22
Inverse Trig Functions Section 4.7
Inverse Trig Functions
Section 4.7
Inverse Sine, Cosine, Tangent Functions
Is a one to one funcsin tion?y x
Consider the function sin( ) restricted over the interval , ...2 2
y x Do you remember the requirement?
It's graph must pass both the horizontal and vertical line tests.
and it’s inverse1sin( ) or si nx y y xGraph of sin( )y x
, by [-2 , 2]2 2
2
2
2
2
22
2
2
1 1
-The unique angle, y, in the interval [ , ] such that sin( ) is the inverse sine (or arcsine) of x, 2 2-denoted by sin or arcsin . The domain of sin is [-1, 1] and the range is [ , ] .2 2
T
y x
x x y x
hink of the range of this function as being to the RIGHT of the Y-AXIS. (Right half of unit circle)
( 0 , - 1 )
( 0 , 1 )
( - 1, 0 )
(- 1
2,
3
2)
(2
2,
- 2
2)
(- 2
2,
- 2
2)
(1
2,
- 3
2)
(2
2,
2
2)
(- 1
2,
- 3
2)
(1
2,
3
2)
(3
2,
- 1
2)
(- 3
2,
1
2)
(- 3
2,
-1
2)
(3
2,
1
2)
(- 2
2,
2
2)
( 1, 0)
2
3
2
11
6
5
6
7
6
5
4
7
4
3
4
5
3
2
3
4
3
360
330
315
300
270
180
240
225
210
150
135
120
90
6
4
360
45
30
0
, by [-2 , 2]2 2
2
2
[-2 , 2] by , 2 2
2
2
y = sin x
y = arcsin x
EXAMPLES 1 1sin2
1arcsin2
OR
These expressions are related to 1sin2
angle ratio
2
1
1
45
A C
B
2
3
1
60
30
A
C
B6
3
4
1 1sin2 6
The angle between and 2 2
1whose sine is , is the angle 2 6
EXAMPLES
1 5sin sin6
angleratio
12
1 1sin sin2
angle
1 5sin6
sin
1si 1s n2
n i
n
6si
5? ?6
6
( 0 , - 1 )
( 0 , 1 )
( - 1, 0 )
(- 1
2,
3
2)
(2
2,
- 2
2)
(- 2
2,
- 2
2)
(1
2,
- 3
2)
(2
2,
2
2)
(- 1
2,
- 3
2)
(1
2,
3
2)
(3
2,
- 1
2)
(- 3
2,
1
2)
(- 3
2,
-1
2)
(3
2,
1
2)
(- 2
2,
2
2)
( 1, 0)
2
3
2
11
6
5
6
7
6
5
4
7
4
3
4
5
3
2
3
4
3
360
330
315
300
270
180
240
225
210
150
135
120
90
6
4
360
45
30
0
ratio
ratio
1 1 in2
s
angle
Q: How do we explain the inconsistency??
A: It’s a matter of Domain and Range
1 1
The unique angle, y, in the interval [0 , ] such that cos( ) is the inverse cosine (or arccosine) of x,
denoted by cos or arccos . The domain of cos is [-1, 1] and the range is [0 , ] .Thi
y x
x x y x
nk of the range of this function as being to the ABOVE of the X-AXIS. (Top half of unit circle)
[0 , ] by [-2 , 2]
[-2 , 2] by [0 , ]
0
y = cos x
y = arccos x
( 0 , - 1 )
( 0 , 1 )
( - 1, 0 )
(- 1
2,
3
2)
(2
2,
- 2
2)
(- 2
2,
- 2
2)
(1
2,
- 3
2)
(2
2,
2
2)
(- 1
2,
- 3
2)
(1
2,
3
2)
(3
2,
- 1
2)
(- 3
2,
1
2)
(- 3
2,
-1
2)
(3
2,
1
2)
(- 2
2,
2
2)
( 1, 0)
2
3
2
11
6
5
6
7
6
5
4
7
4
3
4
5
3
2
3
4
3
360
330
315
300
270
180
240
225
210
150
135
120
90
6
4
360
45
30
0 2
cos( )y x
EXAMPLES 1 1cos2
1arccos2
OR
These expressions are related to 1cos2
angle ratio
2
1
1
45
A C
B
2
3
1
60
30
A
C
B6
3
4
1 1cos2
The angle between 0 and 1whose cosine is is the angle 2 3
3
EXAMPLES
1cos cos3
angleratio
12
1 1cos cos2
angle
1 cos3
cos
1co 1c s2
s o
s
3co
? ?3
3
( 0 , - 1 )
( 0 , 1 )
( - 1, 0 )
(- 1
2,
3
2)
(2
2,
- 2
2)
(- 2
2,
- 2
2)
(1
2,
- 3
2)
(2
2,
2
2)
(- 1
2,
- 3
2)
(1
2,
3
2)
(3
2,
- 1
2)
(- 3
2,
1
2)
(- 3
2,
-1
2)
(3
2,
1
2)
(- 2
2,
2
2)
( 1, 0)
2
3
2
11
6
5
6
7
6
5
4
7
4
3
4
5
3
2
3
4
3
360
330
315
300
270
180
240
225
210
150
135
120
90
6
4
360
45
30
0
ratio
ratio
1 1 os2
c
angle
Q: How do we explain the inconsistency??
A: It’s a matter of Domain and Range
1 1
-The unique angle, y, in the interval ( , ) such that tan( ) is the inverse tangent (or arctan) of x, 2 2-denoted by tan or arctan . The domain of tan is (- , ) and the range is ( , ).2 2
y x
x x y x
Think of the range of this function as being to the RIGHT of the Y-AXIS. (Right half of unit circle, excluding the top and bottom points)
[ , ] by [-2 , 2]2 2
( 0 , - 1 )
( 0 , 1 )
( - 1, 0 )
(- 1
2,
3
2)
(2
2,
- 2
2)
(- 2
2,
- 2
2)
(1
2,
- 3
2)
(2
2,
2
2)
(- 1
2,
- 3
2)
(1
2,
3
2)
(3
2,
- 1
2)
(- 3
2,
1
2)
(- 3
2,
-1
2)
(3
2,
1
2)
(- 2
2,
2
2)
( 1, 0)
2
3
2
11
6
5
6
7
6
5
4
7
4
3
4
5
3
2
3
4
3
360
330
315
300
270
180
240
225
210
150
135
120
90
6
4
360
45
30
0
2
2
-[ 2 , 2] by [ , ]2 2
2
2
y = tan x
y = arctan x
EXAMPLES
1 3tan tan4
angleratio
1 1tan tan 1
angle
1 3tan4
tan
1tant n 1a an4
t
3? ?4
4
( 0 , - 1 )
( 0 , 1 )
( - 1, 0 )
(- 1
2,
3
2)
(2
2,
- 2
2)
(- 2
2,
- 2
2)
(1
2,
- 3
2)
(2
2,
2
2)
(- 1
2,
- 3
2)
(1
2,
3
2)
(3
2,
- 1
2)
(- 3
2,
1
2)
(- 3
2,
-1
2)
(3
2,
1
2)
(- 2
2,
2
2)
( 1, 0)
2
3
2
11
6
5
6
7
6
5
4
7
4
3
4
5
3
2
3
4
3
360
330
315
300
270
180
240
225
210
150
135
120
90
6
4
360
45
30
0
ratio
ratio
1 1 ta n angle
Q: How do we explain the inconsistency??
A: It’s a matter of Domain and Range
Let’s talk DOMAIN and RANGE…
sin( )y x
: ,DOMAIN ANGLES
: 1,1RANGE RATIOS
1sin ( )y x:[ 1,1]DOMAIN
RATIOS
: ,2 2
RANGE
ANGLES
Let’s talk DOMAIN and RANGE…
cos( )y x
: ,DOMAIN ANGLES
: 1,1RANGE RATIOS
1cos ( )y x:[ 1,1]DOMAIN
RATIOS : 0,RANGE
ANGLES
Let’s talk DOMAIN and RANGE…
tan( )y x*
: ,2 2
DOMAIN
ANGLES: ( , )RANGE
RATIOS
1tan ( )y x: ( , )DOMAIN RATIOS
: ,2 2
RANGE
ANGLES
Now some examples…Evaluate w/out Calc
1 1cos(sin )2
RATIO
Find the cosine of the angle whose sine is ½
Start with innermost expression 1 1sin 2
1 1sincos( )2
6
cos6
Work your way out
Evaluate
32
Now some examples…Evaluate w/out Calc
arcsin(cos )3
ANGLE
Find the angle whose sine is the cosine of
Start with innermost expression cos 3
arcsin(cos )3
12
12
arcsin
Work your way out
Evaluate 6
3
Now some examples…Evaluate w/out Calc
7arccos(tan )4
ANGLE
Find the angle whose cos is the tangent of
Start with innermost expression 7tan 4
7taarcc s( no )4
1
arccos 1Work your way out
Evaluate
74
Example: #41 & #42*, P. 422
ca orc scos( )x
,2 2
ANGLES
arcsinsin( ) 1x
sin ? 1
[ 1,1]RATIOS
So 1x
sin 12
arccos ?
( , )ANGLES
[ 1,1]RATIOS
arccos 1 So can be ANY angle in ( , )that has a cosine of 1...x
x ,3 ,5 , , 3
(2 1) where is an integerx n n
... odd multiples of x
State the DOMAIN and RANGE the composite function
sin(arccos )x
RATIOS
Decide the domain of innermost function This is your DOMAIN
:[0,1]RANGE
arccos[sin( )1,1]
sin[0, ]
Find RANGE of innermost function
This is the DOMAIN of the outer function
Determine the RANGE based on “new”
DOMAIN
:[ 1,1]DOMAIN
State the DOMAIN and RANGE the composite function
arccos(sin )x
ANGLES
Decide the domain of innermost function This is your DOMAIN
:[0, ]RANGE
sin(a , )rccos( )
[ar cos ]c 1,1
Find RANGE of innermost function
This is the DOMAIN of the outer function
Determine the RANGE based on “new”
DOMAIN
: ,DOMAIN
State the DOMAIN and RANGE the composite function
1sin(tan (cos ))x
ANGLES
2 2: ,2 2
RANGE
1 cossin(tan ( ( , )) 1s [in(ta ]n )1,1
: ,DOMAIN
sin ,4 4
1s tan [ 1in( ),1]
x
1
1. Let be an angle whose tangent is .x1
opp x xadj
12. Find an .t xAngle whose tan is x =
4. Find the hypotenuse.
2 2 2
2 2
1
1 1
h x
h x x
h
-15. Find sin tan( ( ))x sin( )2 1
x xh x
16. Find cos(tan ( ))x2
1cos1x
-1
Write an ALGEBRAIC expression that represents
sin(tan ( ))x
-13 . Find sin tan( ( ))x sin( ) opphyp
2 1x
550'100
110
#25, p. 433: A boat is due east of the shoreline running north/south. The bearings of the boat from two points on the shore that are 550’ apart from each other are 100 and 110 degrees. How far is the boat from the shore?
Consider two of the triangles formed in the diagram.
d
x
80
d
550 + x
70
x
tan80
tan805.671
dx
d xd x
tan 70550
( 550) tan 702.747 1511.11
dx
d xd x
5.671 2.747 1511.112.924 1511.11
516.83'
x xxx
So 516.83(5.671)2930.94
dd
