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Lecture 6 : Inverse Trigonometric Functions
Inverse Sine Function (arcsin x = sin−^1 x) The trigonometric function sin x is not one-to-one
functions, hence in order to create an inverse, we must restrict its domain.
The restricted sine function is given by
f (x) =
sin x − π 2 ≤^ x^ ≤^
π 2
undefined otherwise
We have Domain(f) = [− π 2 ,^
π 2 ] and Range(f) = [−^1 ,^ 1].
HΠê6, 1ê 2 L H 5 Πê6, 1ê 2 L
-Π (^) - Π 2 Π 2 Π
y= sin x
y= fHxL
We see from the graph of the restricted sine function (or from its derivative) that the function is
one-to-one and hence has an inverse, shown in red in the diagram below.
HΠê2, 1L
H
-Π 4
,
2
L
H1, Πê 2 L
H
,
-Π 4
L
This inverse function, f − 1 (x), is denoted by
f − 1 (x) = sin − 1 x or arcsin x.
Properties of sin−^1 x.
Domain(sin − 1 ) = [− 1 , 1] and Range(sin − 1 ) = [− π 2 ,^
π 2 ].
Since f − 1 (x) = y if and only if f (y) = x, we have:
sin−^1 x = y if and only if sin(y) = x and −
π
2
≤ y ≤
π
2
Since f (f − 1 )(x) = x f − 1 (f (x)) = x we have:
sin(sin − 1 (x)) = x for x ∈ [− 1 , 1] sin − 1 (sin(x)) = x for x ∈
[
π
2
π
2
]
from the graph: sin − 1 x is an odd function and sin − 1 (−x) = − sin − 1 x.
Example Evaluate sin − 1
√−^1 2
using the graph above.
Example Evaluate sin − 1 (
3 /2), sin − 1 (−
Example Evaluate sin − 1 (sin π).
Example Evaluate cos(sin − 1 (
Example Give a formula in terms of x for tan(sin − 1 (x))
Derivative of sin−^1 x.
d
dx
sin − 1 x =
1 − x^2
, − 1 ≤ x ≤ 1.
Proof We have sin − 1 x = y if and only if sin y = x. Using implicit differentiation, we get cos y dy dx = 1 or dy
dx
cos y
Now we know that cos^2 y + sin 2 y = 1, hence we have that cos^2 y + x^2 = 1 and
cos y =
1 − x^2
HΠê4, 1L
2
4
6
y= hHxL
H1, Πê 4 L
5 - 4 - 3 - 2 - 1 1 2 3 4 5
Π 2
Π 4
Π 2
y= arctanHxL
Properties of tan−^1 x.
Domain(tan − 1 ) = (−∞, ∞) and Range(tan − 1 ) = (− π 2 ,^
π 2 ).
Since h − 1 (x) = y if and only if h(y) = x, we have:
tan − 1 x = y if and only if tan(y) = x and −
π
2
< y <
π
2
Since h(h − 1 (x)) = x and h − 1 (h(x)) = x, we have:
tan(tan − 1 (x)) = x for x ∈ (−∞, ∞) tan − 1 (tan(x)) = x for x ∈
π
2
π
2
Frpm the graph, we have: tan − 1 (−x) = − tan − 1 (x).
Also, since lim x→( π 2 −)
tan x = ∞ and lim x→(− π 2 +)
tan x = −∞,
we have lim x→∞
tan − 1 x =
π
2
and lim x→−∞
tan − 1 x = −
π
2
Example Find tan − 1 (1) and tan − 1 ( √^1 3
Example Find cos(tan − 1 ( √^1 3
Derivative of tan−^1 x.
d
dx
tan−^1 x =
x^2 + 1
, −∞ < x < ∞.
Proof We have tan − 1 x = y if and only if tan y = x. Using implicit differentiation, we get sec 2 y dy dx = 1 or dy
dx
sec^2 y
= cos 2 y.
Now we know that cos^2 y = cos^2 (tan−^1 x) = 1 1+x^2
. proving the result.
If we use the chain rule in conjunction with the above derivative, we get
d
dx
tan − 1 (k(x)) =
k′(x)
1 + (k(x))^2
, x ∈ Dom(k)
Example Find the domain and derivative of tan−^1 (ln x)
Domain = (0, ∞)
d
dx
tan − 1 (ln x) =
1 x 1 + (ln x)^2
x(1 + (ln x)^2 )
Integration formulas Reversing the derivative formulas above, we get
∫ 1 √ 1 − x^2
dx = sin − 1 x + C,
x^2 + 1
dx = tan − 1 x + C,
Example (^) ∫ 1 √ 9 − x^2
dx =
1 − x
2 9
dx =
1 − x
2 9
dx =
1 − x
2 9
dx
Let u = x 3 , then^ dx^ = 3du^ and ∫ 1 √ 9 − x^2
dx =
1 − u^2
du = sin − 1 u + C = sin − 1 x 3
+ C
Example
∫ (^1) / 2
0
1 + 4x^2
dx
Let u = 2x, then du = 2dx, u(0) = 0, u(1/2) = 1 and
∫ (^1) / 2
0
1 + 4x^2
dx =
0
1 + u^2
dx =
tan − 1 u| 1 0 =
[tan − 1 (1) − tan − 1 (0)]
[
π
4
− 0] =
π
8
Domain(cos − 1 ) = [− 1 , 1] and Range(cos − 1 ) = [0, π].
Recall from the definition of inverse functions:
g − 1 (x) = y if and only if g(y) = x.
cos − 1 x = y if and only if cos(y) = x and 0 ≤ y ≤ π.
g(g − 1 (x)) = x g − 1 (g(x)) = x
cos(cos − 1 (x)) = x for x ∈ [− 1 , 1] cos − 1 (cos(x)) = x for x ∈
[
0 , π
]
Note from the graph that cos − 1 (−x) = π − cos − 1 (x).
cos − 1 (
3 /2) = and cos − 1 (−
You can use either chart below to find the correct angle between 0 and π.:
tan(cos − 1 (
tan(cos−^1 (x)) =
Must draw a triangle with correct proportions:
1
x
cos θ = x
θ
tan(cos-1x) = tan θ =
1-x^2 x
cos-1x = θ
1-x^2
1
x
cos θ = x
θ
d
dx
cos − 1 x =
1 − x^2
, − 1 ≤ x ≤ 1.
Proof We have cos−^1 x = y if and only if cos y = x. Using implicit differentiation, we get − sin y dy dx = 1 or dy
dx
sin y
Now we know that cos^2 y + sin 2 y = 1, hence we have that sin 2 y + x^2 = 1 and
sin y =
1 − x^2
and d
dx
cos − 1 x =
1 − x^2
Note that d dx cos
− 1 x = − d dx sin
− 1 x. In fact we can use this to prove that sin − 1 x + cos − 1 x =
π
2
If we use the chain rule in conjunction with the above derivative, we get
d
dx
cos − 1 (k(x)) =
−k′(x) √ 1 − (k(x))^2
, x ∈ Dom(k) and − 1 ≤ k(x) ≤ 1.
