Lesson 15: Inverse Trigonometric Functions

Section 3.5
Inverse Trigonometric
Functions
V63.0121.006/016, Calculus I

March 11, 2010

Announcements
Exams returned in recitation
There is WebAssign due Tuesday March 23 and written HW
due Thursday March 25 . . . . . .

Announcements

Exams returned in recitation
There is WebAssign due Tuesday March 23 and written HW
due Thursday March 25
next quiz is Friday April 2

. . . . . .

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.

. . . . . .

Outline

Inverse Trigonometric Functions

Derivatives of Inverse Trigonometric Functions
Arcsine
Arccosine
Arctangent
Arcsecant

Applications

. . . . . .

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

y
.

. . . x
.
π π s
. in
−
. .
2 2

. . . . . .

arcsin
[−π/2, π/2].

y
.

.
. . . x
.
π π s
. in
−
. . .
2 2

. . . . . .

arcsin
[−π/2, π/2].

y
.
y
. =x
.
. . . x
.
π π s
. in
−
. . .
2 2

. . . . . .

arcsin
[−π/2, π/2].

y
.
. . rcsin
a
.
. . . x
.
π π s
. in
−
. . .
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 function after restriction to
[0, π]

y
.

c
. os
. . x
.
0
. .
π

. . . . . .

arccos
[0, π]

y
.

.
c
. os
. . x
.
0
. .
π
.

. . . . . .

arccos
[0, π]

y
.
y
. =x
.
c
. os
. . x
.
0
. .
π
.

. . . . . .

arccos
[0, π]

. . rccos
a
y
.

.
c
. os
. . . 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 function after restriction to
[−π/2, π/2].
y
.

. x
.
3π π π 3π
−
. −
. . .
2 2 2 2

t
.an

. . . . . .

arctan
y
. =x
[−π/2, π/2].
y
.

. x
.
3π π π 3π
−
. −
. . .
2 2 2 2

t
.an

. . . . . .

arctan
[−π/2, π/2].
y
.

π
. a
. rctan
2

. x
.

π
−
.
2

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

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

. x
.
3π π π 3π
−
. −
. . .
2 2 2 2

s
. ec

. . . . . .

arcsec
[0, π/2) ∪ (π, 3π/2].
y
.

.
. x
.
3π π π 3π
−
. −
. . . .
2 2 2 2

s
. ec

. . . . . .

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

.
. x
.
3π π π 3π
−
. −
. . . .
2 2 2 2

s
. ec

. . . . . .

arcsec 3π
.
2
[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

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

Check: Values of inverse trigonometric functions

Example
Find
arcsin(1/2)
arctan(−1)
( √ )
2
arccos −
2

. . . . . .


Example
Find
arcsin(1/2)
arctan(−1)
( √ )
2
arccos −
2

Solution
π
6

. . . . . .

What is arctan(−1)?

.

3
. π/4
.

. .

.
−
. π/4

. . . . . .


.

( )
3
. π/4 3π
. Yes, tan = −1
4
√
2
s
. in(3π/4) =
2
.
√ .
2
. os(3π/4) = −
c
2

.
−
. π/4

. . . . . .


.

( )
3
. π/4 3π
. Yes, tan = −1
4
√ But, the range of arctan
( π π)
2
s
. in(3π/4) = is − ,
2 2 2
.
√ .
2
. os(3π/4) = −
c
2

.
−
. π/4

. . . . . .


.

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

. . . . . .


.

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

. . . . . .


Example
Find
arcsin(1/2)
arctan(−1)
( √ )
2
arccos −
2

Solution
π
6
π
−
4

. . . . . .


Example
Find
arcsin(1/2)
arctan(−1)
( √ )
2
arccos −
2

Solution
π
6
π
−
4
3π
4

. . . . . .

Caution: Notational ambiguity

. in2 x =.(sin x)2
s . in−1 x = (sin x)−1
s

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
. . . . . .

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

1
(f−1 )′ (b) = ′ −1
f (f (b))

. . . . . .

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

1
(f−1 )′ (b) = ′ −1
f (f (b))

“Proof”.
If y = f−1 (x), then
f(y ) = x ,
So by implicit differentiation

dy dy 1 1
f′ (y) = 1 =⇒ = ′ = ′ −1
dx dx f (y) f (f (x))

. . . . . .

The derivative of arcsin

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 =⇒ = =

triangle:

1
.
x
.

.

. . . . . .



dy dy 1 1
cos y = 1 =⇒ = =

triangle:

1
.
x
.

y
. = arcsin x
.

. . . . . .



dy dy 1 1
cos y = 1 =⇒ = =

triangle:

1
.
x
.

y
. = arcsin x
. √
. 1 − x2

. . . . . .



dy dy 1 1
cos y = 1 =⇒ = =

triangle:
√
cos(arcsin x) = 1 − x2 1
.
x
.

y
. = arcsin x
. √
. 1 − x2

. . . . . .



dy dy 1 1
cos y = 1 =⇒ = =

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

. . . . . .

Graphing arcsin and its derivative

1
.√
1 − x2
The domain of f is
[−1, 1], but the domain . . rcsin
a
of f′ is (−1, 1)
lim f′ (x) = +∞
x →1 −
lim f′ (x) = +∞ .
| . .
|
x→−1+ −
. 1 1
.

.

. . . . . .

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)

triangle:
√
sin(arccos x) = 1 − x2 1
. √
. 1 − x2
So
d 1 y
. = arccos x
arccos(x) = − √ .
dx 1 − x2 x
.

. . . . . .

Graphing arcsin and arccos

. . rccos
a

. . rcsin
a

.
| . |.
.
−
. 1 1
.

.

. . . . . .

Graphing arcsin and arccos

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

.

. . . . . .

The derivative of arctan

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

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

. . . . . .



dy dy 1
dx dx sec2 y

triangle:

.

. . . . . .



dy dy 1
dx dx sec2 y

triangle:

x
.

.
1
.

. . . . . .



dy dy 1
dx dx sec2 y

triangle:

x
.

y
. = arctan x
.
1
.

. . . . . .



dy dy 1
dx dx sec2 y

triangle:

√
. 1 + x2 x
.

y
. = arctan x
.
1
.

. . . . . .



dy dy 1
dx dx sec2 y

triangle:

1
cos(arctan x) = √
1 + x2 √
. 1 + x2 x
.

y
. = arctan x
.
1
.

. . . . . .



dy dy 1
dx dx sec2 y

triangle:

1
cos(arctan x) = √
1 + x2 √
. 1 + x2 x
.
So
d 1 y
. = arctan x
arctan(x) = .
dx 1 + x2
1
.

. . . . . .

Graphing arctan and its derivative

y
.
. /2
π
a
. rctan
1
.
1 + x2
. x
.

−
. π/2

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

. . . . . .

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

. . . . . .

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

Solution

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 =⇒ = =

triangle:

.

. . . . . .



dy dy 1 1
sec y tan y = 1 =⇒ = =

triangle:

x
.

.
1
.

. . . . . .



dy dy 1 1
sec y tan y = 1 =⇒ = =

triangle:

x
.

y
. = arcsec x
.
1
.

. . . . . .



dy dy 1 1
sec y tan y = 1 =⇒ = =

triangle:
√
x2 − 1
tan(arcsec x) = √
1 x
. . x2 − 1

y
. = arcsec x
.
1
.

. . . . . .



dy dy 1 1
sec y tan y = 1 =⇒ = =

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

. . . . . .

Another Example

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

. . . . . .

Another Example

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

Solution
1
f′ (x) = earcsec x · √
x x2 − 1

. . . . . .

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?

. . . . . .

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.

y
.

1
. 30 ft/sec

.
θ
.
. 2
. ft

. . . . . .

y = 0.
We have θ = arctan(y/2).
Thus
dθ 1 1 dy
= ·
2 2 dt
dt 1 + ( y /2 )

y
.

1
. 30 ft/sec

.
θ
.
. 2
. ft

. . . . . .

y = 0.
Thus
dθ 1 1 dy
= ·
2 2 dt
dt 1 + ( y /2 )

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

. . . . . .

y = 0.
Thus
dθ 1 1 dy
= ·
2 2 dt
dt 1 + ( y /2 )

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

. . . . . .

Recap

y y′

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

Lesson 15: Inverse Trigonometric Functions

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