inverse circular functions
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Inverse Circular Functions and Equations Involving Them
Institute of Mathematics, University of the Philippines
Lecture 26
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1
Inverse Circular Functions
2
Equations involving Inverse Circular Functions
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Recall: Circular functions are not one-to-one
We restrict each of their domains such that:
the circular function is one-to-one on the restricted domain
the range of the restricted circular function is the same
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Definition
Inverse sine function:
For any x 2 [�1, 1],
y = sin
�1
x , x = sin y, y 2h
� ⇡
2
,
⇡
2
i
⇡2
�⇡2
�1 1
dom sin
�1
: [�1, 1] ran sin
�1
=
h
�⇡
2
,
⇡
2
i
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1. sin
�1
p3
2
!
y = sin
�1
p3
2
!
) sin y =
p3
2
, y 2h
� ⇡
2
,
⇡
2
i
y =
⇡
3
2. sin
�1
✓
�1
2
◆
y = sin
�1
✓
�1
2
◆
) sin y = �1
2
, y 2h
� ⇡
2
,
⇡
2
i
y = �⇡
6
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3. sin
�1
�p2
2
!
= �⇡
4
4. sin
�1
0 = 0
5. sin
�1
(�1) = �⇡
2
6. sin
�1
2: undefined
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Definition
Inverse cosine function:
For any x 2 [�1, 1],
y = cos
�1
x , x = cos y, y 2 [0,⇡]
⇡2
⇡
�1 1
domcos
�1
: [�1, 1] ran cos
�1
= [0,⇡]
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1. cos
�1
p3
2
!
y = cos
�1
p3
2
!
) cos y =
p3
2
, y 2 [0,⇡]
y =
⇡
6
2. cos
�1
�p2
2
!
=
3⇡
4
3. cos
�1
(�1) = ⇡
4. cos
�1
✓
�1
2
◆
=
2⇡
3
5. cos
�1
0 =
⇡
2
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The other inverse circular functions are defined similarly:
y = tan
�1
x , x = tan y, x 2 , y 2⇣
�⇡
2
,
⇡
2
⌘
y = cot
�1
x , x = cot y, x 2 , y 2 (0,⇡)
y = sec
�1
x , x = sec y, x 2 (�1,�1] [ [1,+1), y 2 [0,⇡] \n
⇡
2
o
y = csc
�1
x , x = csc y, x 2 (�1,�1] [ [1,+1), y 2h
�⇡
2
,
⇡
2
i
\ {0}
The sin
�1
, cos
�1
, tan
�1
, . . . functions are also denoted
Arcsin,Arccos,Arctan, . . ..
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Inverse tangent function:
x =
⇡2 x = �⇡
2
y =
⇡2
y = �⇡2
dom tan
�1
= ran tan
�1
=
⇣
�⇡
2
,
⇡
2
⌘
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Inverse cotangent function:
y = ⇡
domcot
�1
= ran cot
�1
= (0,⇡)
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Inverse secant function:
y =
⇡2
(1, 0)
(�1,⇡)
dom sec
�1
= (�1,�1] [ [1,+1) ran sec
�1
= [0,⇡] \n
⇡
2
o
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Inverse cosecant function:
(1,
⇡2 )
(�1,�⇡2 )
domcsc
�1
= (�1,�1] [ [1,+1) ran csc
�1
=
h
�⇡
2
,
⇡
2
i
\ {0}
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⇡2
sin
�1, tan
�1, csc
�1
�⇡2
0
cos
�1, cot
�1, sec
�1
⇡
1
tan
�1
p3
3
!
=
⇡
6
2
tan
�1
(�1) = �⇡
4
3
cot
�1
�
�p3
�
=
5⇡
6
4
Arccot 0 =
⇡
2
5
sec
�1
�
�p2
�
=
3⇡
4
6
sec
�1
(�2) =
2⇡
3
7
Arccsc
p2 =
⇡
4
8
csc
�1
�2
p3
3
!
= �⇡
3
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Recall: (f
�1 � f)(x) = x for every x 2 dom f
Thus,
sin
�1
(sinx) = x for every x 2h
�⇡
2
,
⇡
2
i
cos
�1
(cosx) = x for every x 2 [0,⇡]
tan
�1
(tanx) = x for every x 2⇣
�⇡
2
,
⇡
2
⌘
cot
�1
(cotx) = x for every x 2 (0,⇡)
sec
�1
(secx) = x for every x 2 [0,⇡] \n
⇡
2
o
csc
�1
(cscx) = x for every x 2h
�⇡
2
,
⇡
2
i
\ {0}
Circ
�1
(Circx) = x for every x in the restricted domain of Circ
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1
tan
�1
⇣
tan
⇡
12
⌘
=
⇡
12
2
cos
�1
✓
cos
7⇡
8
◆
=
7⇡
8
3
sin
�1
✓
sin
5⇡
6
◆
6= 5⇡
6
since
5⇡
6
/2h
�⇡
2
,
⇡
2
i
Rather, sin
�1
✓
sin
5⇡
6
◆
= sin
�1
✓
1
2
◆
=
⇡
6
Note that
⇡
6
2h
�⇡
2
,
⇡
2
i
.
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In general, to obtain Circ
�1
(Circ ✓), find ↵ in the restricted domain of
Circ such that Circ↵ = Circ ✓, so that:
Circ
�1
(Circ ✓) = Circ
�1
(Circ↵) = ↵
⇡2
sin
�1, tan
�1, csc
�1
�⇡2
0
cos
�1, cot
�1, sec
�1
⇡
1
csc
�1
�
csc
3⇡
5
�
= csc
�1
�
csc
2⇡
5
�
=
2⇡
5
2
sec
�1
�
sec
8⇡
7
�
= sec
�1
�
sec
6⇡
7
�
=
6⇡
7
3
tan
�1
�
tan
8⇡
11
�
= �3⇡
11
4
cos
�1
�
cos
19⇡
10
�
=
⇡
10
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Recall: (f � f�1
)(x) = x for every x 2 dom f
�1
Thus,
sin(sin
�1
x) = x for every x 2 [�1, 1]
cos(cos
�1
x) = x for every x 2 [�1, 1]
tan(tan
�1
x) = x for every x 2cot(cot
�1
x) = x for every x 2sec(sec
�1
x) = x for every x 2 (�1,�1] [ [1,+1)
csc(csc
�1
x) = x for every x 2 (�1,�1] [ [1,+1)
Circ(Circ
�1
x) = x for every x in the range of Circ
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1
cos
�
cos
�1
2
3
�
=
2
3
2
tan(tan
�1
(�4)) = �4
3
cos
⇣
sin
�1
p3
2
⌘
= cos
�
⇡
3
�
=
1
2
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Evaluate: sin
⇥
cos
�1
�
�3
5
�⇤
Let ✓ = cos
�1
�
�3
5
�
. Then, cos ✓ = �3
5
, ✓ 2⇥
⇡
2
,⇡
⇤
.
Since ✓ lies in QII, sin ✓ =
p1� cos
2
✓ =
q
1� 9
25
=
q
16
25
=
4
5
Alternative Solution: Take x = �3, r = 5.
y =
p
5
2 � (�3)
2
=
p16 = 4
sin ✓ =
4
5
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Evaluate: tan
�
csc
�1
(�p5)� tan
�1
2
3
�
Let ↵ = csc
�1
(�p5), � = tan
�1
2
3
.
csc↵ = �p5, ↵ 2
⇥
�⇡
2
, 0
�
, tan� =
2
3
, � 2⇥
0,
⇡
2
�
tan(↵� �) =
tan↵� tan�
1 + tan↵ tan�
For ↵: take r =
p5, y = �1.
) x =
p5� 1 = 2 ) tan↵ = � 1
2
tan(↵� �) =
�1
2
� 2
3
1 +
�
�1
2
� �
2
3
�
= �7
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Solve for x: cos
�1
⇣
x
4
⌘
+ tan
�1
�p3
3
!
= cot
�1
0
cos
�1
⇣
x
4
⌘
+
⇣
�⇡
6
⌘
=
⇡
2
cos
�1
⇣
x
4
⌘
=
⇡
6
+
⇡
2
cos
�1
⇣
x
4
⌘
=
2⇡
3
x
4
= cos
2⇡
3
x = 4
✓
�1
2
◆
x = �2
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Solve for x: sec
�1
(�p2)� tan
�1
(x� 2) = csc
�1
(�2)
3⇡
4
� tan
�1
(x� 2) = � ⇡
6
3⇡
4
+
⇡
6
= tan
�1
(x� 2)
11⇡
12
= tan
�1
(x� 2)
But ran tan
�1
=
⇣
�⇡
2
,
⇡
2
⌘
.
No solution
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Solve for x: sin
�1
(x)� cos
�1
(x) =
⇡
6
Let ↵ = sin
�1
x, � = cos
�1
x.
Then sin↵ = x, ↵ 2h
�⇡
2
,
⇡
2
i
, cos� = x, � 2 [0,⇡].
↵ =
⇡
6
+ �
sin↵ = sin
⇣
⇡
6
+ �
⌘
sin↵ = sin
⇡
6
cos� + cos
⇡
6
sin�
x =
1
2
(x) +
p3
2
sin�
Since � 2 [0,⇡], sin� =
p
1� cos
2
� =
p1� x
2
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x =
1
2
(x) +
p3
2
p
1� x
2
2x = x+
p3
p
1� x
2
x =
p3
p
1� x
2
x
2
= 3(1� x
2
)
x
2
= 3� 3x
2
4x
2
= 3
x
2
=
3
4
x = ±p3
2
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Check:
x =
p3
2
: sin
�1
p3
2
� cos
�1
p3
2
=
⇡
3
� ⇡
6
=
⇡
6
X
x = �p3
2
: sin
�1
�p3
2
!
� cos
�1
�p3
2
!
= �⇡
3
� 5⇡
6
= �7⇡
6
extraneous
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