§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 1 / 11

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

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 followsthat 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 inverserelationships:

sin−1(x))

= x − 1 ≤ x ≤ 1

sin−1(

sin(x))

= x − π/2 ≤ x ≤ π/2

Definition

sin−1(x))

= x − 1 ≤ x ≤ 1

sin−1(

sin(x))

= x − π/2 ≤ x ≤ π/2

Definition

sin−1(x))

= x − 1 ≤ x ≤ 1

sin−1(

sin(x))

= x − π/2 ≤ x ≤ π/2

Definition

sin−1(x))

= x − 1 ≤ x ≤ 1

sin−1(

sin(x))

= x − π/2 ≤ x ≤ π/2

Problem

Evaluate the following:

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− x2. Since cosine is nonnegativefor θ ∈ [−π/2, π/2], it follows that cos(sin−1(x)) =

√1− x2.

Problem

sin−1(.3))

sin−1(

sin(14π/3)).

Problem

Evaluate cos(

sin−1(x)).

Solution

√1− x2.

Problem

sin−1(.3))

sin−1(

sin(14π/3)).

Problem

Evaluate cos(

sin−1(x)).

Solution

√1− x2.

Problem

sin−1(.3))

sin−1(

sin(14π/3)).

Problem

Evaluate cos(

sin−1(x)).

Solution

√1− x2.

Problem

sin−1(.3))

sin−1(

sin(14π/3)).

Problem

Evaluate cos(

sin−1(x)).

Solution

√1− x2.

Problem

sin−1(.3))

sin−1(

sin(14π/3)).

Problem

Evaluate cos(

sin−1(x)).

Solution

√1− x2.

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 ′ = 1 hence y ′ = 1cos(y) .

However, cos(y) =√

1− x2, which gives the desired result.

Problem

Find the derivative of y = x sin−1(x2).

Theorem

Dx sin−1(x) =1√

1− x2.

Proof.

Let y = sin−1(x).

Problem

Theorem

Dx sin−1(x) =1√

1− x2.

Proof.

Let y = sin−1(x).

Problem

Theorem

Dx sin−1(x) =1√

1− x2.

Proof.

Let y = sin−1(x).

Problem

Theorem

Dx sin−1(x) =1√

1− x2.

Proof.

Let y = sin−1(x).

Problem

Theorem

Dx sin−1(x) =1√

1− x2.

Proof.

Let y = sin−1(x).

Problem

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 thedomain [0, π]. Thus the domain of cos−1 is [−1, 1] and its range is[0, π].

The cancellation equations

cos−1(x))

= x − 1 ≤ x ≤ 1

cos−1(

cos(x))

= x 0 ≤ x ≤ π

Definition

cos−1(x))

= x − 1 ≤ x ≤ 1

cos−1(

cos(x))

= x 0 ≤ x ≤ π

Definition

cos−1(x))

= x − 1 ≤ x ≤ 1

cos−1(

cos(x))

= x 0 ≤ x ≤ π

Definition

cos−1(x))

= x − 1 ≤ x ≤ 1

cos−1(

cos(x))

= x 0 ≤ x ≤ π

Problem

Evaluate cos−1(

cos(14π/3))

Show that sin(

cos−1(x))

1− x2

Theorem

Dx cos−1(x) = − 1√1− x2

Proof.

Let y = cos−1(x). Then cos(y) = x and, by taking derivatives,− sin(y)y ′ = 1; hence y ′ = −1/ sin(y). Since sin(y) =

√1− x2,

y ′ = −1/√

1− x2, as was to be shown.

Problem

Evaluate cos−1(

cos(14π/3))

Show that sin(

cos−1(x))

1− x2

Theorem

Dx cos−1(x) = − 1√1− x2

Proof.

√1− x2,

y ′ = −1/√

Problem

Evaluate cos−1(

cos(14π/3))

Show that sin(

cos−1(x))

1− x2

Theorem

Dx cos−1(x) = − 1√1− x2

Proof.

√1− x2,

y ′ = −1/√

Problem

Evaluate cos−1(

cos(14π/3))

Show that sin(

cos−1(x))

1− x2

Theorem

Dx cos−1(x) = − 1√1− x2

Proof.

√1− x2,

y ′ = −1/√

Problem

Evaluate cos−1(

cos(14π/3))

Show that sin(

cos−1(x))

1− x2

Theorem

Dx cos−1(x) = − 1√1− x2

Proof.

√1− x2,

y ′ = −1/√

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 restricteddomain. Thus the domain of tan−1 is (−∞,∞) and its range is(−π/2, π/2).

tan−1(x))

= x −∞ < x <∞tan−1

(tan(x)

)= x − π/2 < x < π/2.

Definition

tan−1(x))

= x −∞ < x <∞tan−1

(tan(x)

)= x − π/2 < x < π/2.

Definition

tan−1(x))

= x −∞ < x <∞tan−1

(tan(x)

)= x − π/2 < x < π/2.

Definition

tan−1(x))

= x −∞ < x <∞tan−1

(tan(x)

)= x − π/2 < x < π/2.

Problem

Show that sec2(

tan−1(x))

= 1 + x2

Theorem

Dx tan−1(x) =1

1 + x2, x ∈ R.

Proof.

Let y = tan−1(x). Then tan(y) = x and, by the chain rule, sec2(y)y ′ = 1.Thus y ′ = 1/ sec2(y) = 1/(1 + x2), as was to be shown.

Problem

Show that sec2(

tan−1(x))

= 1 + x2

Theorem

Dx tan−1(x) =1

1 + x2, x ∈ R.

Proof.

Problem

Show that sec2(

tan−1(x))

= 1 + x2

Theorem

Dx tan−1(x) =1

1 + x2, x ∈ R.

Proof.

The other inverse trig functions

The remaining inverse trig functions (cot−1, sec−1, and csc−1) are definedin similar ways. See the text for a summary of these functions, theirdefinitions, and derivatives.

Miscellaneous problems

Problem

Find y ′ in each case:

y = tan−1(ex)

y =√

1− x2 sin−1(x)

y = sin−1(x) + cos−1(x)

Problem

y = tan−1(ex)

y =√

1− x2 sin−1(x)

y = sin−1(x) + cos−1(x)

Problem

y = tan−1(ex)

y =√

1− x2 sin−1(x)

y = sin−1(x) + cos−1(x)

Problem

y = tan−1(ex)

y =√

1− x2 sin−1(x)

y = sin−1(x) + cos−1(x)

§7.6–The inverse trigonometric functions

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Transcript of §7.6–The inverse trigonometric functions