§7.6–the inverse trigonometric functions
TRANSCRIPT
§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
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 2 / 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 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(
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 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 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(
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 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 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(
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 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 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(
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 3 / 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− x2. Since cosine is nonnegativefor θ ∈ [−π/2, π/2], it follows that cos(sin−1(x)) =
√1− x2.
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− x2. Since cosine is nonnegativefor θ ∈ [−π/2, π/2], it follows that cos(sin−1(x)) =
√1− x2.
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− x2. Since cosine is nonnegativefor θ ∈ [−π/2, π/2], it follows that cos(sin−1(x)) =
√1− x2.
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− x2. Since cosine is nonnegativefor θ ∈ [−π/2, π/2], it follows that cos(sin−1(x)) =
√1− x2.
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− x2. Since cosine is nonnegativefor θ ∈ [−π/2, π/2], it follows that cos(sin−1(x)) =
√1− x2.
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− x2. Since cosine is nonnegativefor θ ∈ [−π/2, π/2], it follows that cos(sin−1(x)) =
√1− x2.
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 4 / 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 ′ = 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).
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 ′ = 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).
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 ′ = 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).
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 ′ = 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).
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 ′ = 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).
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 ′ = 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).
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 5 / 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 thedomain [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 − 1 ≤ x ≤ 1
cos−1(
cos(x))
= x 0 ≤ x ≤ 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 thedomain [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 − 1 ≤ x ≤ 1
cos−1(
cos(x))
= x 0 ≤ x ≤ 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 thedomain [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 − 1 ≤ x ≤ 1
cos−1(
cos(x))
= x 0 ≤ x ≤ 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 thedomain [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 − 1 ≤ x ≤ 1
cos−1(
cos(x))
= x 0 ≤ x ≤ π
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 6 / 11
The inverse cosine function
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.
Mark R. Woodard (Furman U) §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− 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.
Mark R. Woodard (Furman U) §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− 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.
Mark R. Woodard (Furman U) §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− 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.
Mark R. Woodard (Furman U) §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− 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.
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 7 / 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 restricteddomain. Thus the domain of tan−1 is (−∞,∞) and its range is(−π/2, π/2).
The cancellation equations
tan(
tan−1(x))
= x −∞ < x <∞tan−1
(tan(x)
)= x − π/2 < x < 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 restricteddomain. Thus the domain of tan−1 is (−∞,∞) and its range is(−π/2, π/2).
The cancellation equations
tan(
tan−1(x))
= x −∞ < x <∞tan−1
(tan(x)
)= x − π/2 < x < 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 restricteddomain. Thus the domain of tan−1 is (−∞,∞) and its range is(−π/2, π/2).
The cancellation equations
tan(
tan−1(x))
= x −∞ < x <∞tan−1
(tan(x)
)= x − π/2 < x < 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 restricteddomain. Thus the domain of tan−1 is (−∞,∞) and its range is(−π/2, π/2).
The cancellation equations
tan(
tan−1(x))
= x −∞ < x <∞tan−1
(tan(x)
)= x − π/2 < x < π/2.
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 8 / 11
The inverse tangent function
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.
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 9 / 11
The inverse tangent function
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.
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 9 / 11
The inverse tangent function
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.
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 9 / 11
The other inverse trig functions
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.
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 10 / 11
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)
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 11 / 11
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)
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 11 / 11
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)
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 11 / 11
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)
Mark R. Woodard (Furman U) §7.6–The inverse trigonometric functions Fall 2010 11 / 11