Inverse Trig Functions
c Math Support Center Capsule
A
February 12, 2009
Introduction
Just as trig functions arise in many applications, so do the inverse trig functions. What
may be most surprising is that they are useful not only in the calculation of angles given
the lengths of the sides of a right triangle, but they also give us solutions to some common
integrals. For example, suppose you need to evaluate the following integral:
Z
a
b
√
1
dx
1 − x2
for some appropriate values of a and b. You can use the inverse sine function to solve
it! In this capsule we do not attempt to derive the formulas that we will use; you should
look at your textbook for derivations and complete explanations. This material will simply
summarize the key results and go through some examples of how to use them. As usual,
all angles used here are in radians.
Restrictions on the Domains of the Trig Functions
A function must be one-to-one for it to have an inverse. As we are sure you know, the
trig functions are not one-to-one and in fact they are periodic (i.e. their values repeat
themselves periodically). So in order to define inverse functions we need to restrict the
domain of each trig function to a region in which it is one-to-one but also attains all of
its values. We do this by selecting a specific period for each function and using this as a
domain on which an inverse can be defined. Clearly there are an infinite number of different
restrictions we could chose but the following are choices that are normally used.
1
Standard Restricted Domains
Function
Domain
Range
sin(x)
π
[ −π
2 , 2]
[−1, 1]
cos(x)
[0, π]
[−1, 1]
tan(x)
π
( −π
2 , 2)
(−∞, ∞)
cot(x)
(0, π)
(−∞, ∞)
sec(x)
[0, π2 ) ∪ ( π2 , π]
(−∞, −1] ∪ [1, ∞)
csc(x)
[− π2 , 0) ∪ (0, π2 ]
(−∞, −1] ∪ [1, ∞)
Definitions of the Inverse Functions
When the trig functions are restricted to the domains above they become one-to-one functions, so we can define the inverse functions. For the sine function we use the notation
sin−1 (x) or arcsin(x). Both are read “arc sine” . Look carefully at where we have placed
the -1. Written this way it indicates the inverse of the sine function. If, instead, we write
1
(sin(x))−1 we mean the fraction sin(x)
. The other functions are similar.
The following table summarizes the domains and ranges of the inverse trig functions.
Note that for each inverse trig function we have simply swapped the domain and range for
the corresponding trig function.
Standard Restricted Domains
Function
Domain
Range
sin−1 (x)
[−1, 1]
π
[ −π
2 , 2]
cos−1 (x)
[−1, 1]
[0, π]
tan−1 (x)
(−∞, ∞)
π
( −π
2 , 2)
cot−1 (x)
(−∞, ∞)
(0, π)
sec−1 (x)
(−∞, −1] ∪ [1, ∞)
[0, π2 ) ∪ ( π2 , π]
csc−1 (x)
(−∞, −1] ∪ [1, ∞)
[− π2 , 0) ∪ (0, π2 ]
We can now define the inverse functions more clearly. For the arcsin function we define
π
y = sin−1 (x) if −1 ≤ x ≤ 1, y is in [ −π
2 , 2 ], and sin(y) = x
2
Note that this is only defined when x is in the interval [−1, 1]. The other inverse functions
are similarly defined using the corresponding trig functions.
Some Useful Identities
Here are a few identities that you may find helpful.
cos−1 (x) + cos−1 (−x) = π
sin−1 (x) + cos−1 (x) =
π
2
tan−1 (−x) = −tan−1 (x)
Practicing with the Inverse Functions
Example 1: Find the value of tan(sin−1 ( 51 ).
Solution: The best way to solve this sort of problem is to draw a triangle for yourself
using the Pythagorian Theorem.
5
θ
1
√
2 6
Here we use θ for the value of sin−1 ( 51 ). Notice that we labeled the hypotenuse and the
side opposite θ by using the value of the sin of the angle. We then used the Pythagorian
Theorem to get the remaining side. We now have the information that is needed to find
opposite
1
tan(θ). Since tan(θ) = adjacent
, the answer is √124 = 2 √
6
Example 2: Find the value of sin(cos−1 (− 53 )).
Solution: Look at the following picture:
4
S
S
S 5
S
S
S
S
θ
-3
In this picture we let θ = cos−1 (− 35 ). Then 0 ≤ θ ≤ π and cosθ = − 53 . Because cos(θ)
is negative, θ must be in the second quadrant, i.e. π2 ≤ θ ≤ π. Using the Pythagorean
3
√
Theorem and the fact that θ is in the second quadrant we get that sin(θ) =
√
25−9
5
52 −32
5
=
4
5.
=
Note that although θ does not lie in the restricted domain we used to define
the arcsin function, the unrestricted sin function is defined in the second quadrant and so
we are free to use this fact.
Derivatives of Inverse Trig Functions
The derivatives of the inverse trig functions are shown in the following table.
Function
Derivatives
Derivative
sin−1 (x)
d
−1
dx (sin x)
=
cos−1 (x)
d
−1
dx (cos x)
1
= − √1−x
, |x| < 1
2
tan−1 (x)
d
−1
dx (tan x)
=
1
1+x2
cot−1 (x)
d
−1
dx (cot x)
=
−1
1+x2
sec−1 (x)
d
−1
dx (sec x)
=
√1
,
|x| x2 −1
|x| > 1
csc−1 (x)
d
−1
dx (csc x)
=
√−1
,
|x| x2 −1
|x| > 1
√ 1
,
1−x2
|x| < 1
In practice we often are interested in calculating the derivatives when the variable x is
replaced by a function u(x). This requires the use of the chain rule. For example,
du
d
du
1
(sin−1 u) = √
= √ dx , |u| < 1
dx
1 − u2 dx
1 − u2
The other functions are handled in a similar way.
Example 1: Find the derivative of y = cos−1 (x3 ) for |x3 | < 1
Solution: Note that |x3 | < 1 if and only if |x| < 1, so the derivative is defined whenever
|x| < 1.
4
d
(cos−1 (x3 )) = − q
dx
= −q
1
1 − (x3 )2
1
1−
(x3 )2
d 3
(x )
dx
(3x2 )
3x2
1 − x6
√
Example 2: Find the derivative of y = tan−1 ( 3x).
= −√
Solution:
√
d
1
d √
(tan−1 ( 3x)) =
( 3x)
√ 2 dx
dx
1 + ( 3x)
1
1
=
3
√ 2 √
2 3x
1 + ( 3x)
3
= √
2 3x (1 + 3x)
Exercise 1: For each of the following, find the derivative of the given function with respect
to the independent variable.
(a) y = tan−1 t4
(b) z = t cot−1 (1 + t2 )
√
(c) x = sin−1 1 − t4
+ cos−1 t
√
(e) y = sin−1 x
(d) s =
√ t
1−t2
y
(f) z = cot−1 ( 1−y
2)
5
Solutions:
(a) y = tan−1 t4
dy
d
= tan−1 (t4 )
dt
dt
d 4
1
=
2 dt (t )
4
1 + (t )
4t3
=
1 + t8
(b) z = t cot−1 (1 + t2 )
dz
d
=
t cot−1 (1 + t2 )
dt
dt
= cot−1 (1 + t2 ) + t −1
1 + (1 + t2 )2
2t2
= cot−1 (1 + t2 ) − 4
t + 2t2 + 2
(2t)
√
(c) x = sin−1 1 − t4
p
dx
d
=
sin−1 1 − t4
dt
dt
d p
1
( 1 − t4 )
=q
√
2
dt
4
1−( 1−t )
−1
1
1
(1 − t4 ) 2 (−4t3 )
=p
1 − (1 − t4 ) 2
1
1
=√
√
(−2t3 )
4
4
1−1+t
1−t
1
1
= 2 √
(−2t3 )
t
1 − t4
−2t
=√
1 − t4
6
(d) s =
√ t
1−t2
+ cos−1 t
ds
t
d
d
√
=
+
cos−1 t
dt
dt 1 − t2
dt
−1
√
( 1 − t2 ) 1 − t 21 (1 − t2 ) 2 (−2t)
−1
√
=
+√
( 1 − t2 )2
1 − t2
√
2
t
1 − t2 + √1−t
2
1
− √
=
(1 − t2 )
1 − t2
√
√
2
( 1 − t2 )( 1 − t2 ) + t
(1 − t2 )
1
√
=
−
√
2
(1 − t ) ( 1 − t2 )
( 1 − t2 )(1 − t2 )
2
2
2
(1 − t ) + t − (1 − t )
√
=
( 1 − t2 )(1 − t2 )
t2
=
3
(1 − t2 ) 2
√
(e) y = sin−1 x
√
dy
d
=
sin−1 x
dx
dx
1
d √
=p
x
√ 2 dx
1 − ( x)
1 −1
1
x2
=√
1−x 2
1
= p
2 x(1 − x)
7
y
(f) z = cot−1 ( 1−y
2)
dz
dy
=
y
d
−1
dy cot ( 1−y 2 )
=
1+(
=
1
−1
y
d
dy ( 1−y 2 )
−1
y
d
dy ( 1−y 2 )
y
)2
1−y 2
(1−y 2 )2 +y 2
(1−y 2 )2
=
−(1−y 2 )2
(1−y 2 )2 +y 2
(1−y 2 ) 1 − y (−2y)
(1−y 2 )2
=
−1
(1−y 2 )2 +y 2
(1−y 2 ) 1 − y (−2y)
1
=
−1(1−y 2 +2y 2 )
1−2y 2 +y 4 +y 2
=
−(1+y 2 )
1−y 2 +y 4
Solving Integrals
The formulas listed above for the derivatives lead us to some nice ways to solve some
common integrals. The following is a list of useful ones. These formulas hold for any
constant a 6= 0
= sin−1 ( ua ) + C
R
√ du
a2 −u2
R
du
a2 +u2
R
√ du
u u2 −a2
= a1 tan−1 ( ua ) + C
= a1 sec−1 | ua | + C
for u2 < a2
for all u
for |u| > a > 0
Exercise 2: Verify each of the equations above by taking the derivative of the right hand
side.
We now want to use these formulas to solve some common integrals.
R
dx
Example 1: Evaluate the integral √9−16x
2
Solution: Let a = 3 and u = 4x. Then 16x2 = (4x)2 = u2 and du = 4dx. We get the
following for 16x2 < 9:
8
R
√ dx
9−16x2
=
1
4
R
√ du
a2 −u2
= 14 sin−1 ( ua ) + C
= 14 sin−1 ( 4x
3 )+C
= 14 sin−1 ( 34 x) + C
Exercise 3: Evaluate the following integrals.
R
dx
(a) √25−4x
2
R
dy
(b)
36+4y 2
R z dz
(c)
5+2z 4
R sin x dx
(d) √10−cos2 x
R
dx
(e) √5+4x−x
2
R
7 dx
(f) 25−12x+4x
2
Solutions:
R
R
dx
(a) √25−4x
. For this problem use the formula √adu
= sin−1 ua + C with
2
2 −u2
R
R
dx
= 12 √adu
= 21 sin−1 ( 2x
a = 5, u = 2x and du = 2 dx, giving you √25−4x
2
2 −u2
5 )+C
R
R
dy
(b)
. Use the formula a2du
= a1 tan−1 ( ua ) + C with a = 6, u = 2y
36+4y 2
+u2
and du = 2 dy.
R
R
dy
1
−1 y
This gives us
= 12 a2du
= ( 12 )( 16 ) tan−1 ( 2y
6 ) + C = 12 tan ( 3 ) + C
36+4y 2
+u2
R z dz
(c)
. In order to make the calculations a bit simpler, it is useful to multiply
5+2z 4
the numerator and denominator
2 in order
get the term 4z 4 instead of 2z 4 in the
R z by
R 2z to
dz
dz
denominator. This gives us
=
.
5+2z 4
√10+4z 4
2
Now let u = 2z , du = 4z dz and a = 10 and we have
R z dz
R 4 z dz
R
1
du
2z 2
= 21 10+4z
= 2√110 tan−1 ( √
)+C
√ 2
4 = 2
5+2z 4
2
10
( 10) +u
x dx
(d) √sin
. Let u = cos x, du = −sin x dx and a =
10−cos2 x
R sin x dx
R − sin x dx
√
√
√ x) + C
=
(−1)
= − sin−1 ( cos
2
10
10−cos x
10−cos2 x
R
9
√
10. Then
(e)
R
form
√
dx
.
5+4x−x2
We want to transform this expression into something with the
R
√ du
.
a2 −u2
To do this we need to complete the square of the expression in the
denominator as follows:
5 + 4x − x2
= 5 + 4 − 4 + 4x − x2
= 9 − 4 + 4x − x2
= 9 − (x2 − 4x + 4)
= (3)2 − (x − 2)2
This gives us
R
dx
√
5+4x−x2
=
R
√
dx
(3)2 −(x−2)2
= sin−1 ( x−2
3 ) + C
R
7 dx
(f) 25−12x+4x
2 . Again we need to complete the square. This time we want to transform
R
the expression into something with the form a2du
. We rewrite the denominator as
+u2
follows:
25 − 12x + 4x2
= 16 + 9 − 12x + 4x2
= (4)2 + (2x − 3)2
Now, letting u = 2x − 3 and du = 2 dx we get
R
R
7 dx
2 dx
7
−1 ( 2x−3 ) + C
= 72 (4)2 +(2x−3)
2 = 8 tan
4
25−12x+4x2
10