Download Inverse Trig

Dr. Neal, WKU
MATH 117
Inverse Trig Functions
A function y = f (x) is one–to–one if it is always the case that different x values are
assigned to different y values.
3
For example, y = x + 4 is one–to–one, but y = sin x is not one–to–one. The sine
function assigns many different x values to the same y . For instance 0, π, 2π, 3π, etc. all
have a sine of 0.
3
y = sin x
is not one-to-one
y = x +4
is one-to-one
A one–to–one function f (x) always has an inverse denoted by f !1(x) . The
trigonometric functions cos x , sin x , tan x are not one-to-one; thus technically they do
not have inverses. However we can make them one-to-one by restricting the domains
to only two quadrants so that we have only one segment of the graph.
The Inverse Cosine (or Arccos)
With cos x , we restrict the domain to [0, π] (the 1st and 2nd quadrants). Then the
!1
“inverse cosine” exists: For !1 " x " 1 , cos x is the angle whose cosine is x , where the
angle must be in either the 1st or 2nd quadrant and must be written in radians.
!
1
!
–1
f (x) = cos x
Dom: 0 ≤ x ≤ π Range: –1 ≤ y ≤ 1
1
–1
!1
f !1(x) = cos x
Dom: –1 ≤ x ≤ 1 Range: 0 ≤ y ≤ π
We know the cosine values of the standard angles around the unit circle:
5!
6
!
2!
3! 3
4
!
2
!
3
!
4
cos(0) = 1
" !% 1
cos$ ' =
# 3& 2
!
6
0
" 3! %
2
cos$ ' = –
# 4&
2
1
" !%
3
cos$ ' =
# 6&
2
" !%
cos$ ' = 0
# 2&
" !%
2
cos$ ' =
# 4&
2
" 2! %
1
cos$ ' = –
# 3&
2
" 5! %
3
cos$ ' = –
# 6&
2
cos(!) = –
Dr. Neal, WKU
So we can compute the inverse cosine, or arccos, of the values
1,
1
3
2 1
2
3
,
, , 0, ! , !
,–
, –1 :
2
2
2 2
2
2
!
!
3
2
2
2
!
1
2
0
1
2
2
2
3
2
1
!1
cos !1 x is the angle between 0 and π whose cosine is x
" 3% !
cos !1 $$ '' =
# 2 &
6
cos !1(1) = 0
" 2% !
'' =
cos !1 $$
# 2 &
4
"
2 %' 3!
cos !1 $$ !
' =
# 2 &
4
" 1 % 2!
cos !1 $ ! ' =
# 2&
3
" 1%
!
cos !1 $ ' =
# 2&
3
"
3%
5!
cos !1 $$ ! '' =
# 2 &
6
cos !1(0) =
!
2
cos !1(!1) = !
Evaluating cos(cos!1 x) and cos !1 (cos x)
It is always the case that
cos(cos !1 x ) = x for !1 " x " 1 .
!1
This is because cos x is the angle whose cosine is x ; thus, cos(cos !1 x ) = x . We don't
!1
even have to calculate cos x in this case. For example,
(
)
cos cos !1 (!1 / 2) = !
(
!1
)
(
Also, cos cos (5) and cos cos
the interval !1 " x " 1 .
!1
1
2
(
)
and cos cos !1 ( 3 / 2) =
3
2
)
(!4) are undefined because 5 and –4 are not within
On the other hand,
cos !1 (cos x ) = x only for 0 ! x ! " .
In this case, we can start with any angle x . Then !1 " cos x " 1. So cos !1 (cos x ) exists,
but cos !1 (cos x ) must be an angle from 0 to π and may not equal the original angle x .
Dr. Neal, WKU
Example. Compute (a) cos
!1 #%
$ cos
7" &
(
6 '
(b) cos
!1 #%
$ cos
5" &
(
3 '
(c) cos
# 3" & &
% ! (( .
cos
$
$ 4 ''
!1 #%
Solution. We first evaluate the cosine of the inner given angle, then evaluate the inverse
cosine of the result. The answers must all be angles from 0 to π.
(a) cos
(b) cos
!1 #%
$ cos
!1 #%
$ cos
5" &
# 1& "
( = cos !1 % ( =
'
$ 2' 3
3
#
7" &
3 & 5"
( = cos!1 % !
(=
$ 2 '
6 '
6
(c) cos
#
# 3" & &
2 & 3"
% ! ( ( = cos !1 % !
(=
cos
$
$ 4 ''
$ 2 '
4
!1 #%
The Inverse Sine (or Arcsin)
With sin x , we restrict the domain to [–π/2, π/2] (the 1st and 4th quadrants). Now the
!1
“inverse sine” exists: For !1 " x " 1 , sin x is the angle whose sine is x , where the
angle must be in either the 1st or 4th quadrant. If the angle is in the 4th quadrant, then
we write it as a negative angle.
"
2
1
!
"
"
2
2
!1
1
!1
!
f (x) = sin x
"
!
Dom: ! ≤ x ≤
Range: –1 ≤ y ≤ 1
2
2
"
2
"
3
"
4
"
!
2
!
f !1(x) = sin
!1
x
Dom: –1 ≤ x ≤ 1 Range: !
"
!
≤y ≤
2
2
We know the sine values of the standard angles
from –π/2 to π/2:
"
6
0
"
!
3
"
2
"
4
"
!
6
# "&
sin% ! ( = –1
$ 2'
# "&
3
sin% ! ( = !
$ 3'
2
# "&
1
sin% ! ( = !
$ 6'
2
" !%
2
sin$ ' =
# 4&
2
sin(0 ) = 0
" !%
3
sin$ ' =
# 3&
2
# "&
2
sin% ! ( = !
$ 4'
2
" !% 1
sin$ ' =
# 6& 2
" !%
sin$ ' = 1
# 2&
Dr. Neal, WKU
So now we can compute the inverse sine, or arcsin, of the values
–1, !
1
1
3
2
2
3
,!
, ! , 0, ,
,
,1 :
2
2 2
2
2
2
1
3
2
2
2
1
2
0
!
!
!1
3
2
sin !1 x is the angle between !
sin !1( !1) = !
sin !1( 0) = 0
"
3%
"
sin !1 $$ ! '' = !
# 2 &
3
"
2
!
2
2
"
!
and whose sine is x
2
2
"
2 %'
"
sin !1 $$ !
' =!
# 2 &
4
" 2% !
'' =
sin !1 $$
# 2 &
4
"1 % !
sin !1 $ ' =
#2 &
6
1
2
" 1%
"
sin !1 $ – ' = !
# 2&
6
" 3% !
sin !1 $$ '' =
# 2 &
3
sin !1(1) =
!
2
Evaluating sin(sin"1 x) and sin"1 (sin x)
It is always the case that
!sin(sin !1 x ) = x! for !1 " x " 1 .
!1
This is because sin x is the angle whose sine is x ; thus, sin(sin !1 x ) = x . We don't
!1
even have to calculate sin x in this case. For example,
(
)
2
2
!1
)
sin sin !1 ( 2 / 2) =
(
!1
)
(
(
!1
)
and sin sin (!1 / 2) = !
1
2
Also, sin sin (!3) and sin sin (2. 4) are undefined because –3 and 2.4 are not within
the interval !1 " x " 1 .
On the other hand,
"
"
sin !1 (sin x ) = x only for ! # x # .
2
2
In this case, we can start with any angle x . Then !1 " sin x " 1. So sin !1 (sin x ) exists, but
sin !1 (sin x ) must be an angle from –π/2 to π/2 and may not equal the original angle x .
Dr. Neal, WKU
4 "&
!1 #
(
Example. Compute (a) sin % sin
$
3 '
5" &
!1 #
(b) sin % sin (
$
4 '
(c) sin
# 7" & &
%
((.
sin
$ $! 6 ''
!1 #%
Solution. We first evaluate the sine of the inner given angle, then evaluate the inverse
sine of the result. The answers must all be angles from –π/2 to π/2.
#
4 "&
3&
"
!1 #
( = sin !1 % !
(=!
(a) sin %$ sin
$ 2 '
3 '
3
5" &
2&
"
!1 #
!1 #
( =!
(b) sin %$ sin (' = sin % !
$ 2 '
4
4
# 7" & &
"
!1 #
!1 # 1 &
(c) sin % sin % ! ( ( = sin % ( =
$ $ 6 ''
$2' 6
The Inverse Tangent (or Arctan)
With tan x , we restrict the domain to (–π/2, π/2) (the 1st and 4th quadrants). Then the
!1
“inverse tangent” exists: For all x , tan x is the angle whose tangent is x , where the
angle must be in either the 1st or 4th quadrant. If the angle is in the 4th quadrant, then
we write it as a negative angle.
f (x) = tan x
"
!
< x <
2
2
Range: – ∞ < y < ∞
Domain: !
f !1(x) = tan
!1
x
Domain: – ∞ < x < ∞
"
!
Range: ! < y <
2
2
Dr. Neal, WKU
"
2
"
3
We know the tangent values of the standard angles
from –π/2 to π/2:
"
4
"
6
# "&
tan % ! ( = – ∞
$ 2'
# "&
tan % ! ( = ! 3
$ 3'
# "&
tan % ! ( = –1
$ 4'
0
!
!
"
!
3
"
2
!
"
4
# "&
1
tan % ! ( = !
$ 6'
3
"
6
" !%
tan $ ' = 1
# 4&
" !%
1
tan $ ' =
# 6&
3
tan ( 0) = 0
" !%
tan $ ' = 3
# 3&
" !%
tan $ ' = ∞
# 2&
So now we can compute the inverse tangent, or arctan, of the values
–∞, ! 3 , !1 , !
1
1
, 0,
, 1, 3 , ∞ :
3
3
!
3
1
1
3
0
1
3
"
"1
" 3
"!
tan !1 x is the angle between !
tan !1 (!") = !
tan !1 (0) = 0
"
2
tan !1 (! 3 ) = !
" 1 %
!
tan !1 $ ' =
# 3&
6
"
3
"
!
and whose tangent is x
2
2
tan !1 (!1) = !
tan !1 (1) =
!
4
tan
!1
" 1 %
"
tan !1 $ ! ' = !
#
3&
6
"
4
( 3 ) = !3
tan !1 (" ) =
!
2
Evaluating tan(tan"1 x) and tan!1 (tan x)
It is always the case that
!
tan (tan !1 x ) = x for all x .
!1
This is because tan x is the angle whose tangent is x ; thus, tan (tan !1 x ) = x . We don't
!1
even have to calculate tan x in this case. For example,
(
!1
)
(
!1
)
tan tan (20.4) = 20. 4 and tan tan (!0. 56) = !0. 56
Dr. Neal, WKU
On the other hand,
tan !1 (tan x ) = x only for !
"
"
<x< .
2
2
In this case, we can start with any angle x . Then ! " < tan x < " . So tan !1 (tan x ) exists,
but tan !1 (tan x ) must be an angle between –π/2 and π/2 and may not equal the original
angle x .
Example. Compute (a) tan
!1 #%
$ tan
5" &
(
6 '
(b) tan
!1 #%
$ tan
7" &
(
4 '
(c) tan
# 2" & &
%
(( .
tan
$ $ ! 3 ''
!1 #%
Solution. We first evaluate the tangent of the inner given angle, then evaluate the
inverse tangent of the result. The answers must all be angles from –π/2 to π/2.
(a) tan
(b) tan
!1 #%
$
tan
!1 #%
$
tan
5" &
# 1 &
"
( = tan !1 % !
(=!
'
$
'
6
3
6
7" &
"
( = tan !1 ( !1) = !
4 '
4
(c) tan
# 2" & &
"
( ( = tan !1( 3 ) =
tan % !
$ $ 3 ''
3
!1 #%