Download 3.6 Inverse Trigonometric Functions

3.6
Inverse Trigonometric Functions
Inverse functions
Let f be a function. We write y = f (x) if f has the value y at x, where x is in the
domain of f (these are the possible input values for f ). The range of f are the possible
output values y of f .
Example: f (x) = x2 . The domain is the set R = (−∞, ∞) of real numbers, the range is
the interval [0, ∞) of non-negative real numbers.
The graph of f is the set {(x, y) ∈ R2 | y = f (x)} of points in the coordinate plane.
Example: f (x) = x2 .
(Sketch of graph of f (x) = x2 )
We like to reverse the correspondence given by f , that is given u we like to find v with
u = f (v). To achieve this uniquely, u must be in the range of f and f must be one-to-one
(also called: injective), i.e., if a 6= b are in the domain of f then f (a) 6= f (b).
In this case, we write v = f −1 (u) for the uniquely determined v and call f −1 the
inverse function of f .
Theorem 1 (Relationships between f −1 and f ) Let f be an injective function. Then
(1) y = f −1 (x) if and only if x = f (y), where x is in the domain of f −1 and y is in the
domain of f .
(2) Domain of f −1 = range of f .
(3) Range of f −1 = domain of f .
(4) f (f −1 (x)) = x for every x in the domain of f −1 .
(5) f −1 (f (y)) = y for every y in the domain of f .
(6) The point (a, b) is on the graph of f −1 if and only if the point (b, a) is on the graph
of f .
(7) The graphs of f −1 and f are reflections of each other through the line y = x.
Example: f (x) = x2 with the domain restricted to [0, ∞).
1
(Sketch of graph of f (x) = x2 for x ∈ [0, ∞) and the corresponding inverse function
√
= x)
f −1 (x)
The range of f is still [0, ∞). f is now one-to-one. The inverse function is given by
√
f −1 (x) = x with domain and range also [0, ∞).
We have for x, y ≥ 0:
√
(1) y = x ⇐⇒ x = y 2 ,
√
(4) ( x)2 = x.
(5)
p
y 2 = y.
The relation between the graph of f (x) = x2 and f −1 (x) =
of the theorem can be verified in the above figure.
√
x as stated in part (7)
The sine function is not injective, but it becomes so if we restrict it to the interval
because it is there strictly increasing. Its range is then still the interval [−1, 1].
[− π2 , π2 ]
(Sketch of the graph of f (x) = sin x with highlighting of the rescricted part)
Definition 1 (Inverse sine function)
The inverse of the sine function is called the arcsine function and denoted by arcsin (or
sin−1 ). It is characterized y = arcsin x if and only if x = sin y for x in [−1, 1] (its domain)
and y in [− π2 , π2 ] (its range).
The graph of the arcsine function is obtained by reflecting the restricted sine function
along the diagonal line y = x.
(Sketch of the graph of f (x) = arcsin x
The theorem on inverse functions gives the following two formulas:
sin (arcsin x) = sin (sin−1 x) = x
for x in [−1, 1],
arcsin (sin y)) = sin−1 (sin y) = y
for y in [− π2 , π2 ].
2
Problem 1 Find the exact values of
1
(a) sin (arcsin ),
2
(c) sin−1 (sin
π
(b) arcsin (sin ),
4
2π
),
3
(d) sin−1 (tan
3π
).
4
Solution: (a) Using the special values of the sine function and the special values for the
arcsine function (which are obtained from the previous ones) we get
π
1
1
sin (arcsin ) = sin( ) = .
2
6
2
Another solution is the following: Since 12 is in [−1, 1] we are allowed to use the above
inverse formulas and we obtain directly sin (arcsin 12 ) = 21 .
(b) Since
π
4
is in [− π2 , π2 ], one gets directly arcsin (sin π4 ) = π4 .
(c) One has to be careful since
compute
2π
3
is not in [− π2 , π2 ]. Using special values one can
√
π
2π
3
−1
= .
sin (sin ) = sin
3
2
3
Another solution without knowledge of special value uses the reference angle: The reference
2π
π
angle for 2π
3 which is in quadrant II is π − 3 = 3 . Since the sine function is positive in
π
2π
quadrant II one has sin 3 = sin 3 . But the reference angle π3 is in [− π2 , π2 ] and the inverse
formula can now be aplied and gives
−1
sin−1 (sin
2π
π
π
) = sin−1 (sin ) = .
3
3
3
(d) We can use special values:
sin−1 (tan
3π
π
) = sin−1 (−1) = − .
4
2
3