Lesson 16: Inverse Trigonometric Functions (slides)

Sec on 3.6
Inverse Trigonometric Func ons
V63.0121.011: Calculus I
Professor Ma hew Leingang
New York University

March 28, 2011

.

Announcements
Midterm has been returned. Please see
FAQ on Blackboard (under ”Exams and
Quizzes”)
Quiz 3 this week in recita on on
Sec on 2.6, 2.8, 3.1, 3.2
Quiz 4 April 14–15 on Sec ons 3.3, 3.4,
3.5, and 3.7
Quiz 5 April 28–29 on Sec ons 4.1, 4.2,
4.3, and 4.4

Objectives

Know the deﬁni ons, domains, ranges,
and other proper es of the inverse
trignometric func ons: arcsin, arccos,
arctan, arcsec, arccsc, arccot.
Know the deriva ves of the inverse
trignometric func ons.

Outline
Inverse Trigonometric Func ons

Deriva ves of Inverse Trigonometric Func ons
Arcsine
Arccosine
Arctangent
Arcsecant

Applica ons

What is an inverse function?
Deﬁni on
Let f be a func on with domain D and range E. The inverse of f is the
func on f−1 deﬁned by:

f−1 (b) = a,

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

What is an inverse function?
Deﬁni on
Let f be a func on with domain D and range E. The inverse of f is the
func on 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 func on, there must be only one a in D
corresponding to each b in E.
Such a func on is called one-to-one
The graph of such a func on passes the horizontal line test:
any horizontal line intersects the graph in exactly one point if at
all.
If f is con nuous, then f−1 is con nuous.

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

.
x

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

(a, b)
.
x

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

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

arcsin
Arcsin is the inverse of the sine func on a er restric on to
[−π/2, π/2].
y

. x
π π sin
−
2 2

arcsin
[−π/2, π/2].
y
y=x

. x
π π sin
−
2 2

arcsin
[−π/2, π/2].
y
arcsin

. x
π π sin
−
2 2

The domain of arcsin is [−1, 1]
[ π π]
The range of arcsin is − ,
2 2

arccos
Arccos is the inverse of the cosine func on a er restric on to [0, π]

y

cos
. x
0 π

arccos

y
y=x

cos
. x
0 π

arccos
arccos
y

cos
. x
0 π

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

arctan
Arctan is the inverse of the tangent func on a er restric on to
y
(−π/2, π/2).

. x
3π π π 3π
− −
2 2 2 2

tan

arctan
y=x
y
(−π/2, π/2).

. x
3π π π 3π
− −
2 2 2 2

tan

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

arcsec
Arcsecant is the inverse of secant a er restric on to
[0, π/2) ∪ [π, 3π/2). y

. x
3π π π 3π
− −
2 2 2 2

sec

arcsec
Arcsecant is the inverse of secant a er restric on to x
y=
[0, π/2) ∪ [π, 3π/2). y

. x
3π π π 3π
− −
2 2 2 2

sec

arcsec 3π
2
Arcsecant is the inverse of secant a er restric on to
[0, π/2) ∪ [π, 3π/2). y

π
2
. x

The domain of arcsec is (−∞, −1] ∪ [1, ∞)
[ π ) (π ]
The range of arcsec is 0, ∪ ,π
2 2
π 3π
lim arcsec x = , lim arcsec x =
x→∞ 2 x→−∞ 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 Solu on
Find
arcsin(1/2)
arctan(−1)
( √ )
2
arccos −
2


Example Solu on
Find π
arcsin(1/2) 6
arctan(−1)
( √ )
2
arccos −
2

What is arctan(−1)?
3π/4

.

−π/4

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

−π/4

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

−π/4

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

−π/4

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


Example Solu on
Find π
arcsin(1/2) 6
π
arctan(−1) −
( √ ) 4
2
arccos −
2


Example Solu on
Find π
arcsin(1/2) 6
π
arctan(−1) −
( √ ) 4
2 3π
arccos −
2 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 .
1
I use csc x for and arcsin x for the inverse of sin x.
sin x

The Inverse Function Theorem
Theorem (The Inverse Func on Theorem)
Let f be diﬀeren able at a, and f′ (a) ̸= 0. Then f−1 is deﬁned in an
open interval containing b = f(a), and
1
(f−1 )′ (b) =
f′ (f−1 (b))
In Leibniz nota on we have
dx 1
=
dy dy/dx

Illustrating the IFT
Example
Use the inverse func on theorem to ﬁnd the deriva ve of the
square root func on.

Example

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

Example

Solu on (Leibniz nota on)
If the original func on is y = x2 , then the inverse func on is deﬁned
by x = y2 . Diﬀeren ate implicitly:
dy dy 1
1 = 2y =⇒ = √
dx dx 2 x

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

dy dy 1 1
cos y = 1 =⇒ = =
To simplify, look at a right triangle:

.

dy dy 1 1
cos y = 1 =⇒ = =

y = arcsin x
.

dy dy 1 1
cos y = 1 =⇒ = =

1
x

y = arcsin x
.

dy dy 1 1
cos y = 1 =⇒ = =

1
x

y = arcsin x
.√
1 − x2

dy dy 1 1
cos y = 1 =⇒ = =
√
cos(arcsin x) = 1 − x2
1
x

y = arcsin x
.√
1 − x2

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

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

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).

Solu on
We have
d 1 d 3
arcsin(x3 + 1) = √ (x + 1)
dx 1 − (x3 + 1)2 dx
3x2
= √
−x6 − 2x3

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

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

Graphing arcsin and arccos
arccos

arcsin

| . |
−1 1

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

The derivative of arctan
Let y = arctan x, so x = tan y. Then
dy dy 1
sec2 y = 1 =⇒ = 2y
= cos2 (arctan x)
dx dx sec

dy dy 1
sec2 y = 1 =⇒ = 2y
= cos2 (arctan x)
dx dx sec

.

dy dy 1
sec2 y = 1 =⇒ = 2y
= cos2 (arctan x)
dx dx sec

y = arctan x
.

dy dy 1
sec2 y = 1 =⇒ = 2y
= cos2 (arctan x)
dx dx sec

x

y = arctan x
.
1

dy dy 1
sec2 y = 1 =⇒ = 2y
= cos2 (arctan x)
dx dx sec

√
1 + x2 x

y = arctan x
.
1

dy dy 1
sec2 y = 1 =⇒ = 2y
= cos2 (arctan x)
dx dx sec
1
cos(arctan x) = √
1 + x2
√
1 + x2 x

y = arctan x
.
1

dy dy 1
sec2 y = 1 =⇒ = 2y
= cos2 (arctan x)
dx dx sec
1
cos(arctan x) = √
1 + x2
√
So 1 + x2 x
Fact
d 1 y = arctan x
.
arctan(x) = 1
dx 1 + x2

Graphing arctan and its derivative
y
π/2
arctan
1
1 + x2
. x

−π/2

The domain of f and f′ are both (−∞, ∞)
Because of the horizontal asymptotes, lim f′ (x) = 0
x→±∞

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

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

Solu on

d √ 1 d√ 1 1
arctan x = (√ )2 x= · √
dx 1+ x dx 1+x 2 x
1
= √ √
2 x + 2x x

The derivative of arcsec
Try this ﬁrst.

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

dy dy 1 1
sec y tan y = 1 =⇒ = =


.

dy dy 1 1
sec y tan y = 1 =⇒ = =


y = arcsec x
.

dy dy 1 1
sec y tan y = 1 =⇒ = =


x

y = arcsec x
.
1

dy dy 1 1
sec y tan y = 1 =⇒ = =

√
x2 − 1
tan(arcsec x) =
1
√
x x2 − 1

y = arcsec x
.
1

dy dy 1 1
sec y tan y = 1 =⇒ = =

√
x2 − 1
tan(arcsec x) =
1
So √
x x2 − 1
Fact
d 1 y = arcsec x
arcsec(x) = √ .
dx x x2 − 1
1

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

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

Solu on

1
f′ (x) = earcsec 3x · √ ·3
3x (3x)2 − 1
3earcsec 3x
= √
3x 9x2 − 1

Application
Example
One of the guiding principles of most
sports is to “keep your eye on the
ball.” In baseball, a ba er 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 ba er’s angle of gaze need to
change to follow the ball as it crosses
home plate?

Solu on
Let y(t) be the distance from the ball to home plate, and θ the angle
the ba er’s eyes make with home plate while following the ball. We
know y′ = −130 and we want θ′ at the moment that y = 0.

y

130 ft/sec
θ
.
2

Solu on
We have θ = arctan(y/2). Thus
dθ 1 1 dy
= ·
dt 1 + (y/2)2 2 dt
y

130 ft/sec
θ
.
2

Solu on
dθ 1 1 dy
= ·
dt 1 + (y/2)2 2 dt
When y = 0 and y′ = −130, then y

dθ 1 1 130 ft/sec
= · (−130) = −65 rad/sec
dt y=0 1+0 2 θ
.
2

Solu on
dθ 1 1 dy
= ·
dt 1 + (y/2)2 2 dt
When y = 0 and y′ = −130, then y

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

Summary
y y′ y y′
1 1
arcsin x √ arccos x − √
1 − x2 1 − x2
1 1
arctan x arccot x −
1 + x2 1 + x2
1 1
arcsec x √ arccsc x − √
x x2 − 1 x x2 − 1

Remarkable that the deriva ves of these transcendental
func ons are algebraic (or even ra onal!)

Lesson 16: Inverse Trigonometric Functions (slides)

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Lesson 16: Inverse Trigonometric Functions (slides)