Lesson 16: Inverse Trigonometric Functions

Section 3.5Inverse Trigonometric

Functions

V63.0121.002.2010Su, Calculus I

New York University

June 7, 2010

Announcements

I Exams not graded yet

I Written HW due tomorrow

I Quiz 3 on Thursday covering 3.3, 3.4, 3.5, 3.7

Announcements

I Exams not graded yet

I Written HW due tomorrow

I Quiz 3 on Thursday covering3.3, 3.4, 3.5, 3.7

V63.0121.002.2010Su, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions June 7, 2010 2 / 30

Objectives

I Know the definitions,domains, ranges, and otherproperties of the inversetrignometric functions:arcsin, arccos, arctan,arcsec, arccsc, arccot.

I Know the derivatives of theinverse trignometricfunctions.

What is an inverse function?

Definition

Let 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?

Definition

Let 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-one

I The graph of such a function passes the horizontal line test: anyhorizontal 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].

sin−π

y = xarcsin

I The domain of arcsin is [−1, 1]

I The range of arcsin is[−π

arcsin

sin−π

y = xarcsin

arcsin

sin−π

arcsin

sin−π

arcsin

arccos

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

arccos

I The domain of arccos is [−1, 1]

I The range of arccos is [0, π]

arccos

arctan

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

−3π

2−π

arctan

−π2

I The domain of arctan is (−∞,∞)

I The range of arctan is(−π

)I lim

x→∞arctan x =

2, limx→−∞

arctan x = −π2

arctan

−3π

2−π

arctan

−π2

)I lim

x→∞arctan x =

2, limx→−∞

arctan x = −π2

arctan

−3π

2−π

arctan

−π2

)I lim

x→∞arctan x =

2, limx→−∞

arctan x = −π2

arctan

−3π

2−π

arctan

−π2

)I lim

x→∞arctan x =

2, limx→−∞

arctan x = −π2

arcsec

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

−3π

2−π

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

I The range of arcsec is[0,π

)∪(π

2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π

arcsec

−3π

2−π

)∪(π

2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π

arcsec

−3π

2−π

)∪(π

2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π

arcsec

−3π

2−π

)∪(π

2, π]

I limx→∞

arcsec x =π

2, limx→−∞

arcsec x =3π

Values of Trigonometric Functions

sin x 01

cos x 1

tan x 01√3

3 undef

cot x undef√

3 11√3

sec x 12√3

2 undef

csc x undef 22√2

Check: Values of inverse trigonometric functions

Example

I arcsin(1/2)

I arctan(−1)

I arccos

(−√

Solution

I −π4

Example

I arcsin(1/2)

I arctan(−1)

I arccos

(−√

Solution

I −π4

What is arctan(−1)?

−π/4

sin(3π/4) =

cos(3π/4) = −√

2sin(π/4) = −√

cos(π/4) =

I Yes, tan

)= −1

I But, the range of arctan is(−π

)I Another angle whose

tangent is −1 is −π4

this is in the right range.

I So arctan(−1) = −π4

−π/4

sin(3π/4) =

cos(3π/4) = −√

sin(π/4) = −√

cos(π/4) =

I Yes, tan

)= −1

−π/4

sin(3π/4) =

cos(3π/4) = −√

sin(π/4) = −√

cos(π/4) =

I Yes, tan

)= −1

I Another angle whose

−π/4

sin(3π/4) =

cos(3π/4) = −√

sin(π/4) = −√

cos(π/4) =

I Yes, tan

)= −1

−π/4

sin(3π/4) =

cos(3π/4) = −√

sin(π/4) = −√

cos(π/4) =

I Yes, tan

)= −1

Example

I arcsin(1/2)

I arctan(−1)

I arccos

(−√

Solution

I −π4

Example

I arcsin(1/2)

I arctan(−1)

I arccos

(−√

Solution

I −π4

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) 6= 0. Then f −1 is defined in an openinterval containing b = f (a), and

(f −1)′(b) =1

f ′(f −1(b))

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) 6= 0. Then f −1 is defined in an openinterval containing b = f (a), and

(f −1)′(b) =1

f ′(f −1(b))

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

Derivation: The derivative of arcsin

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

cos ydy

dx= 1 =⇒ dy

cos y=

cos(arcsin x)

To simplify, look at a righttriangle:

cos(arcsin x) =√

1− x2

dxarcsin(x) =

1√1− x2

y = arcsin x√1− x2

cos ydy

dx= 1 =⇒ dy

cos y=

cos(arcsin x)

cos(arcsin x) =√

1− x2

dxarcsin(x) =

1√1− x2

cos ydy

dx= 1 =⇒ dy

cos y=

cos(arcsin x)

cos(arcsin x) =√

1− x2

dxarcsin(x) =

1√1− x2

cos ydy

dx= 1 =⇒ dy

cos y=

cos(arcsin x)

cos(arcsin x) =√

1− x2

dxarcsin(x) =

1√1− x2

y = arcsin x

√1− x2

cos ydy

dx= 1 =⇒ dy

cos y=

cos(arcsin x)

cos(arcsin x) =√

1− x2

dxarcsin(x) =

1√1− x2

cos ydy

dx= 1 =⇒ dy

cos y=

cos(arcsin x)

cos(arcsin x) =√

1− x2

dxarcsin(x) =

1√1− x2

cos ydy

dx= 1 =⇒ dy

cos y=

cos(arcsin x)

cos(arcsin x) =√

1− x2

dxarcsin(x) =

1√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√1− x2

Derivation: The derivative of arccos

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

− sin ydy

dx= 1 =⇒ dy

− sin y=

− sin(arccos x)

sin(arccos x) =√

1− x2

dxarccos(x) = − 1√

1− x2

1 √1− x2

y = arccos x

Derivation: The derivative of arccos

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

− sin ydy

dx= 1 =⇒ dy

− sin y=

− sin(arccos x)

sin(arccos x) =√

1− x2

dxarccos(x) = − 1√

1− x2

1 √1− x2

y = arccos x

Graphing arcsin and arccos

arcsin

arccos

cos θ = sin(π

2− θ)

=⇒ arccos x =π

2− arcsin x

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

Graphing arcsin and arccos

arcsin

arccos

cos θ = sin(π

2− 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 ydy

dx= 1 =⇒ dy

sec2 y= cos2(arctan x)

cos(arctan x) =1√

1 + x2

dxarctan(x) =

1 + x2

y = arctan x

√1 + x2

sec2 ydy

dx= 1 =⇒ dy

cos(arctan x) =1√

1 + x2

dxarctan(x) =

1 + x2

y = arctan x

√1 + x2

sec2 ydy

dx= 1 =⇒ dy

cos(arctan x) =1√

1 + x2

dxarctan(x) =

1 + x2

y = arctan x

√1 + x2

sec2 ydy

dx= 1 =⇒ dy

cos(arctan x) =1√

1 + x2

dxarctan(x) =

1 + x2

y = arctan x

√1 + x2

sec2 ydy

dx= 1 =⇒ dy

cos(arctan x) =1√

1 + x2

dxarctan(x) =

1 + x2

y = arctan x

√1 + x2

sec2 ydy

dx= 1 =⇒ dy

cos(arctan x) =1√

1 + x2

dxarctan(x) =

1 + x2

y = arctan x

√1 + x2

sec2 ydy

dx= 1 =⇒ dy

cos(arctan x) =1√

1 + x2

dxarctan(x) =

1 + x2

y = arctan x

√1 + x2

Graphing arctan and its derivative

arctan

1 + x2

−π/2

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

I Because of the horizontal asymptotes, limx→±∞

f ′(x) = 0

Example

Let f (x) = arctan√

x . Find f ′(x).

Solution

dxarctan

√x =

1 +(√

√x =

1 + x· 1

x + 2x√

Example

Let f (x) = arctan√

x . Find f ′(x).

Solution

dxarctan

√x =

1 +(√

√x =

1 + x· 1

x + 2x√

Derivation: The derivative of arcsec

Try this first.

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

sec y tan ydy

dx= 1 =⇒ dy

sec y tan y=

x tan(arcsec(x))

tan(arcsec x) =

√x2 − 1

dxarcsec(x) =

x2 − 1

y = arcsec x

√x2 − 1

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

sec y tan ydy

dx= 1 =⇒ dy

sec y tan y=

x tan(arcsec(x))

tan(arcsec x) =

√x2 − 1

dxarcsec(x) =

x2 − 1

y = arcsec x

√x2 − 1

sec y tan ydy

dx= 1 =⇒ dy

sec y tan y=

x tan(arcsec(x))

tan(arcsec x) =

√x2 − 1

dxarcsec(x) =

x2 − 1

y = arcsec x

√x2 − 1

sec y tan ydy

dx= 1 =⇒ dy

sec y tan y=

x tan(arcsec(x))

tan(arcsec x) =

√x2 − 1

dxarcsec(x) =

x2 − 1

y = arcsec x

√x2 − 1

sec y tan ydy

dx= 1 =⇒ dy

sec y tan y=

x tan(arcsec(x))

tan(arcsec x) =

√x2 − 1

dxarcsec(x) =

x2 − 1

y = arcsec x

√x2 − 1

sec y tan ydy

dx= 1 =⇒ dy

sec y tan y=

x tan(arcsec(x))

tan(arcsec x) =

√x2 − 1

dxarcsec(x) =

x2 − 1

y = arcsec x

√x2 − 1

sec y tan ydy

dx= 1 =⇒ dy

sec y tan y=

x tan(arcsec(x))

tan(arcsec x) =

√x2 − 1

dxarcsec(x) =

x2 − 1

y = arcsec x

√x2 − 1

sec y tan ydy

dx= 1 =⇒ dy

sec y tan y=

x tan(arcsec(x))

tan(arcsec x) =

√x2 − 1

dxarcsec(x) =

x2 − 1

y = arcsec x

√x2 − 1

Another Example

Example

Let f (x) = earcsec x . Find f ′(x).

Solution

f ′(x) = earcsec x · 1

x2 − 1

Another Example

Example

Let f (x) = earcsec x . Find f ′(x).

Solution

f ′(x) = earcsec x · 1

x2 − 1

Outline

Applications

Application

Example

One of the guiding principles ofmost sports is to “keep your eyeon the ball.” In baseball, a batterstands 2 ft away from home plateas a pitch is thrown with avelocity of 130 ft/sec (about90 mph). At what rate does thebatter’s angle of gaze need tochange to follow the ball as itcrosses home plate?

Solution

Let 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· 1

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!

130 ft/sec

Solution

1 + (y/2)2· 1

∣∣∣∣y=0

1 + 0·12

(−130) = −65 rad/sec

130 ft/sec

Solution

1 + (y/2)2· 1

∣∣∣∣y=0

1 + 0·12

(−130) = −65 rad/sec

130 ft/sec

Solution

1 + (y/2)2· 1

∣∣∣∣y=0

1 + 0·12

(−130) = −65 rad/sec

The human eye can only trackat 3 rad/sec! 2 ft

130 ft/sec

Summary

y y ′

arcsin x1√

1− x2

arccos x − 1√1− x2

arctan x1

1 + x2

arccot x − 1

1 + x2

arcsec x1

x2 − 1

arccsc x − 1

x2 − 1

I Remarkable that thederivatives of thesetranscendental functions arealgebraic (or even rational!)

Lesson 16: Inverse Trigonometric Functions

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Transcript of Lesson 16: Inverse Trigonometric Functions