y be x, then
solve the equation for y.
 Inverse function notation f¯¹(x)
 For a function to have an inverse it has to be one-to-one. One x for one
y value, and one y for one x value. It will pass the vertical and the
horizontal line test.
 Two inverse functions on the graph reflect over y=x
f(x)  f¯¹(x) (x,y)  (y, x)
Sometimes you just don’t have a nice or convenient
algebraic process that will give you an inverse function.
Many functions need a special, new rule for their
inverse. Some examples of these functions are:
FINDING INVERSE OF A TRIG FUNCTION :
Given f(x) = sin (x)
y = sin (x)
x = sin (y)
y = arcsin (x)
f¯¹(x) = arcsin (x)
change f(x) for y
switch x’s and y’s
solve for y
write using function notation
f¯¹(x) = sin¯¹(x)
Inverse Trigonometric function notation:
Inverse sine
arcsin x
or
sin-1 x
Inverse cosine
arccos x
or
cos-1x
Inverse tangent
arctan x
or
tan-1x
FINDING INVERSE OF A TRIG FUNCTION ALGEBRAICALLY :
Given
y = sin (x)
- ½ = sin ( -π/6)
sin (-π/6) = - ½
(-π/6) = sin (- ½ )
switch x and y values
(-π/6) = arcsin ( - ½ )
solve for y
(-π/6) = sin¯¹( - ½ )
sin¯¹( - ½ ) = (-π/6)
How does this look on the graph ??
Graphing Inverse Trigonometric functions:
The graph of y = sin x
D: all reals R: [-1,1] Period: 2π
Y-int.(0,0)
- this function is not one-to-one (different x values yield the same y)
- with domain restricted to [-π/2 ; π/2] y = sin x is one-to one
- therefore, we can use this “piece to finds its inverse
Using the domain restricted to [-π/2 ; π/2] let’s graph y = arcsin x
Remember that y = arcsin x is equivalent to sin y = x
X
y-values of sin x
y = arcsin x
x – values of sin x
Graphs of inverse functions
The graph of y = arc sin x
Domain:
[ 1,1]
  
 , 

Range:  2 2 
NEXT ON THE LIST
• ARCCOSINE
• Sketch the common curve of y = cos x
• Select interval of the cosine that will be oneto-one
• Sketch the graph of arccosine
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)
1.0

5/6
2/3
/2

2/3
/3
/3
2/3

4/3
5/3
x
2
/3
/6
1.0
1.0
1.0
x
Graphs of inverse functions
The graph of y = arccos x
Domain:
Range:
[ 1,1]
[0,  ]
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)
4.0
/2
y=tan(x)
3.0
2.0
/4
1.0
/2
/4
/4
/2
x
4.0
2.0
2.0
1.0
2.0
/4
3.0
/2
4.0
  
D    ,  and R   ,  
 2 2
  
D   ,   and R    , 
 2 2
4.0
x
6.0
Graphs of inverse functions
The graph of y = arctan x
Domain:
(, )
Range:    
 , 
 2 2
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
When x<0, y=arcsin(x) will be in which quadrant?

2
y<0 in IV
When x<0, y=arccos(x) will be in which quadrant? y>0 in II
When x<0, y=arctan(x) will be in which quadrant? y<0 in IV
Evaluating Inverse Trigonometric Functions algebraically.
When evaluating inverse trigonometric functions, you are “looking for the
angle whose (insert: sin, cos, or tan) is x.” Also, keep in mind the
domain and range of each function.
Find the exact value for
 3

arcsin
 2 


Solution:
3
For what value of x is sin (x) = 2 ?
3
Sin ( π/3) =
therefore

2
3 

arcsin
 2  3


Evaluating Inverse Trigonometric Functions algebraically.
Find the exact value for
Solution:
For what value of x is cos (x) =  2 ?
2
Cos ( 3π/4) =  2 ; therefore
2

2  3

cos 



2
4


1