Inverse trigonometric functions

Inverse Trigonometric Functions

Mathematics 4

October 24, 2011

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If sinx = 35 , what is x?

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If sinx = 35 , what is x?

How do we isolate x from the equation above?

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Let us recall inverses!

• f(x) = y = 2x− 1

•

•

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• f(x) = y = 2x− 1

• f−1(x)→

•

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• f(x) = y = 2x− 1

• f−1(x)→ x = 2y − 1 The variables are interchanged.

•

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• f(x) = y = 2x− 1

• f−1(x)→ x = 2y − 1 The variables are interchanged.

• f−1(x) = y =x+ 1

2The y-variable is isolated.

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Given the graph of g(x):

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The inverse of g(x) can be flipping the graph along the diagonal:

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This is the graph of g−1(x)

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Does the function f(x) = sinx have an inverse?

f(x) = sinx

No!The function f(x) = sinx is NOT one-to-one!

It does not pass the Horizontal Line Test!

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f(x) = sinx

No!

The function f(x) = sinx is NOT one-to-one!It does not pass the Horizontal Line Test!

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f(x) = sinx

No!The function f(x) = sinx is NOT one-to-one!

It does not pass the Horizontal Line Test!

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Do any of the six trigonometric functions have inverses?

f(x) = sinx

f(x) = cosx

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f(x) = tanx

f(x) = cotx

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f(x) = secx

f(x) = cscx

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How can we isolate x in f(x) = sin x if f(x) is not one-to-one?

f(x) = sinx

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f(x) = Sinx, x ∈ [−π2 ,

π2 ]

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f(x) = Sinx, x ∈ [−π2 ,

π2 ]

Restrict the domain so that it becomes one-to-one.

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The inverse of f(x) = Sin x, x ∈ [−π2, π2]

f(x) = Sinx, x ∈ [−π2 ,

π2 ]

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Find the graph of the inverse by flipping along the diagonal

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f(x) = sin−1 x = Arcsinx = inverse sine of x

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The Inverse Sine Function

Properties of f(x) = sin−1 x:

Domain:Range:

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Domain: x ∈ [−1, 1]Range:

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Domain: x ∈ [−1, 1]Range: y ∈ [−π

2 ,π2 ]

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Determine the following values:

1. sin−1 12 =

2. Arcsin 1 =

3. sin−1(sin π4 ) =

4. Arcsin(sin 7π6 ) =

5. sin−1(sin 4π3 ) =

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1. sin−1 12 = π

6

2. Arcsin 1 =

3. sin−1(sin π4 ) =


5. sin−1(sin 4π3 ) =

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1. sin−1 12 = π

6

2. Arcsin 1 = π2

3. sin−1(sin π4 ) =


5. sin−1(sin 4π3 ) =

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1. sin−1 12 = π

6

2. Arcsin 1 = π2

3. sin−1(sin π4 ) = π

4


5. sin−1(sin 4π3 ) =

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1. sin−1 12 = π

6

2. Arcsin 1 = π2

3. sin−1(sin π4 ) = π

4

4. Arcsin(sin 7π6 ) = −π

6

5. sin−1(sin 4π3 ) =

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1. sin−1 12 = π

6

2. Arcsin 1 = π2

3. sin−1(sin π4 ) = π

4

4. Arcsin(sin 7π6 ) = −π

6

5. sin−1(sin 4π3 ) = −π

3

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The Inverse Cosine Function

Graphing the inverse cosine function

f(x) = cosx

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f(x) = Cosx, x ∈ [0, π]

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f(x) = Cosx, x ∈ [0, π]


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The inverse of f(x) = Cos x, x ∈ [0, π]

f(x) = Cosx, x ∈ [0, π]

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f(x) = cos−1 x = Arccosx = inverse cosine of x

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Properties of f(x) = cos−1 x:

Domain:Range:

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Domain: x ∈ [−1, 1]Range:

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Domain: x ∈ [−1, 1]Range: y ∈ [0, π]

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The Inverse cosine Function


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1. cos−1 12 =

2. Arccos 0 =

3. Arccos(cos 7π6 ) =

4. cos−1(cos 7π4 ) =

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1. cos−1 12 = π

3

2. Arccos 0 =


4. cos−1(cos 7π4 ) =

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1. cos−1 12 = π

3

2. Arccos 0 = π2


4. cos−1(cos 7π4 ) =

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1. cos−1 12 = π

3

2. Arccos 0 = π2

3. Arccos(cos 7π6 ) = 5π

6

4. cos−1(cos 7π4 ) =

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1. cos−1 12 = π

3

2. Arccos 0 = π2

3. Arccos(cos 7π6 ) = 5π

6

4. cos−1(cos 7π4 ) = π

4

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Pick-up quiz: Quiz # 1

Evaluate the following values:

1. Arcsin(−√22 )

2. Arccos(−√32 )

3. sin−1(sin 2π3 )

4. cos−1(cos 11π6 )

5. Arcsin(cos 5π3 )

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Pick-up quiz: Quiz # 1

Evaluate the following values:

1. Arcsin(−√22 ) = −π

4

2. Arccos(−√32 ) = 5π

6

3. sin−1(sin 2π3 ) = π

3

4. cos−1(cos 11π6 ) = π

6

5. Arcsin(cos 5π3 ) = π

6

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Inverse Trigonometric Functions of Non-Special Angles

Example 1: Evaluate cos(sin−1 45)

Let θ = sin−1 45

sin θ = 45

cos θ = 35

cos θ > 0 because θ = sin−1 45 can only be in Q1 or Q4.

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Let θ = sin−1 45

sin θ =

45

cos θ = 35


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Let θ = sin−1 45

sin θ = 45

cos θ = 35


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Let θ = sin−1 45

sin θ = 45

cos θ =

35


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Let θ = sin−1 45

sin θ = 45

cos θ = 35


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Example 2: Evaluate cos(sin−1−23)

Let θ = sin−1−23

sin θ = −23

cos θ =

√1−

(−2

3

)2=√53


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sin θ =

−23

cos θ =

√1−

(−2

3

)2=√53


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sin θ = −23

cos θ =

√1−

(−2

3

)2=√53


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sin θ = −23

cos θ =

√1−

(−2

3

)2

=√53


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sin θ = −23

cos θ =

√1−

(−2

3

)2=√53


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Example 3: Evaluate cos[sin−1(−2

3) + cos−1(1

6)]

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3) + cos−1(1

6)]

α = sin−1(−23) β = cos−1(16)

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3) + cos−1(1

6)]

α = sin−1(−23)

sinα = −23

β = cos−1(16)

cosβ = 16

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3) + cos−1(1

6)]

α = sin−1(−23)

sinα = −23

cosα =√53

β = cos−1(16)

cosβ = 16

sinβ =√356

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3) + cos−1(1

6)]

α = sin−1(−23)

sinα = −23

cosα =√53

β = cos−1(16)

cosβ = 16

sinβ =√356

cos(α+ β) = cosα cosβ − sinα sinβ

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3) + cos−1(1

6)]

α = sin−1(−23)

sinα = −23

cosα =√53

β = cos−1(16)

cosβ = 16

sinβ =√356


=(√

53

) (16

)−(−2

3

) (√356

)

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3) + cos−1(1

6)]

α = sin−1(−23)

sinα = −23

cosα =√53

β = cos−1(16)

cosβ = 16

sinβ =√356


=(√

53

) (16

)−(−2

3

) (√356

)=

√5 + 2

√35

18

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The Inverse Tangent Function

Finding the graph of f(x) = tan−1 x

f(x) = tanx

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f(x) = Tanx, x ∈ (−π2 ,

π2 )

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f(x) = Tanx, x ∈ (−π2 ,

π2 )


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The inverse of f(x) = Tanx, x ∈ (−π2, π2)

f(x) = Tanx, x ∈ (−π2 ,

π2 )

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f(x) = tan−1 x = Arctanx = inverse tangent of x

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Domain: Range:

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Domain: x ∈ RRange: y ∈ (−π

2 ,π2 )

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Other Inverse Trigonometric Functions

f(x) = cotx

f(x) = secx

f(x) = cscx

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f(x) = Cotx, x ∈ (0, π)

f(x) = Secx, x ∈ [0, π]

f(x) = Cscx, x ∈ (−π2 ,

π2 )

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f(x) = Arccotx

f(x) = Arcsecx

f(x) = Arccscx

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f(x) = Arccotx

f(x) = Arcsecx

f(x) = Arccscx

Domain: x ∈ RRange: {0 < y < π}

Domain: {x ≤ −1} ∪ {x ≥ 1}Range: {0 ≤ y ≤ π, y 6= π

2 }

Domain: {x ≤ −1} ∪ {x ≥ 1}Range: {−π

2 ≤ y ≤π2 , y 6= 0}

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Ranges of the Inverse Trigonometric Functions

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Ranges of the Inverse Trigonometric Functions

f(x) = Arcsin(x)f(x) = Arctan(x)f(x) = Arccsc(x)

f(x) = Arccos(x)f(x) = Arccot(x)f(x) = Arcsec(x)

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Any questions?

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Inverse trigonometric functions

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