M27 Inverse Circ Functions and Equations.pdf

(1)

Inverse Circular

Functions and Equations

Involving Them

At the end of this lecture, a student must be able to:

1 _{Evaluate values involving inverse circular functions}

(2)

Inverse Circular Functions

Recall:

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Function Domain Range

sin−1x [−1,1] [−π

2,

π

2]

cos−1x [−1,1] [0, π]

tan−1x _R (−π

2,

π

2)

cot−1_x

R (0, π) sec−1x (−∞,−1]∪[1,+∞) [0, π]\ {π

2}

csc−1x (−∞,−1]∪[1,+∞) [−π

2,

π

(3)

Note: The ranges of the inverse circular functions correspond to the restricted domains of the circular functions.

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(4)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(5)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(6)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(7)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) =

− π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(8)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(9)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1 4)

is undefined

cossin−1

√

3 2

=

cos π₃

(10)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(11)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(12)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(13)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

Circ−1x=y m

Circy=x,

y ∈ranCirc−1

cos−1₍₋√2 2 ) =

3π

4

tan−1(−√3) = − π

3

csc−1₍₋1

4) is undefined

cossin−1

√

3 2

=

cos π₃

(14)

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

1 tan−1 tan π

12

= ₁₂π

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

Rather,sin−1 sin5₆π

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(15)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(16)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(17)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(18)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(19)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π

since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(20)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(21)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(22)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(23)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(24)

1 tan−1 tan π

12

= π 12

2 cos−1 cos7π

8

= 7π₈

3 sin−1 sin5π

6

= 5₆π since 5₆π ∈/

−π

2,

π

2

= sin−1 1₂

= π₆

Note that π₆ ∈

−π

2,

π

2

(25)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

sin−1_, _tan−1_,_csc−1

−π 2

To obtainCirc−1(Circ θ) whenθ is not in the restricted domain of Circ:

(26)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

(27)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

Useθ’s reference angle

(28)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

Useθ’s reference angle to findα in the restricted domain ofCirc

(29)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

Useθ’s reference angle to findα in the restricted domain ofCircsuch that Circα= Circθ.

(30)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

Useθ’s reference angle to findα in the restricted domain ofCircsuch that Circα= Circθ. Thus Circ−1( Circ θ)

(31)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

Useθ’s reference angle to findα in the restricted domain ofCircsuch that Circα= Circθ. Thus Circ−1(Circ θ) = Circ−1(Circ α)

(32)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

(33)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

(34)

csc−1 csc3₅π

= csc−1 csc2₅π

= 2₅π

π 2

−π 2

(35)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

1 sec−1 sec8π

7

= sec−1 _sec6π

7

= 6₇π

2 tan−1 tan8π

11

=−3π

11

3 cos−1 cos19π

10

(36)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

1 sec−1 sec8π

7

= sec−1 _sec6π

7

= 6₇π

2 tan−1 tan8π

11

=−3π

11

3 cos−1 cos19π

10

(37)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

1 sec−1 sec8π

7

= sec−1 _sec6π

7

= 6₇π

2 tan−1 tan8π

11

=−3π

11

3 cos−1 cos19π

10

(38)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

1 sec−1 sec8π

7

= sec−1 _sec6π

7

= 6₇π

2 tan−1 tan8π

11

=−3π

11

3 cos−1 cos19π

10

(39)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

1 sec−1 sec8π

7

= sec−1 _sec6π

7

= 6₇π

2 tan−1 tan8π

11

=−3π

11

3 cos−1 cos19π

10

(40)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

1 sec−1 sec8π

7

= sec−1 _sec6π

7

= 6₇π

2 tan−1 tan8π

11

=−3π

11

3 cos−1 cos19π

10

(41)

π 2

sin−1_, _tan−1_, _csc−1

−π 2

0

cos−1_, _cot−1_, _sec−1

π

1 sec−1 sec8π

7

= sec−1 _sec6π

7

= 6₇π

2 tan−1 tan8π

11

=−3π

11

3 cos−1 cos19π

10

(42)

Example: Find the numerical value ofsincos−1 ₋3 5

.

Solution:

If we let θ= cos−1 −3 5

, we are computing forsinθ. Now, cosθ =−3

5 and θ∈[0, π].

SinceP(θ) lies in QII,sinθ=√1−cos2_θ ₌q₁₋ 9 25 =

q

16 25 =

(43)

.

Solution:

,

we are computing forsinθ. Now, cosθ =−3

5 and θ∈[0, π].

q

16 25 =

(44)

.

Solution:

, we are computing forsinθ.

Now, cosθ =−3

5 and θ∈[0, π].

q

16 25 =

(45)

.

Solution:

5 and θ∈[0, π].

q

16 25 =

(46)

.

Solution:

5 and θ∈[0, π].

SinceP(θ) lies in QII,

sinθ=√1−cos2_θ ₌q₁₋ 9 25 =

q

16 25 =

(47)

.

Solution:

5 and θ∈[0, π].

SinceP(θ) lies in QII,sinθ=

√

1−cos2_θ ₌q₁₋ 9 25 =

q

16 25 =

(48)

.

Solution:

5 and θ∈[0, π].

SinceP(θ) lies in QII,sinθ=√1−cos2_θ

=

q

1− 9 25 =

q

16 25 =

(49)

.

Solution:

5 and θ∈[0, π].

SinceP(θ) lies in QII,sinθ=√1−cos2_θ ₌q₁₋ 9 25

=

q

16 25 =

(50)

.

Solution:

5 and θ∈[0, π].

q

16 25

(51)

.

Solution:

5 and θ∈[0, π].

q

16 25 =

(52)

Compute for the exact numerical value ofsin1₂(sec−1(−3 2))

.

Solution:

If we let θ= sec−1 −3 2

, we are computing forsinθ₂.

Nowsecθ=−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

Sinceθ ∈[0, π]\ {π

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(53)

.

Solution:

,

we are computing forsinθ₂.

Nowsecθ=−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(54)

.

Solution:

Nowsecθ=−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(55)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}.

Thuscosθ=−2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(56)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(57)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(58)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}.

Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(59)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(60)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2

=

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(61)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=−

2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2

=

q

5 3

1 2

= √

(62)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(63)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(64)

.

Solution:

Nowsecθ =−3

2 and θ∈[0, π]\ {0}. Thus cosθ=− 2 3.

2},

θ

2 ∈[0,

π

2]\ {

π

4}. Hence,

sinθ 2 =

r

1−cosθ

2 =

s

1−(−2 3)

2 =

q

5 3

1 2

= √

(65)

Example: Find the numerical value oftan csc−1(−√5)− tan−1 2₃.

Solution:

If we let α= csc−1(−√5)and β = tan−1 2₃, we are computing for

tan(α−β) = tanα−tanβ 1 + tanαtanβ.

Now, cscα =−√5, α∈

−π

2,

π

2

\ {0}, and tanβ = 2₃,β ∈[0, π). SinceP(α) lies in QIV,

cosα=p1−sin2α =

q

1− 1 5 =

q

4 5 =

2

√

5. Thus, tanα=− 1 2.

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(66)

Example: Find the numerical value oftan csc−1(−√5)− tan−1 2₃. Solution:

If we let α= csc−1(−√5)and β = tan−1 2₃,

we are computing for

−π

2,

π

2

cosα=p1−sin2α =

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(67)

Example: Find the numerical value oftan csc−1(−√5)− tan−1 2₃. Solution:

tan(α−β)

= tanα−tanβ 1 + tanαtanβ.

−π

2,

π

2

cosα=p1−sin2α =

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(68)

Solution:

−π

2,

π

2

cosα=p1−sin2α =

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(69)

Solution:

−π

2,

π

2

\ {0}, and

tanβ = 2₃,β ∈[0, π). SinceP(α) lies in QIV,

cosα=p1−sin2α =

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(70)

Solution:

−π

2,

π

2

\ {0}, and tanβ = 2₃,β ∈[0, π).

SinceP(α) lies in QIV, cosα=p1−sin2α =

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(71)

Solution:

−π

2,

π

2

\ {0}, and tanβ = 2₃,β ∈[0, π). SinceP(α)lies in QIV,

cosα=p1−sin2α =

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(72)

Solution:

−π

2,

π

2

cosα=p1−sin2α

=

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(73)

Solution:

−π

2,

π

2

cosα=p1−sin2α=

q

1− 1 5

=

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(74)

Solution:

−π

2,

π

2

cosα=p1−sin2α=

q

1− 1 5 =

q

4 5 =

2

√

5.

Thus, tanα=−1 2.

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(75)

Solution:

−π

2,

π

2

cosα=p1−sin2α=

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(76)

Solution:

−π

2,

π

2

cosα=p1−sin2α=

q

1− 1 5 =

q

4 5 =

2

√

5. Thus, tanα=−

1 2.

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−7

(77)

Solution:

−π

2,

π

2

cosα=p1−sin2α=

q

1− 1 5 =

q

4 5 =

2

√

tan(α−β) = −

1 2 −

2 3

1 + −1 2

₂

3

=−

(78)

Equations involving Inverse Circular Functions

Solve forx: cos−1 x

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄ + −π

6

= π₂ cos−1 x

4

= π

6 +

π

2

cos−1 x₄= 2₃π x

4 = cos 2π

3

x= 4 −1 2

(79)

Equations involving Inverse Circular Functions

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄

+ −π

6

= π₂ cos−1 x

4

= π

6 +

π

2

4 = cos 2π

3

x= 4 −1 2

(80)

Equations involving Inverse Circular Functions

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄ + −π

6

=

π

2

cos−1 x

4

= π

6 +

π

2

4 = cos 2π

3

x= 4 −1 2

(81)

Equations involving Inverse Circular Functions

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄ + −π

6

= π₂

cos−1 x

4

= π

6 +

π

2

4 = cos 2π

3

x= 4 −1 2

(82)

Equations involving Inverse Circular Functions

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄ + −π

6

= π₂ cos−1 x

4

= π

6 +

π

2

4 = cos 2π

3

x= 4 −1 2

(83)

Equations involving Inverse Circular Functions

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄ + −π

6

= π₂ cos−1 x

4

= π

6 +

π

2

cos−1 x₄= 2₃π

x

4 = cos 2π

3

x= 4 −1 2

(84)

Equations involving Inverse Circular Functions

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄ + −π

6

= π₂ cos−1 x

4

= π

6 +

π

2

4 = cos 2π

3

x= 4 −1 2

(85)

Equations involving Inverse Circular Functions

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄ + −π

6

= π₂ cos−1 x

4

= π

6 +

π

2

4 = cos 2π

3

x= 4 −1 2

(86)

Equations involving Inverse Circular Functions

4

+ tan−1₋√3 3

= cot−1₀

Solution:

cos−1 x₄ + −π

6

= π₂ cos−1 x

4

= π

6 +

π

2

4 = cos 2π

3

x= 4 −1 2

(87)

Solve forx: sec−1₍₋√₂₎₋_tan−1₍_x₋_{2) = csc}−1₍₋₂₎

Solution:

3π

4 −tan

−1₍_x₋_{2) =} ₋ π

6 3π

4 +

π

6 = tan

−1₍_x₋₂₎

11π

12 = tan

−1₍_x₋₂₎

Butran tan−1 = −π

2,

π

2

.

(88)

Solution:

3π

4

−tan−1(x−2) = − π 6 3π

4 +

π

6 = tan

−1₍_x₋₂₎

11π

12 = tan

−1₍_x₋₂₎

2,

π

2

.

(89)

Solution:

3π

4 −tan

−1₍_x₋_{2) =}

− π 6 3π

4 +

π

6 = tan

−1₍_x₋₂₎

11π

12 = tan

−1₍_x₋₂₎

2,

π

2

.

(90)

Solve forx: sec−1₍₋√₂₎₋_tan−1₍_x₋_{2) =} _csc−1₍₋₂₎

Solution:

3π

4 −tan

−1₍_x₋_{2) =} ₋ π

6

3π

4 +

π

6 = tan

−1₍_x₋₂₎

11π

12 = tan

−1₍_x₋₂₎

2,

π

2

.

(91)

Solution:

3π

4 −tan

−1₍_x₋_{2) =} ₋ π

6 3π

4 +

π

6 = tan

−1₍_x₋₂₎

11π

12 = tan

−1₍_x₋₂₎

2,

π

2

.

(92)

Solution:

3π

4 −tan

−1₍_x₋_{2) =} ₋ π

6 3π

4 +

π

6 = tan

−1₍_x₋₂₎

11π

12 = tan

−1₍_x₋₂₎

2,

π

2

.

(93)

Solution:

3π

4 −tan

−1₍_x₋_{2) =} ₋ π

6 3π

4 +

π

6 = tan

−1₍_x₋₂₎

11π

12 = tan

−1₍_x₋₂₎

2,

π

2

.

(94)

Solution:

3π

4 −tan

−1₍_x₋_{2) =} ₋ π

6 3π

4 +

π

6 = tan

−1₍_x₋₂₎

11π

12 = tan

−1₍_x₋₂₎

2,

π

2

.

(95)

Solve forx: sin−1(x)− cos−1(x) = π₆

Solution:

If we let α= sin−1x and β = cos−1_x_{, then we have} _α₋_β ₌ π

6

Thensinα=x with α∈

−π

2,

π

2

and cosβ =xwith β ∈[0, π].

α= π₆ +β

sinα= sin π₆ +β

sinα= sinπ₆ cosβ+ cosπ₆sinβ

x= 1₂(x) +

√

3 2 sinβ

(96)

Solve forx: sin−1(x)− cos−1(x)= π₆ Solution:

If we let α= sin−1x and β = cos−1_x_{, then we have}

α−β = π₆

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(97)

Solution:

6

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(98)

Solution:

6

Thensinα=xwith α∈

−π

2,

π

2

and

cosβ =xwith β ∈[0, π].

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(99)

Solution:

6

−π

2,

π

2

and cosβ =x with β∈[0, π].

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(100)

Solution:

6

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(101)

Solution:

6

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(102)

Solution:

6

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(103)

Solution:

6

−π

2,

π

2

α= π₆ +β

sinα= sinπ₆ cosβ+ cosπ₆sinβ x=

1 2(x) +

√

3 2 sinβ

(104)

Solution:

6

−π

2,

π

2

α= π₆ +β

sinα= sinπ₆ cosβ+ cosπ₆sinβ x= 1₂

(x) +

√

3 2 sinβ

(105)

Solution:

6

−π

2,

π

2

α= π₆ +β

sinα= sinπ₆ cosβ+ cosπ₆sinβ x= 1₂(x)

+

√

3 2 sinβ

(106)

Solution:

6

−π

2,

π

2

α= π₆ +β

sinα= sinπ₆ cosβ+ cosπ₆sinβ x= 1₂(x) +

√

3 2

sinβ

(107)

Solution:

6

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(108)

Solution:

6

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

Sinceβ ∈[0, π],sinβ =

p

(109)

Solution:

6

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

Sinceβ ∈[0, π],sinβ =p1−cos2_β

(110)

Solution:

6

−π

2,

π

2

α= π₆ +β

x= 1₂(x) +

√

3 2 sinβ

(111)

Solve forx: sin−1(x)−cos−1₍_x_{) =} π

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(112)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(113)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(114)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(115)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(116)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(117)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(118)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(119)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(120)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(121)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 −

5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(122)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6

=−7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos

(123)

6

Solution (cont.):

x= 1₂(x) +

√

3 2

√ 1−x2

2x=x+√3√1−x2

x=√3√1−x2

x2 = 3(1−x2)

x2 = 3−3x2

4x2 = 3

x2 = 3₄

x=±

√

3 2

Checking:

Forx=−

√

3 2 :

sin−1

− √ 3 2

−cos−1

− √ 3 2

=−π

3 − 5π

6 =− 7π

6

extraneous

Forx=

√

3 2 :

sin−1

√

3 2 −cos