Inverse Circular Functions.pdf

7/24/2019 Inverse Circular Functions.pdf

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Inverse Circular Functions and Equations Involving Them

Institute of Mathematics, University of the Philippines

Lecture 26

(IMath, UP) Inverse Circular Functions Lec. 26 1 / 28

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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


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We restrict each of their domains such that:


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We restrict each of their domains such that:

the circular function is one-to-one on the restricted domainthe range of the restricted circular function is the same


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Inverse sine function

−1

1

−π−2π π 2π−

π

2−

3π

2

π

2


http://find/




−1

1

−π−2π π 2π−

π

2−

3π

2

π

2


http://find/

http://goback/




−1

1

−π−2π π 2π−

π

2−

3π

2

π

2


http://find/



Definition

Inverse sine function:

For any x ∈ [−1, 1],

y = sin−1 x ⇔ x = sin y, y ∈− π

2, π

2

π

2

−

π

2

−1 1

dom sin−1 : [−1, 1] ran sin−1 =−π

2, π

2


http://find/



1. sin−1

√ 3

2


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1. sin−1

√ 3

2

y = sin−1

√ 3

2


http://find/



1. sin−1

√ 3

2

y = sin−1

√ 3

2

⇒ sin y =

√ 3

2 , y ∈

− π

2, π

2

y = π

3


http://find/

http://goback/



1. sin−1

√ 3

2

y = sin−1

√ 3

2

⇒ sin y =

√ 3

2 , y ∈

− π

2, π

2

y = π

32. sin−1

−1

2


http://find/



1. sin−1

√ 3

2

y = sin−1

√ 3

2

⇒ sin y =

√ 3

2 , y ∈

− π

2, π

2

y = π

32. sin−1

−1

2

y = sin−1−1

2


http://find/

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1. sin−1

√ 3

2

y = sin−1

√ 3

2

⇒ sin y =

√ 3

2 , y ∈

− π

2, π

2

y = π

32. sin−1

−1

2

y = sin−1−1

2 ⇒ sin y = −1

2, y ∈ −

π

2, π

2


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http://goback/



1. sin−1

√ 3

2

y = sin−1

√ 3

2

⇒ sin y =

√ 3

2 , y ∈

− π

2, π

2

y = π

32. sin−1

−1

2

y = sin−1

−1

2 ⇒ sin y = −1

2, y ∈ −

π

2, π

2 y = −π

6


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3. sin−1

−√

2

2


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3. sin−1

−√

2

2

= −π

4


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3. sin−1

−√

2

2

= −π

4

4. sin−1 0


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3. sin−1

−√

2

2

= −π

4

4. sin−1 0 = 0


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3. sin−1

−√

2

2

= −π

4

4. sin−1 0 = 0

5. sin−1(−1)


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3. sin−1

−√

2

2

= −π

4

4. sin−1 0 = 0

5. sin−1(−1) = −

π

2


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3. sin−1

−√

2

2

= −π

4

4. sin−1 0 = 0

5. sin−1(−1) = −

π

26. sin−1 2


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3. sin−1

−√

2

2

= −π

4

4. sin−1 0 = 0

5. sin−1(−1) = −

π

26. sin−1 2: undefined


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−1

1

−π−2π π 2π−

π

2−

3π

2

π

2


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http://goback/



−1

1

−π−2π π 2π−

π

2−

3π

2

π

2


http://find/



Definition

Inverse cosine function:

For any x ∈ [−1, 1],

y = cos−1 x ⇔ x = cos y, y ∈ [0, π]

π

2

π

−1 1

dom cos−1 : [−1, 1] ran cos−1 = [0, π]


http://goforward/

http://find/

http://goback/



1. cos−1

√ 3

2


http://goforward/

http://find/

http://goback/



1. cos−1

√ 3

2

y = cos−1√ 3

2


http://find/



1. cos−1

√ 3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]


http://find/



1. cos−1

√ 3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6


http://find/



1. cos−1

√ 3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6

2. cos−1−√ 2

2


http://find/



1. cos−1

√ 3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6

2. cos−1−√ 2

2

= 3π

4


√

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1. cos−1

√3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6

2. cos−1−√ 2

2

= 3π

4

3. cos−1(−1)


√

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1. cos−1

√3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6

2. cos−1−√ 2

2

= 3π

4

3. cos−1(−1) = π


√

http://find/



1. cos−1

√3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6

2. cos−1−√ 2

2

= 3π

4

3. cos−1(−1) = π

4. cos−1−

1

2


√ 3

http://goforward/

http://find/

http://goback/



1. cos−1

√3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6

2. cos−1−√ 2

2

= 3π

4

3. cos−1(−1) = π

4. cos−1−

1

2 =

2π

3


√ 3

http://find/



1. cos−1

√3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6

2. cos−1−√ 2

2

= 3π

4

3. cos−1(−1) = π

4. cos−1−

1

2 =

2π

3

5. cos−1 0


√ 3

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1. cos−1

√3

2

y = cos−1√ 3

2

⇒ cos y = √ 3

2 , y ∈ [0, π]

y = π

6

2. cos−1−√ 2

2

= 3π

4

3. cos−1(−1) = π

4. cos−1−

1

2 =

2π

3

5. cos−1 0 = π

2


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The other inverse circular functions are defined similarly:

y = tan−1 x ⇔ x = tan y,


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y = tan−1 x ⇔ x = tan y, x ∈ ,


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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2


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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y,


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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ ,


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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)


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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)

y = sec−1

x ⇔ x = sec y,


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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)

y = sec−1

x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞),


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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)

y = sec−1

x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π

2


http://find/




y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)

y = sec−1

x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π

2

y = csc−1 x ⇔ x = csc y,


http://find/




y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)

y = sec−1

x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π

2

y = csc−1 x ⇔ x = csc y, x ∈ (−∞,−1] ∪ [1, +∞),


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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)

y = sec−1

x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π

2

y = csc−1 x ⇔ x = csc y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈−π

2, π

2

\ {0}


Th h i i l f i d fi d i il l

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y = tan−1 x ⇔ x = tan y, x ∈ , y ∈

−π

2, π

2

y = cot−1 x ⇔ x = cot y, x ∈ , y ∈ (0, π)

y = sec−1

x ⇔ x = sec y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈ [0, π] \ π

2

y = csc−1 x ⇔ x = csc y, x ∈ (−∞,−1] ∪ [1, +∞), y ∈−π

2, π

2

\ {0}

The sin−1, cos−1, tan−1, . . . functions are also denoted

Arcsin, Arccos, Arctan, . . ..


Inverse tangent function:

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x = π

2 x = −π

2



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x = π

2 x = −π

2


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y = π

2

y = −π

2



http://find/



y = π

2

y = −π

2

dom tan−1 = ran tan−1 =−π

2, π

2


http://find/



Inverse cotangent function:

y = π

dom cot−1 = ran cot−1 = (0, π)


http://find/



Inverse secant function:

y = π

2

(1, 0)

(−1, π)

dom sec−1 = (−∞,−1] ∪ [1, +∞) ran sec−1 = [0, π] \π

2


http://find/



Inverse cosecant function:

(1, π2)

(−1,−π

2)

dom csc−1 = (−∞,−1] ∪ [1, +∞) ran csc−1 =−π

2, π

2

\ {0}


π

2

sin−1 tan−1 csc−1

http://find/



sin , tan , csc

−π

2

0

cos−1, cot−1, sec−1

π


π

2


http://find/



sin , tan , csc

−π

2

0


π

1 tan−1√ 3

3


π

2


http://find/



sin , tan , csc

−π

2

0


π

1 tan−1√ 3

3

= π

6


π

2

sin−1, tan−1, csc−1

http://find/



sin , tan , csc

−π

2

0


π

1 tan−1√ 3

3

= π

6

2 tan−1(−1)


π

2


http://find/

http://goback/



, ,

−π

2

0


π

1 tan−1√ 3

3

= π

6

2 tan−1(−1) = −π

4


π

2

sin−1, tan−1, csc−11 1 1

http://find/

http://goback/



, ,

−π

2

0


π

1 tan−1√ 3

3

= π6

2 tan−1(−1) = −π

4

3 cot−1 −√ 3


π

2

sin−1, tan−1, csc−11 1 1

http://find/



−π

2

0


π

1 tan−1√ 3

3

= π6

2 tan−1(−1) = −π

4

3 cot−1 −√ 3 =

5π

6


π

2

sin−1, tan−1, csc−11 1 1

http://find/



−π

2

0


π

1 tan−1√ 3

3

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

6

4 Arccot 0


π

2

sin−1, tan−1, csc−11 t 1 1

http://find/



−π

2

0


π

1 tan−1√ 33

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

6

4 Arccot 0 = π

2


π

2

sin−1, tan−1, csc−1−1 t−1 −1

http://find/



−π

2

0

cos 1, cot 1

, sec 1

π

1 tan−1√

33

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

6

4 Arccot 0 = π

2

5 sec−1 −√ 2


π

2


cos−1 cot−1 sec−1

http://find/



−π

2

0

cos 1, cot 1

, sec 1

π

1 tan−1√

33

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

6

4 Arccot 0 = π

2

5 sec−1 −√ 2 =

3π

4


π

2



http://find/



−π

2

0

cos , cot , sec

π

1 tan−1√

33

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

6

4 Arccot 0 = π

2

5 sec−1 −√ 2 =

3π

4

6 sec−1(−2)


π

2



http://find/



−π

2

0

cos , cot , sec

π

1 tan−1√

33

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

6

4 Arccot 0 = π

2

5 sec−1 −√ 2 =

3π

4

6 sec−1(−2) = 2π

3


π

2



http://find/



−π

2

0

cos , cot , sec

π

1 tan−1√

33

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

6

4 Arccot 0 = π

2

5 sec−1 −√ 2 =

3π

4

6 sec−1(−2) = 2π

3

7 Arccsc√

2


π

2



http://find/



−π

2

0

cos , cot , sec

π

1 tan−1√

33

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

64 Arccot 0 =

π

2

5 sec−1 −√ 2 =

3π

4

6 sec−1(−2) = 2π

3

7 Arccsc√

2 = π

4


π

2



http://find/



−π

2

0

cos , cot , sec

π

1 tan−1√

33

= π6

2 tan−1(−1) = −π

4

3

cot−1 −√ 3 =

5π

64 Arccot 0 =

π

2

5 sec−1 −√ 2 =

3π

4

6 sec−1(−2) = 2π

3

7 Arccsc√

2 = π

4

8 csc−1

−2√ 3

3


http://find/



Recall: (f −1 ◦ f )(x) = x for every x ∈ dom f



(IMath UP) Inverse Circular Functions Lec 26 17 / 28

http://find/




Thus



Thus,

sin−1(sin x) = x for every x ∈ −π

2 , π

2

cos−1(cos x) = x for every x ∈ [0, π]



Thus,

http://goforward/

http://find/

http://goback/



Thus,


2 , π

2


tan−1(tan x) = x for every x ∈−π

2, π

2

cot−1

(cot x) = x for every x ∈ (0, π)sec−1(sec x) = x for every x ∈ [0, π] \

π

2

csc−1(csc x) = x for every x ∈

−π

2, π

2

\ {0}



Thus,

http://find/



,


2 , π

2


tan−1(tan x) = x for every x ∈−π

2, π

2

cot−1

(cot x) = x for every x ∈ (0, π)sec−1(sec x) = x for every x ∈ [0, π] \

π

2

csc−1(csc x) = x for every x ∈

−π

2, π

2

\ {0}

Circ−1(Circ x) = x for every x in the restricted domain of Circ


1 tan−1 tan π

http://find/



1 tan tan12


1 tan−1 tan π

= π

http://find/



tan tan12 =

12


1 tan−1 tan π

= π

http://find/



tan

tan

12

12

2 cos−1

cos 7π8


1 tan−1 tan π

= π

http://find/



ta

ta

12 12

2 cos−1

cos 7π8

= 7π

8


1 tan−1 tan π

12 =

π

12

http://find/



12 12

2 cos−1

cos 7π8

= 7π

8

3 sin−1

sin

5π

6


1 tan−1 tan π

12 =

π

12

http://find/



12 12

2 cos−1

cos 7π8

= 7π

8

3 sin−1

sin

5π

6

= 5π

6


1 tan−1 tan π

12 =

π

12

http://find/



12 12

2 cos−1

cos 7π8

= 7π

8

3 sin−1

sin

5π

6

= 5π

6 since

5π

6 /∈

−π

2, π

2


1 tan−1 tan π

12 =

π

12

http://find/



12 12

2 cos−1

cos 7π8

= 7π

8

3 sin−1

sin

5π

6

= 5π

6 since

5π

6 /∈

−π

2, π

2

Rather, sin−1

sin 5π6


1 tan−1 tan π

12 =

π

12

http://find/



12 12

2 cos−1

cos 7π8

= 7π

8

3 sin−1

sin

5π

6

= 5π

6 since

5π

6 /∈

−π

2, π

2

Rather, sin−1

sin 5π6

= sin−1

12

(IM th UP) I s Ci l F ti s L 26 18 / 28

1 tan−1 tan π

12 =

π

12

http://find/



12 12

2 cos−1

cos 7π8

= 7π

8

3 sin−1

sin

5π

6

= 5π

6 since

5π

6 /∈

−π

2, π

2

Rather, sin−1

sin 5π6

= sin−1

12

= π

6

(IM th UP) I Ci l F ti L 26 18 / 28

1 tan−1 tan π

12 =

π

12

http://find/



12 12

2 cos−1

cos 7π8

= 7π

8

3 sin−1

sin

5π

6

= 5π

6 since

5π

6 /∈

−π

2, π

2

Rather, sin−1

sin 5π6

= sin−1

12

= π

6

Note that π

6 ∈

−π

2, π

2

.

(IM th UP) I Ci l F ti L 26 18 / 28

http://find/



In general, to obtain Circ−1(Circ θ), find α in the restricted domain of

Circ such that Circ α = Circ θ,



(IM h UP) I Ci l F i L 26 19 / 28


Circ such that Circ α = Circ θ, so that:

Circ−1(Circ θ)

http://find/






Circ−1(Circ θ) = Circ−1(Circ α) = α

http://find/







http://find/



π

2


−π

2

0


π


http://find/



π

2


−π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5





http://find/



π

2


−π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5

= 2π

5





http://find/



π

2


−π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5

= 2π

5

2 sec−1

sec 8π

7





http://find/



π

2


−π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5

= 2π

5

2 sec−1

sec 8π7

= sec−1

sec 6π

7





http://find/



π

2


−π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5

= 2π

5

2 sec−1

sec 8π7

= sec−1

sec 6π

7

= 6π

7





http://find/



π

2


−π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5

= 2π

5

2 sec−1

sec 8π7

= sec−1

sec 6π

7

= 6π

7

3 tan−1 tan 8π

11





π

http://find/



π

2


−π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5

= 2π

5

2 sec−1

sec 8π7

= sec−1

sec 6π

7

= 6π

7

3 tan−1 tan 8π

11 =

−3π

11





π

http://find/



π

2


−

π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5

= 2π

5

2 sec−1

sec 8π7

= sec−1

sec 6π

7

= 6π

7

3 tan−1 tan 8π

11 =

−3π

114 cos−1

cos 19π10





π

http://find/



π

2


−

π

2

0


π

1 csc−1

csc 3π5

= csc−1

csc 2π

5

= 2π

5

2 sec−1

sec 8π7

= sec−1

sec 6π

7

= 6π

7

3 tan−1 tan 8π

11 =

−3π

114 cos−1

cos 19π10

= π

10


Recall: (f ◦ f −1)(x) = x for every x ∈ dom f −1

http://find/

http://goback/





Thus,

http://find/



sin(sin−1 x) = x for every x ∈ [−1, 1]



Thus,

http://find/




cos(cos−1 x) = x for every x ∈ [−1, 1]



Thus,

http://goforward/

http://find/

http://goback/





tan(tan−1 x) = x for every x ∈

cot(cot−1 x) = x for every x

∈

sec(sec−1 x) = x for every x ∈ (−∞,−1] ∪ [1, +∞)

csc(csc−1 x) = x for every x ∈ (−∞,−1] ∪ [1, +∞)



Thus,

http://find/





tan(tan−1 x) = x for every x ∈

cot(cot−1 x) = x for every x

∈

sec(sec−1 x) = x for every x ∈ (−∞,−1] ∪ [1, +∞)

csc(csc−1 x) = x for every x ∈ (−∞,−1] ∪ [1, +∞)

Circ(Circ−1 x) = x for every x in the range of Circ


http://find/



1 cos

cos−1 23

= 2

3


http://find/



1 cos

cos−1 23

= 2

3

2 tan(tan−1(−4))


http://find/



1 cos

cos−1 23

= 2

3

2 tan(tan−1(−4)) = −4


http://goforward/

http://find/

http://goback/



1 cos

cos−1 23

= 2

3

2 tan(tan−1(−4)) = −4

3 cossin−1√ 32


http://find/



1 cos

cos−1 23

= 2

3

2 tan(tan−1(−4)) = −4

3 cossin−1√ 32 = cos

π

3


http://find/



1 cos

cos−1 23

= 2

3

2 tan(tan−1(−4)) = −4

3 cossin−1√ 32 = cos

π

3 = 12


Evaluate: sin

cos−1−3

5

http://find/




Evaluate: sin

cos−1−3

5

Let θ = cos

−1 3

5.

http://find/



−5


Evaluate: sin

cos−1−3

5

Let θ = cos

−1 3

5. Then, cos θ = 3

5,

http://find/



−5 −5


Evaluate: sin

cos−1−3

5

Let θ = cos

−1 3

5. Then, cos θ = 3

5, θ

∈ π2 , π.

http://find/



−5 −5 ∈ 2


Evaluate: sin

cos−1−3

5

Let θ = cos

−1 3

5. Then, cos θ = 3

5, θ

∈ π2 , π.

http://find/



−5 −5 ∈ 2

Since θ lies in QII,


Evaluate: sin

cos−1−3

5

Let θ = cos

−1 −3

5. Then, cos θ =

−3

5, θ

∈ π2 , π.

http://find/



−5 −5 ∈ 2

Since θ lies in QII, sin θ =


Evaluate: sin

cos−1−3

5

Let θ = cos−1 −3

5. Then, cos θ =

−3

5, θ

∈ π2 , π.

http://find/



5 5 ∈ 2

Since θ lies in QII, sin θ =√

1− cos2 θ


Evaluate: sin

cos−1−3

5

Let θ = cos−1 −3

5. Then, cos θ =

−3

5, θ

∈ π2 , π.

http://find/



∈ Since θ lies in QII, sin θ =

√ 1− cos2 θ =

1− 9

25


Evaluate: sin

cos−1−3

5

Let θ = cos−1 −

3

5. Then, cos θ =

−3

5

, θ∈

π

2

, π.

http://find/




√ 1− cos2 θ =

1− 9

25 =

1625


Evaluate: sin

cos−1−3

5


3

5. Then, cos θ =

−3

5

, θ∈

π

2

, π.

http://goforward/

http://find/

http://goback/




√ 1− cos2 θ =

1− 9

25 =

1625 = 4

5


Evaluate: sin

cos−1−3

5


3

5. Then, cos θ =

−3

5

, θ∈

π

2

, π.

http://goforward/

http://find/

http://goback/




√ 1− cos2 θ =

1− 9

25 =

1625 = 4

5

Alternative Solution:


Evaluate: sin

cos−1−3

5


3

5. Then, cos θ =

−3

5

, θ ∈

π

2

, π.

http://find/




√ 1− cos2 θ =

1− 9

25 =

1625 = 4

5

Alternative Solution: Take x =

−3, r = 5.


Evaluate: sin

cos−1−3

5


3

5. Then, cos θ =

−3

5

, θ ∈

π

2

, π.

http://find/




√ 1− cos2 θ =

1− 9

25 =

1625 = 4

5


−3, r = 5.

y =

52 − (−3)2


Evaluate: sin

cos−1−3

5


3

5. Then, cos θ =

−3

5

, θ ∈

π

2

, π.

http://find/




√ 1− cos2 θ =

1− 9

25 =

1625 = 4

5


−3, r = 5.

y =

52 − (−3)2 =√

16


http://find/



Evaluate: sin

cos−1−3

5


3

5. Then, cos θ =

−3

5

, θ ∈

π

2

, π.




√ 1− cos2 θ =

1− 9

25 =

1625 = 4

5


−3, r = 5.

y =

52 − (−3)2 =√

16 = 4

sin θ


Evaluate: sin

cos−1−3

5


3

5. Then, cos θ =

−3

5

, θ ∈

π

2

, π.

http://find/




√ 1− cos2 θ =

1− 9

25 =

1625 = 4

5


−3, r = 5.

y =

52 − (−3)2 =√

16 = 4

sin θ = 45


Evaluate: tan

csc−1(−√ 5)− tan−1 23

http://find/




Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

http://find/




Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√ 5, α ∈

http://find/



√


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√ 5, α ∈ −π

2 , 0

, tan β =

2

3 , β ∈ 0,

π

2

http://find/



√tan(α− β ) =


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2 , 0

, tan β =

2

3 , β ∈ 0,

π

2

http://find/



tan(α− β ) = tan α− tan β

1 + tan α tan β


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2 , 0

, tan β =

2

3 , β ∈ 0,

π

2

http://find/




1 + tan α tan β

For α: take r


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2 , 0

, tan β =

2

3 , β ∈ 0,

π

2

β

http://find/




1 + tan α tan β

For α: take r = √ 5,


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2 , 0

, tan β =

2

3 , β ∈ 0,

π

2

t t β

http://find/




1 + tan α tan β

For α: take r =

√ 5, y = −1.


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2 , 0

, tan β =

2

3 , β ∈ 0,

π

2

t t β

http://find/




1 + tan α tan β

For α: take r =

√ 5, y = −1.

⇒ x =√


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2 , 0,

tan β =

2

3, β ∈ 0,

π

2

t t β

http://find/




1 + tan α tan β

For α: take r =

√ 5, y = −1.

⇒ x =√

5


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2 , 0,

tan β =

2

3, β ∈ 0,

π

2

tan α tan β

http://find/




1 + tan α tan β

For α: take r =

√ 5, y = −1.

⇒ x =√

5− 1


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2, 0, tan β = 2

3, β ∈ 0, π

2

tan α tan β

http://find/




1 + tan α tan β

For α: take r =

√ 5, y = −1.

⇒ x =√

5− 1 = 2


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2, 0, tan β = 2

3, β ∈ 0, π

2

tan α− tan β

http://find/




1 + tan α tan β

For α

: take r =

√ 5

, y = −1

.

⇒ x =√

5− 1 = 2 ⇒ tan α =


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2, 0, tan β = 2

3, β ∈ 0, π

2

tan α− tan β

http://find/



tan(α− β ) = tan α tan β

1 + tan α tan β

For α

: take r =

√ 5

, y = −1

.

⇒ x =√

5− 1 = 2 ⇒ tan α = − 12


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2, 0, tan β = 2

3, β ∈ 0, π

2

tan α− tan β

http://find/




1 + tan α tan β

For α: take r =√

5, y = −

1.

⇒ x =√

5− 1 = 2 ⇒ tan α = − 12

tan(α

−β ) =

−12 − 2

3

1 +−1

2 2

3


Evaluate: tan

csc−1(−√ 5)− tan−1 23

Let α = csc−1(−√ 5), β = tan−1 2

3 .

csc α = −√

5, α ∈ −π

2, 0, tan β = 2

3, β ∈ 0, π

2

( )tan α− tan β

http://find/




1 + tan α tan β

For α: take r =√

5, y = −

1.

⇒ x =√

5− 1 = 2 ⇒ tan α = − 12

tan(α

−β ) =

−12 − 2

3

1 +−1

2 2

3 =

−7


Solve for x: cos−1x

4

+ tan−1

−√ 3

3

= cot−1 0

http://find/





4

+ tan−1

−√ 3

3

= cot−1 0

cos−1 x

4

http://find/





4

+ tan−1

−√ 3

3

= cot−1 0

cos−1 x

4 + −

π

6 =

http://find/





4

+ tan−1

−√ 3

3

= cot−1 0

cos−1 x

4 + −

π

6 =

π

2

http://find/

http://goback/





4

+ tan−1

−√ 3

3

= cot−1 0

cos−1 x

4 + −

π

6 =

π

2cos−1

x

4

=

π

6 +

π

2

http://find/



4

6 2



4

+ tan−1

−√ 3

3

= cot−1 0

cos−1 x

4 + −

π

6 =

π

2cos−1

x

4

=

π

6 +

π

2

http://find/



4

6 2

cos−1

x

4 =

2π

3



4

+ tan−1

−√ 3

3

= cot−1 0

cos−1 x

4 + −

π

6 =

π

2cos−1

x

4

=

π

6 +

π

2

http://find/



4

6 2

cos−1

x

4 =

2π

3

x4

= cos 2π3



4

+ tan−1

−√ 3

3

= cot−1 0

cos−1 x

4 + −

π

6 =

π

2cos−1

x

4

=

π

6 +

π

2

http://find/



4

6 2

cos−1

x

4 =

2π

3

x4

= cos 2π3

x = 4

−1

2



4

+ tan−1

−√ 3

3

= cot−1 0

cos−1 x

4 + −

π

6 =

π

2cos−1

x

4

=

π

6 +

π

2

http://find/



4

6 2

cos−1

x

4 =

2π

3

x4

= cos 2π3

x = 4

−1

2

x = −2


Solve for x: sec−1(−√ 2)− tan−1(x− 2) = csc−1(−2)

http://find/





3π

4

http://find/

http://goback/





3π

4 −tan−1(x

−2) =

http://find/





3π

4 −tan−1(x

−2) =

− π

6

http://find/

http://goback/





3π

4 −tan−1(x

−2) =

− π

63π

4 +

π

6 = tan−1(x − 2)

http://find/



11π

12 = tan−1(x − 2)



3π

4 −tan−1(x

−2) =

− π

63π

4 +

π

6 = tan−1(x − 2)

http://find/



11π

12 = tan−1(x − 2)

But ran tan−1 =−π

2, π

2

.



3π

4 −tan−1(x

−2) =

−

π

63π

4 +

π

6 = tan−1(x − 2)

http://goforward/

http://find/

http://goback/



11π

12 = tan−1(x − 2)

But ran tan−1 =−π

2, π

2

.

No solution


Solve for x: sin−1(x)− cos

−1(x) = π

6

http://find/





−1(x) = π

6

Let α = sin−1 x, β = cos−1 x.

http://find/





−1(x) = π

6


Then sin α = x

,

http://find/

http://goback/





−1(x) = π

6


Then sin α = x

, α ∈ −

π

2 ,

π

2,

http://find/

http://goback/





−1(x) = π

6


Then sin α = x, α ∈ −

π

2, π

2, cos β = x,

http://goforward/

http://find/

http://goback/





−1(x) = π

6



π

2, π

2, cos β = x, β

∈ [0, π].

http://find/





−1(x) = π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β

http://find/





−1(x) = π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β π

http://goforward/

http://find/

http://goback/



sin α = sinπ

6 + β


http://find/




−1(x) = π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β π

β



sin α = sinπ

6 + β

sin α = sin π

6 cos β + cos π

6 sin β

x =


http://find/




−1(x) = π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β

i iπ

+ β



sin α = sin

6 + β

sin α = sin π

6 cos β + cos π

6 sin β

x = 1

2(x)



−1(x) = π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β

i iπ

+ β

http://find/



sin α = sin

6 + β

sin α = sin π

6 cos β + cos π

6 sin β

x = 1

2(x) +

√ 3

2


Solve for x: sin−1

(x)− cos−1(x) =

π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β

si siπ

+ β

http://find/



sin α = sin

6 + β

sin α = sin π

6 cos β + cos π

6 sin β

x = 1

2(x) +

√ 3

2 sin β



(x)− cos−1

(x) = π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β

sin α sinπ

+ β

http://find/



sin α = sin

6 + β

sin α = sin π

6 cos β + cos π

6 sin β

x = 1

2(x) +

√ 3

2 sin β

Since β ∈ [0, π],



(x)− cos−1

(x) =

π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β

sin α = sinπ

+ β

http://find/

http://goback/



sin α = sin

6 + β

sin α = sin π

6 cos β + cos π

6 sin β

x = 1

2(x) +

√ 3

2 sin β

Since β ∈ [0, π], sin β =



(x)− cos−1

(x) =

π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β

sin α = sinπ

+ β

http://find/



sin α = sin

6 + β

sin α = sin

π

6 cos β + cos

π

6 sin β

x = 1

2(x) +

√ 3

2 sin β


1− cos2 β



(x)− cos−1

(x) =

π

6



π

2, π

2, cos β = x, β

∈ [0, π].

α = π

6 + β

sin α = sinπ

+ β

http://find/



sin α = sin

6 + β

sin α = sin

π

6 cos β + cos

π

6 sin β

x = 1

2(x) +

√ 3

2 sin β


1− cos2 β = √ 1− x2


x = 1

2(x) +

√ 3

2

1− x2

http://find/




x = 1

2(x) +

√ 32

1− x2

2x = x +√

3

1− x2

http://find/




x = 1

2(x) +

√ 32

1− x2

2x = x +√

3

1− x2

x = √ 3 1− x

2

http://find/




x = 1

2(x) +

√ 32

1− x2

2x = x +√

3

1− x2

x = √ 3 1− x

2

x2 = 3(1− x2)

http://find/




x = 1

2(x) +

√ 32

1− x2

2x = x +√

3

1− x2

x = √ 3 1− x

2

x2 = 3(1− x2)

x2 = 3− 3x2

http://find/



4x2 = 3


x = 1

2(x) +

√ 32

1− x2

2x = x +√

3

1− x2

x = √ 3 1− x

2

x2 = 3(1− x2)

x2 = 3− 3x2

http://find/



4x2 = 3

x2 = 34


x = 1

2(x) +

√ 32

1− x2

2x = x +√

3

1− x2

x = √ 3 1− x

2

x2 = 3(1− x2)

x2 = 3− 3x2

2

http://find/



4x2 = 3

x2 = 34

x = ±√

3

2


Check:

x = √ 32

: sin−1 √ 32 − cos−1 √ 3

2

http://find/




Check:

x = √ 32

: sin−1 √ 32 − cos−1 √ 3

2 = π

3 −

http://find/




Check:

x = √ 32

: sin−1 √ 32 − cos−1 √ 3

2 = π

3 − π

6

http://find/




Check:

x = √ 32

: sin−1 √ 32 − cos−1 √ 3

2 = π

3 − π

6 = π

6

http://find/




http://find/



Check:

x = √ 32

: sin−1 √ 32 − cos−1 √ 3

2 = π

3 − π

6 = π

6

x = −√

3

2 : sin−1

−√

3

2

− cos−1

−√

3

2



2

2

2


http://find/



Check:

x = √ 32

: sin−1 √ 32 − cos−1 √ 3

2 = π

3 − π

6 = π

6

x = −√

3

2 : sin−1

−√

3

2

− cos−1

−√

3

2

= −π

3 − 5π

6



2

2

2

3 6


Check:

x = √ 32

: sin−1 √ 32 − cos−1 √ 3

2 = π

3 − π

6 = π

6

x = −√

3

2 : sin−1

−√

3

2

− cos−1

−√

3

2

= −π

3 − 5π

6 = −7π

6

http://find/



2

2

2

3 6 6


http://find/

Inverse Circular Functions.pdf

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Transcript of Inverse Circular Functions.pdf