Download Inverse Trigonometric Functions

1.3. Inverse Trigonometric Functions
In this section we look at the derivatives of the inverse trigonometric functions and the corresponding
integrals. The inverse trigonometric functions are the inverses of the regular trigonometric functions.
Restrictions are placed on the trigonometric functions to make the one-to-one. Here is a summary followed
by more details on each one.

y

(1)
y = sin-1 x = that number y such that sin y = x and -
(2)
y = cos-1 x = that number y such that cos y = x and 0  y  
(3)
y = tan-1 x = that number y such that tan y = x and -
(4)
y = cot-1 x = that number y such that cot y = x and 0 < y < 
(5)
y = sec-1 x = that number y such that sec y = x and 0  y <
(6)
y = csc-1 x = that number y such that csc y = x and 0 < y 
2

2
2
<y<

2

2

2
or -   y <
3
2
or -  < y 
3
2


Inverse Sine. y = sin-1x = arcsin x is the inverse function to x = sin y with the restriction - 2  y  2. In
other words

1

y = sin-1 x is that angle y such that - 2  y  2 and sin y = x
(7)

y

Here is the graph of x = sin y with the part where - 2  y  2 shown in red. Also is a table of values of
y = sin-1x and its graph.
x
sin y
y
sin 1 x
1.5

1
1.0
0.5
y = sin-1x
0
x
0
x
1.0
2
0.5
3
2
1
1
2
3
-1
y
1.0
0.5
0.5
1.0
1.5
-
2
2
-
Since x = sin y is an odd function, it follows that y = sin-1x is also, i.e.
sin-1(- x) = - sin-1x
Inverse Cosine. y = cos-1x = arccos x is the inverse function to x = cos y with
the restriction 0  y  . In other words
1
y
x
1.3 - 1
2
4
1.0
1.0


2
2
0.5
0.5
-

4
 1.57
 - 1.57
 0.78
 - 0.78
x
y = cos-1 x is that angle y such that 0  y   and cos y = x
(9)
Here is the graph of x = cos y with the part where 0  y   shown in red. Also a table of values of
y = cos-1x and its graph
x
cos y
y
x
1.0
cos
1
y = cos-1x
x
-1
x
3.0

3
2
2
2.5
0.5
2
2.0
y
1
1
2
3
1.0
0.5
1.0
0.5
2

2
2
0.5
1.0

0
1.5
4
0.5
1.0
4
1
 1.57
 0.78
0
There is a connection between cos-1 and sin-1.
Proposition 1.
cos-1(- x) =  - cos-1x

cos-1x = 2 - sin-1x
Proof. To show the first let y = cos-1x and z = cos-1(- x). Then cos y = x and cos z = - x. We need to show z
=  - y. However, cos( - y) = - cos y = - x. So z =  - y which was to be shown. To show the secind first
consider the case where 0  x  1. If we let y = cos-1x and z = sin-1x then cos y = x and sin z = x. We need



to show y = 2 - z. However, cos(2 - z) = sin z = x. So y = 2 - z which was to be shown. Now consider the



case – 1  x  0. Then 0  - x  1. So cos x =  - cos-1(- x) =  - (2 - sin-1(- x)) =  - (2 + sin-1(x)) = 2
- sin-1x. //


Inverse Tangent. y = tan-1x = arctan x is the inverse function to x = tan y with the restriction - 2 < y < 2. In
other words


y = tan-1 x is that angle y such that - 2 < y < 2 and tan y = x
Here is the derivative of y = sin-1x and the corresponding integral.
Inverse Cotangent. y = cot-1x = arccot x is the inverse function to x = tan y with the restriction 0 < y < .
In other words
y = cot-1 x is that angle y such that 0 < y <  and cot y = x
1.1 - 2
Inverse Secant. y = sec-1x = arcsec x is the inverse function to x = sec y where y lies in one of the two
3

intervals 0  y < 2 and   y < 2 In other words

3
y = sec-1 x is that angle y such that 0  y < 2 or   y < 2 and sec y = x
Inverse Cosecant. y = csc-1x = arccsc x is the inverse function to x = csc y where y lies in one of the two
3

intervals 0  y < 2 and   y < 2 In other words

3
y = csc-1 x is that angle y such that 0  y < 2 or   y < 2 and
csc y = x
(10)
d
sin-1 x =
dx
Here is a summary of the derivatives of the inverse trigonometric
(11)
d
cos-1 x
dx
functions. Let's see why some of
d
tan-1 x
dx
d
cot-1 x
dx
d
sec-1 x
dx
(12)
Proposition 1. Show that
(13)
d
sin-1x) =
dx (
1
1 - x2
d
cos-1x) =
dx (
-1
1 - x2
(14)
d
csc-1 x
dx
(15)
1
1 - x2
-1
=
1 - x2
1
=
1 + x2
-1
=
1 - x2
1
=
x x2 - 1
-1
=
x x2 - 1
Proof. Let y = sin-1x so that x = sin y. Use implicit differentiation. Take
the derivative of both sides of this last formula with respect to x to get 1 = (cos y)
1
=
1 - sin2y

2
- sin-1x.
dy
dy
1
. Thus =
=
dx
dx cos y
1
which proves the sin-1 formula. The cos-1 follows from the sin-1 formula and cos-1x =
1 - x2
//
(16)
d
sin-1 u(x) =
dx
(17)
d
cos-1 u(x)
dx
Using the chain rule we get the following analogues of
(10) – (15) when x is replaced by an expression involving
x.
(18)
Example 1. Find
dy
if y = sin-1(3x – 4)
dx
(19)
dy d
= sin-1(3x – 4) =
dx dx
1
3
 d (3x - 4) =
.
2 dx

1 - (3x - 4)
1 - (3x - 4)2
Solution. Using (16) one has
Example 2. Find
dy
if y = sin-1( sin x)
dx
1.1 - 3
(20)
(21)
d
tan-1 u(x)
dx
d
cot-1 u(x)
dx
d
sec-1 u(x)
dx
d
csc-1 u(x)
dx
1
du
2 dx
1 - u(x)
-1
du
=
2 dx
1 - u(x)
1
du
=
1 + u(x)2 dx
- 1 du
=
1 - u(x)2 dx
1
du
=
2
u(x) u(x) - 1 dx
-1
du
=
u(x) u(x)2 - 1 dx
Solution. Using (16) one has
 d (sin x)1/2 =
dx

1 - ( sin x) 
dy d
= sin-1( sin x) =
dx dx
1
2
cos x
 1  (1/2) (sin x)-1/2  d (sin x) =
.
dx


2 1 - sin x sin x
 1 - sin x
Example 3. Find
dy
1 + 2cos x
if y = cos-1
dx
 2 + cos x 
Solution. Using (17) one has
dy
d
1 + 2cos x
=
cos-1
dx
dx
 2 + cos x  =
=
-1
 d 1 + 2cos x
2
1 + 2cos x dx  2 + cos x 
1-
 2 + cos x 
(2 + cos x) d (1 + 2cos x) - (1 + 2cos x) d (2 + cos x)
dx
dx
- (2 + cos x)


2
2
2
(2
+
cos
x)

(2 + cos x) - (1 + 2cos x)
=
- (2 + cos x)
(2 + cos x)(- 2sin x) - (1 + 2cos x)(- sin x)
2
2 
(2 + cos x)2

4 + 4cos x + cos x - 1 - 4cos x - 4cos x
=
-1
- 4 sin x - 2sin x cos x + sin x + 2sin x cos x
2 
2 + cos x

3 - 3cos x
=
3 sin x
=
3 - 3cos2 x (2 + cos x)
3 sin x
3
=
2
+
cos
x
1 - cos x (2 + cos x)
2
For the last formula we are assuming x lies in an interval where sin x is non-negative. If x lies in an interval
- 3
in which sin x is negative then the answer would be
.
2 + cos x
1
-1
(22)

 1 - x2 dx = sin x
Each of the derivative formulas (10) – (15) has a corresponding
1
(23)
 1 + x2 dx = tan-1 x

integration formula. However the integration formulas for the co
1
functions don't give new integration formulas.
-1
(24)

 x x2 - 1 dx = sec x
Example 4. Find 

Solution.
dx =
1
dx
6 - 2x2
1
1 
 1 dx = 
dx =
2
2

x
6
 6 - 2x
 6 1- 3



3 du and the integral becomes
1 

6
1
1-

2
x 
3
dx. Let u =
x
dx
. Then du =
and
3
3
3
1
1
x
du =
sin-1u =
sin-1 .
2
1-u
2
2
 3
1.1 - 4