§7.6–The inverse trigonometric functions

§7.6–The inverse trigonometric functions Mark R. Woodard Furman U

Fall 2010

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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Outline

1

The inverse sine function

2

The inverse cosine function

3

The inverse tangent function

4

The other inverse trig functions

5

Miscellaneous problems

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse sine function

Definition

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse sine function

Definition The sine function is one-to-one on [−π/2, π/2] and has range [−1, 1] on this domain.

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse sine function

Definition The sine function is one-to-one on [−π/2, π/2] and has range [−1, 1] on this domain. We define sin−1 to be the inverse of sine on this domain. It follows that sin−1 has domain [−1, 1] and range [−π/2, π/2].

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

3 / 11

The inverse sine function

Definition The sine function is one-to-one on [−π/2, π/2] and has range [−1, 1] on this domain. We define sin−1 to be the inverse of sine on this domain. It follows that sin−1 has domain [−1, 1] and range [−π/2, π/2].

Cancellation equations Because of these restrictions, we must be a little careful with the inverse relationships:
sin sin−1 (x) = x − 1 ≤ x ≤ 1
sin−1 sin(x) = x − π/2 ≤ x ≤ π/2

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse sine function

Problem Evaluate the following:

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse sine function

Problem Evaluate the following:
sin sin−1 (.3)

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse sine function

Problem Evaluate the following:
sin sin−1 (.3)
sin−1 sin(14π/3) .

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

4 / 11

The inverse sine function

Problem Evaluate the following:
sin sin−1 (.3)
sin−1 sin(14π/3) .

Problem
Find cos 2 sin−1 (1/4) .

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

4 / 11

The inverse sine function

Problem Evaluate the following:
sin sin−1 (.3)
sin−1 sin(14π/3) .

Problem
Find cos 2 sin−1 (1/4) .

Problem
Evaluate cos sin−1 (x) .

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

4 / 11

The inverse sine function

Problem Evaluate the following:
sin sin−1 (.3)
sin−1 sin(14π/3) .

Problem
Find cos 2 sin−1 (1/4) .

Problem
Evaluate cos sin−1 (x) .

Solution

√ Let θ = sin−1 (x). Then cos(θ) = ± 1 − x 2 . Since cosine is nonnegative √ −1 for θ ∈ [−π/2, π/2], it follows that cos(sin (x)) = 1 − x 2 . Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

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The inverse sine function

Theorem Dx sin−1 (x) = √

Mark R. Woodard (Furman U)

1 . 1 − x2

§7.6–The inverse trigonometric functions

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The inverse sine function

Theorem Dx sin−1 (x) = √

1 . 1 − x2

Proof.

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse sine function

Theorem Dx sin−1 (x) = √

1 . 1 − x2

Proof. Let y = sin−1 (x).

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

5 / 11

The inverse sine function

Theorem Dx sin−1 (x) = √

1 . 1 − x2

Proof. Let y = sin−1 (x). Then sin(y ) = x. By the chain rule, cos(y )y 0 = 1 hence y 0 =

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

1 cos(y ) .

Fall 2010

5 / 11

The inverse sine function

Theorem Dx sin−1 (x) = √

1 . 1 − x2

Proof. Let y = sin−1 (x). Then sin(y ) = x. By the chain rule, cos(y )y 0 = 1 hence y 0 = √ However, cos(y ) = 1 − x 2 , which gives the desired result.

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

1 cos(y ) .

Fall 2010

5 / 11

The inverse sine function

Theorem Dx sin−1 (x) = √

1 . 1 − x2

Proof. Let y = sin−1 (x). Then sin(y ) = x. By the chain rule, cos(y )y 0 = 1 hence y 0 = √ However, cos(y ) = 1 − x 2 , which gives the desired result.

1 cos(y ) .

Problem Find the derivative of y = x sin−1 (x 2 ).

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse cosine function

Definition

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse cosine function

Definition The cosine function is one-to-one on the interval [0, π] and has range [−1, 1] on that domain.

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

6 / 11

The inverse cosine function

Definition The cosine function is one-to-one on the interval [0, π] and has range [−1, 1] on that domain. Let cos−1 denote the inverse of the cosine function restricted to the domain [0, π]. Thus the domain of cos−1 is [−1, 1] and its range is [0, π].

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

6 / 11

The inverse cosine function

Definition The cosine function is one-to-one on the interval [0, π] and has range [−1, 1] on that domain. Let cos−1 denote the inverse of the cosine function restricted to the domain [0, π]. Thus the domain of cos−1 is [−1, 1] and its range is [0, π].

The cancellation equations
cos cos−1 (x) = x
cos−1 cos(x) = x

Mark R. Woodard (Furman U)

−1≤x ≤1 0≤x ≤π

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse cosine function

Problem

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse cosine function

Problem
Evaluate cos−1 cos(14π/3)

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse cosine function

Problem
Evaluate cos−1 cos(14π/3)
√ Show that sin cos−1 (x) = 1 − x 2

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse cosine function

Problem
Evaluate cos−1 cos(14π/3)
√ Show that sin cos−1 (x) = 1 − x 2

Theorem Dx cos−1 (x) = − √

Mark R. Woodard (Furman U)

1 1 − x2

§7.6–The inverse trigonometric functions

Fall 2010

7 / 11

The inverse cosine function

Problem
Evaluate cos−1 cos(14π/3)
√ Show that sin cos−1 (x) = 1 − x 2

Theorem Dx cos−1 (x) = − √

1 1 − x2

Proof. Let y = cos−1 (x). Then cos(y ) = x and, by taking derivatives, √ 0 = 1; hence y 0 = −1/ sin(y ). Since sin(y ) = − sin(y )y √ 1 − x 2, 0 2 y = −1/ 1 − x , as was to be shown.

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

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The inverse tangent function

Definition

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§7.6–The inverse trigonometric functions

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The inverse tangent function

Definition The tangent is one-to-one on the interval (−π/2, π/2) and has range (−∞, ∞) on this domain.

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

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The inverse tangent function

Definition The tangent is one-to-one on the interval (−π/2, π/2) and has range (−∞, ∞) on this domain. Let tan−1 be the inverse of the tangent function on this restricted domain. Thus the domain of tan−1 is (−∞, ∞) and its range is (−π/2, π/2).

Mark R. Woodard (Furman U)

§7.6–The inverse trigonometric functions

Fall 2010

8 / 11

The inverse tangent function

Definition The tangent is one-to-one on the interval (−π/2, π/2) and has range (−∞, ∞) on this domain. Let tan−1 be the inverse of the tangent function on this restricted domain. Thus the domain of tan−1 is (−∞, ∞) and its range is (−π/2, π/2).

The cancellation equations
tan tan−1 (x) = x
tan−1 tan(x) = x

Mark R. Woodard (Furman U)

−∞