Lesson 16: Inverse Trigonometric Functions (Section 041 handout)

Section 3.5
Inverse Trigonometric Functions
V63.0121.041, Calculus I
New York University
November 1, 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
V63.0121.041, Calculus I (NYU) Section 3.5 Inverse Trigonometric Functions November 1, 2010 2 / 31
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.
Notes
Notes
Notes
1
Section 3.5 : Inverse Trigonometric FunctionsV63.0121.041, Calculus I November 1, 2010

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
Derivatives of Inverse Trigonometric Functions
Arcsine
Arccosine
Arctangent
Arcsecant
Applications
Notes
Notes
Notes
2

arcsin
Arcsin is the inverse of the sine function after restriction to [−π/2, π/2].
x
y
sin
−
π
2
π
2
y = x
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
y = x
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
y = x
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
Notes
Notes
Notes
3

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
y = x
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
Solution
π
6
Notes
Notes
Notes
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))
Upshot: Many times the derivative of f −1
(x) can be found by implicit
differentiation and the derivative of f :
Notes
Notes
Notes
5

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)
To simplify, look at a right
triangle:
cos(arcsin x) = 1 − x2
So
d
dx
arcsin(x) =
1
√
1 − x2
1
x
y = arcsin 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).
Solution
Notes
Notes
Notes
6

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
d
dx
arccos(x) = −
1
√
1 − x2
1
1 − x2
x
y = arccos x
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)
triangle:
cos(arctan x) =
1
√
1 + x2
So
d
dx
arctan(x) =
1
1 + x2
x
1
y = arctan x
1 + x2
Notes
Notes
Notes
7

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).
Solution
Derivation: The derivative of arcsec
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))
triangle:
tan(arcsec x) =
√
x2 − 1
1
So
d
dx
arcsec(x) =
1
x
√
x2 − 1
x
1
y = arcsec x
x2 − 1
Notes
Notes
Notes
8

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?
Notes
Notes
Notes
9

Solution
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!)
Notes
Notes
Notes
10

Lesson 16: Inverse Trigonometric Functions (Section 041 handout)

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Lesson 16: Inverse Trigonometric Functions (Section 041 handout)