Lesson 16: Inverse Trigonometric Functions (Section 021 slides)

Section 3.5
Inverse Trigonometric
Functions
V63.0121.021, Calculus I
New York University
November 2, 2010
Announcements
Midterm grades have been submitted
Quiz 3 this week in recitation on Section 2.6, 2.8, 3.1, 3.2
Thank you for the evaluations
. . . . . .

. . . . . .
Announcements
Midterm grades have been
submitted
Quiz 3 this week in
recitation on Section 2.6,
2.8, 3.1, 3.2
Thank you for the
evaluations
V63.0121.021, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 2, 2010 2 / 40

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. . . . . .
Objectives
Know the definitions,
domains, ranges, and
other properties of the
inverse trignometric
functions: arcsin, arccos,
arctan, arcsec, arccsc,
arccot.
Know the derivatives of the
inverse trignometric
functions.

. . . . . .
Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
Arcsine
Arccosine
Arctangent
Arcsecant
Applications

. . . . . .
What is an inverse function?
Definition
Let f be a function with domain D and range E. The inverse of f is the
function f−1
defined by:
f−1
(b) = a,
where a is chosen so that f(a) = b.

. . . . . .
What is an inverse function?
Definition
Let f be a function with domain D and range E. The inverse of f is the
function f−1
defined by:
f−1
(b) = a,
where a is chosen so that f(a) = b.
So
f−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 D
corresponding to each b in E.
Such a function is called one-to-one
The graph of such a function passes the horizontal line test: any
horizontal line intersects the graph in exactly one point if at all.
If f is continuous, then f−1
is continuous.

. . . . . .
Graphing the inverse function
If b = f(a), then f−1
(b) = a.
.
.x
.y

. . . . . .
(b) = a.
So if (a, b) is on the graph
of f, then (b, a) is on the
graph of f−1
.
.
.x
.y
. .(a, b)
.
.(b, a)

. . . . . .
(b) = a.
graph of f−1
.
On the xy-plane, the point
(b, a) is the reflection of
(a, b) in the line y = x.
.
.x
.y
. .(a, b)
.
.(b, a)
.y = x

. . . . . .
(b) = a.
graph of f−1
.
On the xy-plane, the point
(b, a) is the reflection of
(a, b) in the line y = x.
Therefore:
.
.x
.y
. .(a, b)
.
.(b, a)
.y = x
Fact
The graph of f−1
is the reflection of the graph of f in the line y = x.

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

. . . . . .
arcsin
. .x
.y
.sin
.
.
.
.−
π
2
.
.
π
2

. . . . . .
arcsin
. .x
.y
.sin
.
.
.
.−
π
2
.
.
π
2
.y = x

. . . . . .
arcsin
. .x
.y
.sin
.
.
.
.−
π
2
.
.
π
2
.
..arcsin
The domain of arcsin is [−1, 1]
The range of arcsin is
[
−
π
2
,
π
2
]

. . . . . .
arccos
Arccos is the inverse of the cosine function after restriction to [0, π]
. .x
.y
.cos
.
.0
.
.π

. . . . . .
arccos
. .x
.y
.cos
.
.
.
.0
.
.π

. . . . . .
arccos
. .x
.y
.cos
.
.
.
.0
.
.π
.y = x

. . . . . .
arccos
. .x
.y
.cos
.
.
.
.0
.
.π
.
..arccos
The domain of arccos is [−1, 1]
The range of arccos is [0, π]

. . . . . .
arctan
Arctan is the inverse of the tangent function after restriction to
[−π/2, π/2].
. .x
.y
.tan
.−
3π
2
.−
π
2
.
π
2 .
3π
2

. . . . . .
arctan
[−π/2, π/2].
. .x
.y
.tan
.−
3π
2
.−
π
2
.
π
2 .
3π
2
.y = x

. . . . . .
arctan
[−π/2, π/2].
. .x
.y
.arctan
.−
π
2
.
π
2
The domain of arctan is (−∞, ∞)
The range of arctan is
(
−
π
2
,
π
2
)
lim
x→∞
arctan x =
π
2
, lim
x→−∞
arctan x = −
π
2

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

. . . . . .
arcsec
[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−
3π
2
.−
π
2
.
π
2 .
3π
2
.
.

. . . . . .
arcsec
[0, π/2) ∪ (π, 3π/2].
. .x
.y
.sec
.−
3π
2
.−
π
2
.
π
2 .
3π
2
.
.
.y = x

. . . . . .
arcsec
[0, π/2) ∪ (π, 3π/2].
. .x
.y
.
.
.
.
.
π
2
.
3π
2
The domain of arcsec is (−∞, −1] ∪ [1, ∞)
The range of arcsec is
[
0,
π
2
)
∪
(π
2
, π
]
lim
x→∞
arcsec x =
π
2
, lim
x→−∞
arcsec x =
3π
2

. . . . . .
Values of Trigonometric Functions
x 0
π
6
π
4
π
3
π
2
sin x 0
1
2
√
2
2
√
3
2
1
cos x 1
√
3
2
√
2
2
1
2
0
tan x 0
1
√
3
1
√
3 undef
cot x undef
√
3 1
1
√
3
0
sec x 1
2
√
3
2
√
2
2 undef
csc x undef 2
2
√
2
2
√
3
1

. . . . . .
Check: Values of inverse trigonometric functions
Example
Find
arcsin(1/2)
arctan(−1)
arccos
(
−
√
2
2
)

. . . . . .
Example
Find
arcsin(1/2)
arctan(−1)
arccos
(
−
√
2
2
)
Solution
π
6

. . . . . .
What is arctan(−1)?
. .
.
.
.3π/4
.
.−π/4

. . . . . .
. .
.
.
.3π/4
.
.−π/4
.sin(3π/4) =
√
2
2
.cos(3π/4) = −
√
2
2
Yes, tan
(
3π
4
)
= −1

. . . . . .
. .
.
.
.3π/4
.
.−π/4
.sin(3π/4) =
√
2
2
.cos(3π/4) = −
√
2
2
Yes, tan
(
3π
4
)
= −1
But, the range of arctan is(
−
π
2
,
π
2
)

. . . . . .
. .
.
.
.3π/4
.
.−π/4
.sin(π/4) = −
√
2
2
.cos(π/4) =
√
2
2
Yes, tan
(
3π
4
)
= −1
−
π
2
,
π
2
)
Another angle whose
tangent is −1 is −
π
4
, and
this is in the right range.

. . . . . .
. .
.
.
.3π/4
.
.−π/4
.sin(π/4) = −
√
2
2
.cos(π/4) =
√
2
2
Yes, tan
(
3π
4
)
= −1
−
π
2
,
π
2
)
Another angle whose
tangent is −1 is −
π
4
, and
this is in the right range.
So arctan(−1) = −
π
4

. . . . . .
Example
Find
arcsin(1/2)
arctan(−1)
arccos
(
−
√
2
2
)
Solution
π
6
−
π
4

. . . . . .
Example
Find
arcsin(1/2)
arctan(−1)
arccos
(
−
√
2
2
)
Solution
π
6
−
π
4
3π
4

. . . . . .
Caution: Notational ambiguity
..sin2
x = (sin x)2
.sin−1
x = (sin x)−1
sinn
x means the nth power of sin x, except when n = −1!
The book uses sin−1
x for the inverse of sin x, and never for
(sin x)−1
.
I use csc x for
1
sin x
and arcsin x for the inverse of sin x.

. . . . . .
Outline
Arcsine
Arccosine
Arctangent
Arcsecant
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 an
open interval containing b = f(a), and
(f−1
)′
(b) =
1
f′
(f−1
(b))
In Leibniz notation we have
dx
dy
=
1
dy/dx

. . . . . .
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 an
open interval containing b = f(a), and
(f−1
)′
(b) =
1
f′
(f−1
(b))
In Leibniz notation we have
dx
dy
=
1
dy/dx
Upshot: Many times the derivative of f−1
(x) can be found by implicit
differentiation and the derivative of f:

. . . . . .
Illustrating the Inverse Function Theorem
.
.
Example
Use the inverse function theorem to find the derivative of the square root
function.

. . . . . .
.
.
Example
function.
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
2
√
b

. . . . . .
.
.
Example
function.
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
2
√
b
Solution (Leibniz notation)
If the original function is y = x2
, then the inverse function is defined by x = y2
.
Differentiate implicitly:
1 = 2y
dy
dx
=⇒
dy
dx
=
1
2
√
x

. . . . . .
Derivation: The derivative of arcsin
Let y = arcsin x, so x = sin y. Then
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=
1
cos(arcsin x)

. . . . . .
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=
1
cos(arcsin x)
To simplify, look at a right
triangle:
.

. . . . . .
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=
1
cos(arcsin x)
triangle:
.
. .
.
.y = arcsin x

. . . . . .
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=
1
cos(arcsin x)
triangle:
.
. .
.
.y = arcsin x
.1
.x

. . . . . .
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=
1
cos(arcsin x)
triangle:
.
. .
.
.y = arcsin x
.1
.x
.
√
1 − x2

. . . . . .
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=
1
cos(arcsin x)
triangle:
cos(arcsin x) =
√
1 − x2
.
. .
.
.y = arcsin x
.1
.x
.
√
1 − x2

. . . . . .
cos y
dy
dx
= 1 =⇒
dy
dx
=
1
cos y
=
1
cos(arcsin x)
triangle:
cos(arcsin x) =
√
1 − x2
So
Fact
d
dx
arcsin(x) =
1
√
1 − x2
.
. .
.
.y = arcsin x
.1
.x
.
√
1 − x2

. . . . . .
Graphing arcsin and its derivative
The domain of f is [−1, 1],
but the domain of f′
is
(−1, 1)
lim
x→1−
f′
(x) = +∞
lim
x→−1+
f′
(x) = +∞ ..|
.−1
.|
.1
.
..arcsin
.
1
√
1 − x2

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

. . . . . .
Composing with arcsin
Example
Let f(x) = arcsin(x3
+ 1). Find f′
(x).
Solution
We have
d
dx
arcsin(x3
+ 1) =
1
√
1 − (x3 + 1)2
d
dx
(x3
+ 1)
=
3x2
√
−x6 − 2x3

. . . . . .
Derivation: The derivative of arccos
Let y = arccos x, so x = cos y. Then
− sin y
dy
dx
= 1 =⇒
dy
dx
=
1
− sin y
=
1
− sin(arccos x)

. . . . . .
Derivation: The derivative of arccos
Let y = arccos x, so x = cos y. Then
− sin y
dy
dx
= 1 =⇒
dy
dx
=
1
− sin y
=
1
− sin(arccos x)
triangle:
sin(arccos x) =
√
1 − x2
So
Fact
d
dx
arccos(x) = −
1
√
1 − x2
.
.1
.
√
1 − x2
.x
.
.y = arccos x

. . . . . .
Graphing arcsin and arccos
..|
.−1
.|
.1
.
..arcsin
.
..arccos

. . . . . .
Graphing arcsin and arccos
..|
.−1
.|
.1
.
..arcsin
.
..arccos
Note
cos θ = sin
(π
2
− θ
)
=⇒ arccos x =
π
2
− arcsin x
So it’s not a surprise that their
derivatives are opposites.

. . . . . .
Derivation: The derivative of arctan
Let y = arctan x, so x = tan y. Then
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y
= cos2
(arctan x)

. . . . . .
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y
= cos2
(arctan x)
triangle:
.

. . . . . .
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y
= cos2
(arctan x)
triangle:
.
.
..
.y = arctan x

. . . . . .
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y
= cos2
(arctan x)
triangle:
.
.
..
.y = arctan x
.x
.1

. . . . . .
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y
= cos2
(arctan x)
triangle:
.
.
..
.y = arctan x
.x
.1
.
√
1 + x2

. . . . . .
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y
= cos2
(arctan x)
triangle:
cos(arctan x) =
1
√
1 + x2
.
.
..
.y = arctan x
.x
.1
.
√
1 + x2

. . . . . .
sec2
y
dy
dx
= 1 =⇒
dy
dx
=
1
sec2 y
= cos2
(arctan x)
triangle:
cos(arctan x) =
1
√
1 + x2
So
Fact
d
dx
arctan(x) =
1
1 + x2
.
.
..
.y = arctan x
.x
.1
.
√
1 + x2

. . . . . .
Graphing arctan and its derivative
. .x
.y
.arctan
.
1
1 + x2
.π/2
.−π/2
The domain of f and f′
are both (−∞, ∞)
Because of the horizontal asymptotes, lim
x→±∞
f′
(x) = 0

. . . . . .
Composing with arctan
Example
Let f(x) = arctan
√
x. Find f′
(x).

. . . . . .
Composing with arctan
Example
Let f(x) = arctan
√
x. Find f′
(x).
Solution
d
dx
arctan
√
x =
1
1 +
(√
x
)2
d
dx
√
x =
1
1 + x
·
1
2
√
x
=
1
2
√
x + 2x
√
x

. . . . . .
Derivation: The derivative of arcsec
Try this first.

. . . . . .
Try this first. Let y = arcsec x, so x = sec y. Then
sec y tan y
dy
dx
= 1 =⇒
dy
dx
=
1
sec y tan y
=
1
x tan(arcsec(x))

. . . . . .
sec y tan y
dy
dx
= 1 =⇒
dy
dx
=
1
sec y tan y
=
1
x tan(arcsec(x))
triangle:
.

. . . . . .
sec y tan y
dy
dx
= 1 =⇒
dy
dx
=
1
sec y tan y
=
1
x tan(arcsec(x))
triangle:
..
.y = arcsec x

. . . . . .
sec y tan y
dy
dx
= 1 =⇒
dy
dx
=
1
sec y tan y
=
1
x tan(arcsec(x))
triangle:
.
.x
.1
.
.y = arcsec x

. . . . . .
sec y tan y
dy
dx
= 1 =⇒
dy
dx
=
1
sec y tan y
=
1
x tan(arcsec(x))
triangle:
tan(arcsec x) =
√
x2 − 1
1
.
.x
.1
.
.y = arcsec x
.
√
x2 − 1

. . . . . .
sec y tan y
dy
dx
= 1 =⇒
dy
dx
=
1
sec y tan y
=
1
x tan(arcsec(x))
triangle:
tan(arcsec x) =
√
x2 − 1
1
So
Fact
d
dx
arcsec(x) =
1
x
√
x2 − 1
.
.x
.1
.
.y = arcsec x
.
√
x2 − 1

. . . . . .
Another Example
Example
Let f(x) = earcsec 3x
. Find f′
(x).

. . . . . .
Another Example
Example
Let f(x) = earcsec 3x
. Find f′
(x).
Solution
f′
(x) = earcsec 3x
·
1
3x
√
(3x)2 − 1
· 3
=
3earcsec 3x
3x
√
9x2 − 1

. . . . . .
Outline
Arcsine
Arccosine
Arctangent
Arcsecant
Applications

. . . . . .
Application
Example
One of the guiding principles of
most sports is to “keep your
eye on the ball.” In baseball, a
batter stands 2 ft away from
home plate as a pitch is thrown
with a velocity of 130 ft/sec
(about 90 mph). At what rate
does the batter’s angle of gaze
need to change to follow the
ball as it crosses home plate?

. . . . . .
Application
Example
One of the guiding principles of
most sports is to “keep your
eye on the ball.” In baseball, a
batter stands 2 ft away from
home plate as a pitch is thrown
with a velocity of 130 ft/sec
(about 90 mph). At what rate
does the batter’s angle of gaze
need to change to follow the
ball as it crosses home plate?
Solution
Let y(t) be the distance from the ball to home plate, and θ the angle the
batter’s eyes make with home plate while following the ball. We know
y′
= −130 and we want θ′
at the moment that y = 0.

. . . . . .
Solution
y′
.
.2 ft
.y
.130 ft/sec
.
.θ

. . . . . .
Solution
y′
We have θ = arctan(y/2). Thus
dθ
dt
=
1
1 + (y/2)2
·
1
2
dy
dt
.
.2 ft
.y
.130 ft/sec
.
.θ

. . . . . .
Solution
y′
dθ
dt
=
1
1 + (y/2)2
·
1
2
dy
dt
When y = 0 and y′
= −130,
then
dθ
dt y=0
=
1
1 + 0
·
1
2
(−130) = −65 rad/sec
.
.2 ft
.y
.130 ft/sec
.
.θ

. . . . . .
Solution
y′
dθ
dt
=
1
1 + (y/2)2
·
1
2
dy
dt
When y = 0 and y′
= −130,
then
dθ
dt y=0
=
1
1 + 0
·
1
2
(−130) = −65 rad/sec
The human eye can only track
at 3 rad/sec!
.
.2 ft
.y
.130 ft/sec
.
.θ

. . . . . .
Summary
y y′
arcsin x
1
√
1 − x2
arccos x −
1
√
1 − x2
arctan x
1
1 + x2
arccot x −
1
1 + x2
arcsec x
1
x
√
x2 − 1
arccsc x −
1
x
√
x2 − 1
Remarkable that the
derivatives of these
transcendental functions
are algebraic (or even
rational!)

Lesson 16: Inverse Trigonometric Functions (Section 021 slides)

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Lesson 16: Inverse Trigonometric Functions (Section 021 slides)