Download 7.6. Inverse Trigonometric Functions Defining the inverses The

Instructor: Prof. Dr. Ayman H. Sakka
7.6. Inverse Trigonometric Functions
Inverse trigonometric functions arise when we want to calculate angles from side measurements
in triangles. They also provide useful antiderivatives and appear frequently in the solutions of
differential equations. This section shows how these functions are defined, graphed, and evaluated,
how their derivatives are computed, and why they appear as important antiderivatives.
Defining the inverses
The six basic trigonometric functions are not one-to-one in their domains, but we can restrict their
domains to intervals on which they are one-to-one.
The arcsine function
The function f (x) = sin x is one-to-one on any one of the intervals [− π2 , π2 ], [ π2 , 3π
], . . . But we choose
2
[− π2 , π2 ] to satisfy the relationship sin−1 (1/x) = csc−1 x.
Definition. ( The arcsine function)
For any x ∈ [−1, 1], y = sin−1 x is the number y ∈ [− π2 , π2 ] for which sin y = x.
Example 1. Find sin−1 (0).
solution
Remarks.
(1) sin−1 x 6=
2.0
1
.
sin x
1.5
1.0
1
(2)
= (sin x)−1 = csc x.
sin x
0.5
0.0
The graph of y = sin−1 x
The domain of sin−1 x is [−1, 1].
The range of sin−1 x is
[− π2 , π2 ].
-2
-1.5
-1
-0.5
0
-0.5
-1.0
-1.5
-2.0
0.5
1
1.5
2
Remark. The graph of y = sin−1 x is symmetric about the origin. Therefore y = sin−1 x is an odd
function; that is
sin−1 (−x) = − sin−1 x.
Example 1. Find sin
−1
−1
√ .
2
Solution:
3
−1
Example 2. Find the value of tan sin
.
7
Solution:
Remarks.
(a) sin−1 (sin x) = x for x ∈ [− π2 , π2 ].
(b) sin(sin−1 x) = x for x ∈ [−1, 1].
Example 1. Find sin(sin−1 2) and sin−1 (sin 2π).
Solution:
The arccosine function
The function f (x) = cos x is one-to-one on the interval [0, π].
Definition. ( The arccossine function)
For any x ∈ [−1, 1], y = cos−1 x is the number y ∈ [0, π] for which cos y = x.
1
−1
Example 1. Find cos
.
2
Solution:
2
3.5
3.0
The graph of y = cos−1 x
The domain of cos−1 x is
The range of cos−1 x is
2.5
[−1, 1].
[0, π].
2.0
1.5
1.0
0.5
0.0
-3
-2
-1
0
-0.5
Remarks.
(a) The function cos−1 x is neither even nor odd.
(b) cos−1 (cos x) = x for x ∈ [0, π].
(c) cos(cos−1 x) = x for x ∈ [−1, 1].
Example 1. Find cos−1 (cos(π/2)) and cos−1 (cos 5π).
Solution:
Identities involving arcsine and arccosine
For all x ∈ [−1, 1], we have
(1) cos−1 (−x) = π − cos−1 x
(2) sin−1 x + cos−1 x =
3
π
2
1
2
3
proof
√
Example 1. Find cos−1 (− 3/2).
Solution:
√
2
Example 2. Find the value of 2log4 π − cos−1 (−1/ 2).
Solution:
Example 3. Solve for x:
ln(e2x + 2ex + 1) = 2 sin[cos−1 (−x) − sin−1 x].
Solution:
4
Inverses of tan x, cot x, sec x, and csc x
Definition. ( The arctangent and arccotangent functions)
(1) For any x ∈ R, y = tan−1 x is the number y ∈ (−π/2, π/2) for which tan y = x.
(2) For any x ∈ R, y = cot−1 x is the number y ∈ (0, π) for which cot y = x.
Remark. The graph of y = tan−1 x is symmetric about the origin. Therefore y = tan−1 x is an odd
function; that is
tan−1 (−x) = − tan−1 x.
Example 1. (Exam)
Simplify the expression
cos (tan−1 (3/x)) .
Solution:
Example 2. Evaluate lim tan−1 x.
x→∞
Solution:
Definition. ( The arcsecant and arccosecant functions)
(1) For any x ∈ (−∞, −1] ∪ [1, ∞), y = sec−1 x is the number y ∈ [0, π/2) ∪ (π/2, π] for which
sec y = x.
(2) For any x ∈ (−∞, −1] ∪ [1, ∞), y = csc−1 x is the number y ∈ [−π/2, 0) ∪ (0, π/2] for which
csc y = x.
Useful Identities
(1) cot−1 (x) = π/2 − tan−1 x.
(2) sec−1 x = cos−1 (1/x).
(3) csc−1 x = sin−1 (1/x).
5
Proof:
Example 1. Find sec−1 2.
Solution:
Example 2. Find
sin cos−1 (1/3) + csc−1 3 .
Solution:
Example 3. Find csc tan−1
√
4 − x2 .
x
Solution:
Example 4. (Exam)
Solve for x:
2x
−1
√
log5 (25) + tan sin
= cos sec−1 x
4x2 + 1
x
Solution:
6
Example 5. Evaluate lim sec−1 x.
x→−∞
Solution:
The derivatives of inverse trigonometric functions
Inverse trigonometric functions provide antiderivatives for a verity of functions.
The derivatives of y=sin−1 x and y=cos−1 x
f (x) = sin x is differentiable on the interval (−π/2, π/2) and f 0 (x) = cos x > 0 in this interval.
Therefore, f −1 (x) = sin−1 x is differentiable on the interval (−1, 1) and
d
1
sin−1 x = √
,
dx
1 − x2
|x| < 1
This implies
Z
√
1
dx = sin−1 x + C
1 − x2
Since cos−1 x = π/2 − sin−1 x, we have
d
d
−1
cos−1 x = − sin−1 x = √
,
dx
dx
1 − x2
Proof:
7
|x| < 1
2
Example 1. Find y 0 if y = sin−1 (ex + 3x).
Solution:
Example 2. Find y 0 if y = 9sin
−1 (3x)
+ cos−1 (x2 ).
Solution:
Z
dx
√
.
25 − x2
Z
dx
√
.
6x − x2
Example 3. Find
Solution:
Example 4. Find
Solution:
Z
Example 5. Find
−1
sin
dy
p
.
y 1 − y2
Solution:
8
Z
1
Example 6. Find
√
1/2
dx
.
3 + 4x − 4x2
Solution: Solution:
The derivatives of y=sec−1 x and y=csc−1 x
Since sec−1 x = cos−1 (1/x), we have
d
−1
−1
d
sec−1 x =
cos−1 (1/x) = √
( 2 ).
dx
dx
1 − x−2 x
Simplifying we get
1
d
sec−1 x = √
,
dx
|x| x2 − 1
|x| > 1
and hence
Z
1
dx = sec−1 |x| + C
x x2 − 1
√
Since csc−1 x = π/2 − sec−1 x, we have
d
d
−1
csc−1 x = − sec−1 x = √
,
dx
dx
|x| x2 − 1
Example 1. Find y 0 if y = sec−1 (3x).
Solution:
9
|x| > 1
Example 2. (Exam)
Find y 0 if y = 3x + sec−1 (3x).
Solution:
Z
Example 3. Find
1
2
dx
.
x 4x2 − 1
√
Solution:
Z
Example 4. Find
dx
√
.
(x − 4) x2 − 8x + 7
Solution:
sec−1 x
Example 5. Find lim+ √
.
x→1
x2 − 1
Solution:
10
The derivatives of y=tan−1 x and y=cot−1 x
The functions tan−1 x and cot−1 x are differentiable for all x ∈ R and
1
d
tan−1 x =
,
dx
1 + x2
d
−1
d
cot−1 x = − tan−1 x = 2
.
dx
dx
x +1
Moreover
Z
1
dx = tan−1 x + C
2
1+x
Proof:
Example 1. Find y 0 if y = log5 (tan−1 (5x)).
Solution:
−1
Example 2. Find y 0 if y = xcot
x
+ sin(tan−1 (x2 )).
Solution:
Z
Example 3. Find
1
2
cot−1 x
dx.
x2 + 1
11
Solution:
Z
Example 4. Find
4x2
dx
.
+ 10x + 7
Solution:
Z
Example 5. Find
x2 dx
.
1 + x6
Solution:
tan−1 (4x)
.
x→0
x
Example 6. Find lim
Solution:
12