Inverse Circular Functions

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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