Section 3.5
Inverse Trigonometric
Functions
V63.0121.027, Calculus I
October 29, 2009
Announcements
. . . . . .

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

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

arcsin
Arcsin is the inverse of the sine function after restriction to
[−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
Arccos is the inverse of the cosine function after restriction to
[0, π]
y
.
.
c
. os
. . x
.
0
. .
π
.
. . . . . .

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

arccos
Arccos is the inverse of the cosine function after restriction to
[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
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
Arctan is the inverse of the tangent function after restriction to
y
. =x
[−π/2, π/2].
y
.
. x
.
3π π π 3π
−
. −
. . .
2 2 2 2
t
.an
. . . . . .

arctan
Arctan is the inverse of the tangent function after restriction to
[−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
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
Arcsecant is the inverse of secant after restriction to
y
. =x
[0, π/2) ∪ (π, 3π/2].
y
.
.
. x
.
3π π π 3π
−
. −
. . . .
2 2 2 2
s
. ec
. . . . . .

arcsec 3π
.
Arcsecant is the inverse of secant after restriction to
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
. . . . . .

Check: Values of inverse trigonometric functions
Example
Find
arcsin(1/2)
arctan(−1)
( √ )
2
arccos −
2
Solution
π
6
. . . . . .

Check: Values of inverse trigonometric functions
Example
Find
arcsin(1/2)
arctan(−1)
( √ )
2
arccos −
2
Solution
π
6
π
−
4
. . . . . .

Check: Values of inverse trigonometric functions
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.
1
I use csc x for and arcsin x for the inverse of sin x.
sin x
. . . . . .

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

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)
. . . . . .

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)
To simplify, look at a right
triangle:
.
. . . . . .

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)
To simplify, look at a right
triangle:
1
.
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)
To simplify, look at a right
triangle:
1
.
x
.
y
. = arcsin 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)
To simplify, look at a right
triangle:
1
.
x
.
y
. = arcsin x
. √
. 1 − x2
. . . . . .

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)
To simplify, look at a right
triangle:
√
cos(arcsin x) = 1 − x2 1
.
x
.
y
. = arcsin x
. √
. 1 − x2
. . . . . .

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)
To simplify, look at a right
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
. . rcsin
a
.
| . .
|
−
. 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)
To simplify, look at a right
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
a
. rccos
a
. rcsin
.
| . .
|
−
. 1 1
.
. . . . . .

Graphing arcsin and arccos
a
. rccos
Note
(π )
cos θ = sin −θ
a
. rcsin 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
. . . . . .

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
To simplify, look at a right
triangle:
.
. . . . . .

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
To simplify, look at a right
triangle:
x
.
.
1
.
. . . . . .

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
To simplify, look at a right
triangle:
x
.
y
. = arctan x
.
1
.
. . . . . .

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
To simplify, look at a right
triangle:
√
. 1 + x2 x
.
y
. = arctan x
.
1
.
. . . . . .

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
To simplify, look at a right
triangle:
1
cos(arctan x) = √
1 + x2 √
. 1 + x2 x
.
y
. = arctan x
.
1
.
. . . . . .

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
To simplify, look at a right
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
x
.
1 + x2
−
. π/2
. . . . . .

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
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))
. . . . . .

The derivative of arcsec
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))
To simplify, look at a right
triangle:
.
. . . . . .

The derivative of arcsec
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))
To simplify, look at a right
triangle:
.
. . . . . .

The derivative of arcsec
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))
To simplify, look at a right
triangle:
x
.
.
1
.
. . . . . .

The derivative of arcsec
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))
To simplify, look at a right
triangle:
x
.
y
. = arcsec x
.
1
.
. . . . . .

The derivative of arcsec
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))
To simplify, look at a right
triangle:
√
x2 − 1
tan(arcsec x) = √
1 x
. . x2 − 1
y
. = arcsec x
.
1
.
. . . . . .

The derivative of arcsec
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))
To simplify, look at a right
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
. . . . . .

Outline
Inverse Trigonometric Functions
Derivatives of Inverse Trigonometric Functions
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?
. . . . . .

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
. . . . . .

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.
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
. . . . . .

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.
We have θ = arctan(y/2).
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
. . . . . .

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.
We have θ = arctan(y/2).
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 Remarkable that the
arccos x − √
1 − x2 derivatives of these
1
arctan x transcendental functions
1 + x2 are algebraic (or even
1
arccot x − rational!)
1 + x2
1
arcsec x √
x x2 − 1
1
arccsc x − √
x x2 − 1
. . . . . .