Lesson 16: Inverse Trigonometric Functions (Section 041 slides)

Section 3.5Inverse Trigonometric

FunctionsV63.0121.041, Calculus I

New York University

November 1, 2010

Announcements

I Midterm grades have been submittedI Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2I Thank you for the evaluations

. . . . . .

Announcements

I Midterm grades have beensubmitted

I Quiz 3 this week inrecitation on Section 2.6,2.8, 3.1, 3.2

I Thank you for theevaluations

V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 2 / 32

. . . . . .

Objectives

I Know the definitions,domains, ranges, andother properties of theinverse trignometricfunctions: arcsin, arccos,arctan, arcsec, arccsc,arccot.

I Know the derivatives of theinverse trignometricfunctions.

. . . . . .

What is an inverse function?

DefinitionLet f be a function with domain D and range E. The inverse of f is thefunction f−1 defined by:

f−1(b) = a,

where a is chosen so that f(a) = b.

Sof−1(f(x)) = x, f(f−1(x)) = x

. . . . . .

What is an inverse function?

DefinitionLet f be a function with domain D and range E. The inverse of f is thefunction f−1 defined by:

f−1(b) = a,

where a is chosen so that f(a) = b.

Sof−1(f(x)) = x, f(f−1(x)) = x

. . . . . .

What functions are invertible?

In order for f−1 to be a function, there must be only one a in Dcorresponding to each b in E.

I Such a function is called one-to-oneI The graph of such a function passes the horizontal line test: any

horizontal line intersects the graph in exactly one point if at all.I If f is continuous, then f−1 is continuous.

. . . . . .

Outline

Inverse Trigonometric Functions

Derivatives of Inverse Trigonometric FunctionsArcsineArccosineArctangentArcsecant

Applications

. . . . . .

arcsin

Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].

.−π

.y = x

. .arcsin

I The domain of arcsin is [−1,1]I The range of arcsin is

[−π

. . . . . .

arcsin

.−π

.y = x

. .arcsin

[−π

. . . . . .

arcsin

.−π

.y = x

. .arcsin

[−π

. . . . . .

arcsin

.−π

.y = x

. .arcsin

[−π

. . . . . .

arccos

Arccos is the inverse of the cosine function after restriction to [0, π]

.cos..0

.y = x

. .arccos

I The domain of arccos is [−1,1]I The range of arccos is [0, π]

. . . . . .

arccos

.y = x

. .arccos

. . . . . .

arccos

.y = x

. .arccos

. . . . . .

arccos

.y = x

. .arccos

. . . . . .

arctan

Arctan is the inverse of the tangent function after restriction to[−π/2, π/2].

.−3π2

.−π

2 .3π2

.y = x

.arctan

.−π

I The domain of arctan is (−∞,∞)

I The range of arctan is(−π

)I lim

x→∞arctan x =

2, limx→−∞

arctan x = −π

. . . . . .

arctan

.−3π2

.−π

2 .3π2

.y = x

.arctan

.−π

)I lim

x→∞arctan x =

2, limx→−∞

arctan x = −π

. . . . . .

arctan

.−3π2

.−π

2 .3π2

.y = x

.arctan

.−π

)I lim

x→∞arctan x =

2, limx→−∞

arctan x = −π

. . . . . .

arctan

.−3π2

.−π

2 .3π2

.y = x

.arctan

.−π

)I lim

x→∞arctan x =

2, limx→−∞

arctan x = −π

. . . . . .

arcsec

Arcsecant is the inverse of secant after restriction to[0, π/2) ∪ (π,3π/2].

.−3π2

.−π

2 .3π2

.y = x

I The domain of arcsec is (−∞,−1] ∪ [1,∞)

I The range of arcsec is[0,

)∪(π2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π2

. . . . . .

arcsec

.−3π2

.−π

2 .3π2

.y = x

)∪(π2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π2

. . . . . .

arcsec

.−3π2

.−π

2 .3π2

.y = x

)∪(π2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π2

. . . . . .

arcsec

.−3π2

.−π

2 .3π2

.y = x

)∪(π2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π2

. . . . . .

Values of Trigonometric Functions

sin x 0 12

cos x 1√32

tan x 01√3

1√3 undef

cot x undef√3 1

sec x 12√3

2 undef

csc x undef 22√2

. . . . . .

Check: Values of inverse trigonometric functions

Example

FindI arcsin(1/2)I arctan(−1)

I arccos

(−√22

Solution

6I −π

. . . . . .

Example

I arccos

(−√22

Solution

I −π

. . . . . .

What is arctan(−1)?

..3π/4

..−π/4

.sin(3π/4) =√22

.cos(3π/4) = −√22.sin(π/4) = −

.cos(π/4) =√22

I Yes, tan(3π4

)= −1

I But, the range of arctan is(−π

)I Another angle whose

tangent is −1 is −π

4, and

this is in the right range.I So arctan(−1) = −π

. . . . . .

..3π/4

..−π/4

.sin(3π/4) =√22

.cos(3π/4) = −√22

.sin(π/4) = −√22

.cos(π/4) =√22

I Yes, tan(3π4

)= −1

4, and

. . . . . .

..3π/4

..−π/4

.sin(3π/4) =√22

.cos(3π/4) = −√22

.sin(π/4) = −√22

.cos(π/4) =√22

I Yes, tan(3π4

)= −1

I Another angle whosetangent is −1 is −π

4, and

. . . . . .

..3π/4

..−π/4

.sin(3π/4) =√22

.cos(3π/4) = −√22

.sin(π/4) = −√22

.cos(π/4) =√22

I Yes, tan(3π4

)= −1

4, and

this is in the right range.

I So arctan(−1) = −π

. . . . . .

..3π/4

..−π/4

.sin(3π/4) =√22

.cos(3π/4) = −√22

.sin(π/4) = −√22

.cos(π/4) =√22

I Yes, tan(3π4

)= −1

4, and

. . . . . .

Example

I arccos

(−√22

Solution

6I −π

. . . . . .

Example

I arccos

(−√22

Solution

6I −π

. . . . . .

Caution: Notational ambiguity

..sin2 x = (sin x)2 .sin−1 x = (sin x)−1

I sinn x means the nth power of sin x, except when n = −1!I The book uses sin−1 x for the inverse of sin x, and never for

(sin x)−1.

I I use csc x for1

sin xand arcsin x for the inverse of sin x.

. . . . . .

Outline

Applications

. . . . . .

The Inverse Function Theorem

Theorem (The Inverse Function Theorem)

Let f be differentiable at a, and f′(a) ̸= 0. Then f−1 is defined in anopen interval containing b = f(a), and

(f−1)′(b) =1

f′(f−1(b))

In Leibniz notation we have

Upshot: Many times the derivative of f−1(x) can be found by implicitdifferentiation and the derivative of f:

. . . . . .

The Inverse Function Theorem

Theorem (The Inverse Function Theorem)

Let f be differentiable at a, and f′(a) ̸= 0. Then f−1 is defined in anopen interval containing b = f(a), and

(f−1)′(b) =1

f′(f−1(b))

In Leibniz notation we have

Upshot: Many times the derivative of f−1(x) can be found by implicitdifferentiation and the derivative of f:

. . . . . .

Illustrating the Inverse Function Theorem.

Example

Use the inverse function theorem to find the derivative of the square rootfunction.

Solution (Newtonian notation)

Let f(x) = x2 so that f−1(y) =√y. Then f′(u) = 2u so for any b > 0 we have

(f−1)′(b) =1

Solution (Leibniz notation)

If the original function is y = x2, then the inverse function is defined by x = y2.Differentiate implicitly:

1 = 2ydydx

=⇒ dydx

. . . . . .

Example

(f−1)′(b) =1

1 = 2ydydx

=⇒ dydx

. . . . . .

Example

(f−1)′(b) =1

1 = 2ydydx

=⇒ dydx

. . . . . .

Derivation: The derivative of arcsin

Let y = arcsin x, so x = sin y. Then

cos ydydx

= 1 =⇒ dydx

cos y=

1cos(arcsin x)

To simplify, look at a righttriangle:

cos(arcsin x) =√1− x2

arcsin(x) =1√

1− x2 .

. .y = arcsin x

.√1− x2

. . . . . .

cos ydydx

= 1 =⇒ dydx

cos y=

1cos(arcsin x)

arcsin(x) =1√

1− x2

. .y = arcsin x

.√1− x2

. . . . . .

cos ydydx

= 1 =⇒ dydx

cos y=

1cos(arcsin x)

arcsin(x) =1√

1− x2

. .y = arcsin x

.√1− x2

. . . . . .

cos ydydx

= 1 =⇒ dydx

cos y=

1cos(arcsin x)

arcsin(x) =1√

1− x2

. .y = arcsin x

.√1− x2

. . . . . .

cos ydydx

= 1 =⇒ dydx

cos y=

1cos(arcsin x)

arcsin(x) =1√

1− x2

. .y = arcsin x

.√1− x2

. . . . . .

cos ydydx

= 1 =⇒ dydx

cos y=

1cos(arcsin x)

arcsin(x) =1√

1− x2

. .y = arcsin x

.√1− x2

. . . . . .

cos ydydx

= 1 =⇒ dydx

cos y=

1cos(arcsin x)

arcsin(x) =1√

1− x2 .

. .y = arcsin x

.√1− x2

. . . . . .

Graphing arcsin and its derivative

I The domain of f is [−1,1],but the domain of f′ is(−1,1)

I limx→1−

f′(x) = +∞

I limx→−1+

f′(x) = +∞ ..|.−1

. .arcsin

1− x2

. . . . . .

Composing with arcsin

Example

Let f(x) = arcsin(x3 + 1). Find f′(x).

SolutionWe have

arcsin(x3 + 1) =1√

1− (x3 + 1)2ddx

(x3 + 1)

=3x2√

−x6 − 2x3

. . . . . .

Composing with arcsin

Example

Let f(x) = arcsin(x3 + 1). Find f′(x).

SolutionWe have

arcsin(x3 + 1) =1√

1− (x3 + 1)2ddx

(x3 + 1)

=3x2√

−x6 − 2x3

. . . . . .

Derivation: The derivative of arccos

Let y = arccos x, so x = cos y. Then

− sin ydydx

= 1 =⇒ dydx

− sin y=

1− sin(arccos x)

sin(arccos x) =√1− x2

arccos(x) = − 1√1− x2 .

.1.√1− x2

.x. .y = arccos x

. . . . . .

Derivation: The derivative of arccos

Let y = arccos x, so x = cos y. Then

− sin ydydx

= 1 =⇒ dydx

− sin y=

1− sin(arccos x)

sin(arccos x) =√1− x2

arccos(x) = − 1√1− x2 .

.1.√1− x2

.x. .y = arccos x

. . . . . .

Graphing arcsin and arccos

..|.−1

. .arcsin

. .arccos

cos θ = sin(π2− θ)

=⇒ arccos x =π

2− arcsin x

So it’s not a surprise that theirderivatives are opposites.

. . . . . .

Graphing arcsin and arccos

..|.−1

. .arcsin

. .arccosNote

cos θ = sin(π2− θ)

=⇒ arccos x =π

2− arcsin x

So it’s not a surprise that theirderivatives are opposites.

. . . . . .

Derivation: The derivative of arctan

Let y = arctan x, so x = tan y. Then

sec2 ydydx

= 1 =⇒ dydx

sec2 y= cos2(arctan x)

cos(arctan x) =1√

arctan(x) =1

1+ x2 .

.1. .y = arctan x

.√1+ x2

. . . . . .

sec2 ydydx

= 1 =⇒ dydx

cos(arctan x) =1√

arctan(x) =1

.1. .y = arctan x

.√1+ x2

. . . . . .

sec2 ydydx

= 1 =⇒ dydx

cos(arctan x) =1√

arctan(x) =1

. .y = arctan x

.√1+ x2

. . . . . .

sec2 ydydx

= 1 =⇒ dydx

cos(arctan x) =1√

arctan(x) =1

.1. .y = arctan x

.√1+ x2

. . . . . .

sec2 ydydx

= 1 =⇒ dydx

cos(arctan x) =1√

arctan(x) =1

.1. .y = arctan x

.√1+ x2

. . . . . .

sec2 ydydx

= 1 =⇒ dydx

cos(arctan x) =1√

arctan(x) =1

.1. .y = arctan x

.√1+ x2

. . . . . .

sec2 ydydx

= 1 =⇒ dydx

cos(arctan x) =1√

arctan(x) =1

1+ x2 .

.1. .y = arctan x

.√1+ x2

. . . . . .

Graphing arctan and its derivative

.arctan

.−π/2

I The domain of f and f′ are both (−∞,∞)

I Because of the horizontal asymptotes, limx→±∞

f′(x) = 0

. . . . . .

Composing with arctan

Example

Let f(x) = arctan√x. Find f′(x).

Solution

arctan√x =

(√x)2 d

dx√x =

· 12√x

2√x+ 2x

. . . . . .

Composing with arctan

Example

Let f(x) = arctan√x. Find f′(x).

Solution

arctan√x =

(√x)2 d

dx√x =

· 12√x

2√x+ 2x

. . . . . .

Derivation: The derivative of arcsec

Try this first.

Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

sec y tan y=

1x tan(arcsec(x))

tan(arcsec x) =√x2 − 11

arcsec(x) =1

x√x2 − 1 .

.1. .y = arcsec x

x2 − 1

. . . . . .

Try this first. Let y = arcsec x, so x = sec y. Then

sec y tan ydydx

= 1 =⇒ dydx

sec y tan y=

1x tan(arcsec(x))

arcsec(x) =1

x√x2 − 1 .

.1. .y = arcsec x

x2 − 1

. . . . . .

sec y tan ydydx

= 1 =⇒ dydx

sec y tan y=

1x tan(arcsec(x))

arcsec(x) =1

x√x2 − 1

.1. .y = arcsec x

x2 − 1

. . . . . .

sec y tan ydydx

= 1 =⇒ dydx

sec y tan y=

1x tan(arcsec(x))

arcsec(x) =1

x√x2 − 1

.1. .y = arcsec x

x2 − 1

. . . . . .

sec y tan ydydx

= 1 =⇒ dydx

sec y tan y=

1x tan(arcsec(x))

arcsec(x) =1

x√x2 − 1

. .y = arcsec x

x2 − 1

. . . . . .

sec y tan ydydx

= 1 =⇒ dydx

sec y tan y=

1x tan(arcsec(x))

arcsec(x) =1

x√x2 − 1

.1. .y = arcsec x

x2 − 1

. . . . . .

sec y tan ydydx

= 1 =⇒ dydx

sec y tan y=

1x tan(arcsec(x))

arcsec(x) =1

x√x2 − 1

.1. .y = arcsec x

x2 − 1

. . . . . .

sec y tan ydydx

= 1 =⇒ dydx

sec y tan y=

1x tan(arcsec(x))

arcsec(x) =1

x√x2 − 1 .

.1. .y = arcsec x

x2 − 1

. . . . . .

Another Example

Example

Let f(x) = earcsec 3x. Find f′(x).

Solution

f′(x) = earcsec 3x · 13x√

(3x)2 − 1· 3

=3earcsec 3x

3x√9x2 − 1

. . . . . .

Another Example

Example

Let f(x) = earcsec 3x. Find f′(x).

Solution

f′(x) = earcsec 3x · 13x√

(3x)2 − 1· 3

=3earcsec 3x

3x√9x2 − 1

. . . . . .

Outline

Applications

. . . . . .

Application

ExampleOne of the guiding principles ofmost sports is to “keep youreye on the ball.” In baseball, abatter stands 2 ft away fromhome plate as a pitch is thrownwith a velocity of 130 ft/sec(about 90mph). At what ratedoes the batter’s angle of gazeneed to change to follow theball as it crosses home plate?

SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0.

. . . . . .

Application

ExampleOne of the guiding principles ofmost sports is to “keep youreye on the ball.” In baseball, abatter stands 2 ft away fromhome plate as a pitch is thrownwith a velocity of 130 ft/sec(about 90mph). At what ratedoes the batter’s angle of gazeneed to change to follow theball as it crosses home plate?

. . . . . .

We have θ = arctan(y/2). Thus

1+ (y/2)2· 12dydt

When y = 0 and y′ = −130,then

∣∣∣∣y=0

1+ 0·12(−130) = −65 rad/sec

The human eye can only trackat 3 rad/sec!

..2 ft

.130 ft/sec

. . . . . .

SolutionLet y(t) be the distance from the ball to home plate, and θ the angle thebatter’s eyes make with home plate while following the ball. We knowy′ = −130 and we want θ′ at the moment that y = 0.We have θ = arctan(y/2). Thus

1+ (y/2)2· 12dydt

∣∣∣∣y=0

1+ 0·12(−130) = −65 rad/sec

..2 ft

.130 ft/sec

. . . . . .

1+ (y/2)2· 12dydt

∣∣∣∣y=0

1+ 0·12(−130) = −65 rad/sec

..2 ft

.130 ft/sec

. . . . . .

1+ (y/2)2· 12dydt

∣∣∣∣y=0

1+ 0·12(−130) = −65 rad/sec

..2 ft

.130 ft/sec

. . . . . .

Summary

y y′

arcsin x1√

1− x2

arccos x − 1√1− x2

arctan x1

arccot x − 11+ x2

arcsec x1

x√x2 − 1

arccsc x − 1x√x2 − 1

I Remarkable that thederivatives of thesetranscendental functionsare algebraic (or evenrational!)

Lesson 16: Inverse Trigonometric Functions (Section 041 slides)

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Lesson 16: Inverse Trigonometric Functions (Section 041 slides)