Section 8 Inverse Trigonometric Functions
Inverse Sine Function
Recall that for every function y = f (x) ; one may de…ne its INVERSE
FUNCTION y = f ¡1 (x) as the unique solution of
x = f (y) :
In other words, the inverse function y = f ¡1 (x) "undo" the function y =
f (x) :
¡
¢
f f ¡1 (x) = x; f ¡1 (f (x)) = x:
We also know that even if a function y = f (x) is de…ned every, its inverse
function y = f ¡1 (x) may not be de…ned everywhere. The su¢cient condition that guarantees existence of inverse functions is called "horizontal line
test": a horizontal line can intersect the graph of y = f (x) at most once.
For trigonometric functions, for instance the graph of y = sin x intersects
horizontal y = 0:6 in…nite many times:
Therefore, to de…ne inverse function of y = sin x; we consider the restricted
sine function
1
h ¼ ¼i
i.e., the restriction of y = sin x on ¡ ;
: This function has the domain
2 2
h ¼ ¼i
¡ ;
and the range [¡1; 1] ; it captures all information of the sine function
2 2
over all real numbers. Notice that the only di¤erence between the restricted
Sine function and the (unrestricted) Sine function is their domains:
h ¼ ¼i
restricted y = sin x de…ned only for x on ¡ ;
2 2
y = sin x de…ned for all real numbers x
h ¼ ¼i
De…nition: The restricted Sine function de…ned on ¡ ;
is invert2 2
ible, and its inverse function is denoted as
y = sin¡1 x or y = arcsin x:
h ¼ ¼i
The domain of y = arcsin x is [¡1; 1] ; and the range y = arcsin x is ¡ ;
:
2 2
The graph of y = arcsin x is the re‡ection of the graph for the restricted Sine
function about y = x :
Example 1 Find the following
Sine function values:
³ pinverse
´
¡1¢
3
(a) arcsin 2 (b) arcsin ¡ 2
2
Solution: (a) By the general de…nition of inverse functions, y = arcsin
is the solution of the restricted Sine function for y :
¡1 ¢
2
1
= sin y
2
The words "restricted Sine" means that
¡
Therefore
¼
y= ;
6
¼
¼
·y· :
2
2
or
µ ¶
1
¼
arcsin
=
2
6
(Notice that the notation "arcsin" comes from the fact that the length of the
arc whose sine is 1/2 is ¼ =6:)
(b) Similarly, we need to solve
p
3
¡
= sin y
2
for
¼
¼
·y· :
2
2
à p !
¼
3
¼
y = ¡ ; or arcsin ¡
=¡
3
2
3
¡
The solution is
Recall that by the de…nition of inverse function,
¡
¢
¡
¢
f f ¡1 (x) = x for x in D f ¡1
f ¡1 (f (x)) = x for x in D (f)
Therefore,
sin (arcsin x) = x for x in [¡1; 1]
h ¼ ¼i
arcsin (sin x) = x for x in ¡ ;
2 2
Example 2 Find following values:
3
³ ³ ¼ ´´
³ ³ ¼ ´´
(a) arcsin sin
= sin¡1 sin
=
µ µ4 ¶¶
µ 4µ ¶¶
3¼
3¼
(b) arcsin sin
= sin¡1 sin
=
µ µ 4 ¶¶
µ µ 4 ¶¶
7¼
7¼
(c) arcsin sin
= sin¡1 sin
=
6
6
£
¤
¼
Solution: (a) According to the formula above, since x = is in ¡ ¼2 ; ¼2
4
³ ³ ¼ ´´ ¼
arcsin sin
=
4
4
£
¤
3¼
(b) This time, since x =
is not in ¡ ¼2 ; ¼2 ; the above formula that
4
would have lead to
µ µ ¶¶
3¼
3¼
arcsin sin
=
doesn’t apply.
4
4
£
¤
We thus solve this problem, we need to …nd an angle x in ¡ ¼2 ; ¼2 such that
µ ¶
3¼
sin x = sin
:
4
This can be easily done as follows:
µ ¶
³
3¼
¼´
¼
sin
= sin ¼ ¡
= sin :
4
4
4
Thus
µ µ ¶¶
³ ³ ¼ ´´ ¼
3¼
arcsin sin
= arcsin sin
= :
4
4
4
7¼
(c) This time again
is outside of the domain of the restricted sine
6
£ ¼ ¼¤
function ¡ 2 ; 2 : according to the method in (b),
µ ¶
³
³ ¼´
7¼
¼´
¼
sin
= sin ¼ +
= ¡ sin = sin ¡
:
6
6
6
6
£
¤
¼
Now since ¡ is in ¡ ¼2 ; ¼2 ;
6
µ µ ¶¶
³ ³ ¼ ´´
7¼
¼
arcsin sin
= arcsin sin ¡
=¡ :
6
6
6
4
Example 3 Find the following values
µ
µ ¶¶
µ
µ ¶¶
2
2
¡1
(a) sin arcsin
= sin sin
=
3
3
¡
¢
(b) sin (arcsin 2) = sin sin¡1 (2) =
2
Solution: (a) Obviously, x = is in the domain of arcsin x; [¡1; 1] :
3
µ
µ ¶¶
2
2
sin arcsin
=
3
3
(b) x = 2 is not in the [¡1; 1] ; arcsin 2 is unde…ned. So sin (arcsin 2) is
not de…ned.
¡
¢
¡
¢
Example 4 (a) What is cos sin¡1 x ? (b) What is sin 2 sin¡1 x ?
Solution: (a) Set µ = sin¡1 x which is de…ned for any ¡1 · x · 1: This
means
¼
¼
sin µ = x and ¡ · µ ·
2
2
Thus cos µ ¸ 0; and
p
p
cos µ = 1 ¡ sin2 µ = 1 ¡ x2 :
(b)
p
sin 2µ = 2 sin µ cos µ = 2x 1 ¡ x2
Inverse Cosine function
For y = cos x
5
we de…ne the restricted Cosine function as
y = cos x for x in [0; ¼ ]
so that its graph is the piece in [0; ¼ ]
Apparently, the restricted Cosine function passes the horizontal line test
and thus is invertible. We call the inverse function of the restricted Cosine
function inverse Cosine and is denoted by
y = cos¡1 x or y = arccos x:
Analogous to the inverse sine function, there are some basic facts for y =
cos¡1 x :
y = cos¡1 x has domain [¡1; 1] and range [0; ¼]
6
¡
¢
cos cos¡1 x = x for x in [¡1; 1]
cos¡1 (cos x) = x for x in [0; ¼ ]
Example 5 Find the values:
µ ¶
1
¡1
¡1
(a) cos (0) ; (b) cos
¡
2
¡1
Solution: (a) µ = cos (0) is the angle in [0; ¼] satisfying
cos µ = 0 =) µ =
¼
2
µ ¶
1
(b) µ = cos¡1 ¡
is the solution of
2
1
cos µ = ¡ ; µ in [0; ¼]
2
Since cos
¼
1
= ;
3
2
cos
So
³
2¼
¼´
¼
1
= cos ¼ ¡
= ¡ cos = ¡ :
3
3
3
2
¡1
µ = cos
µ ¶
1
2¼
¡
=
2
3
Example 6 compute
µ
¶
¡1 2
(a) cos cos
; (b) arccos (cos 4)
3
2
Solution: (a) Since x = is in the domain of arccos x;
3
µ
¶
2
¡1 2
cos cos
=
3
3
(b) Note that x = 4 > ¼ ' 3:14; which is beyond the domain of the
restricted cosine. So the formula
arccos (cos x) = x doesn’t apply.
7
we need to …nd an angle µ in [0; ¼] such that cos µ = cos 4: To this end
cos 4 = cos (2¼ ¡ 4) ; µ = 2¼ ¡ 4 is in [0; ¼]
and
arccos (cos 4) = arccos (cos µ) = µ = 2¼ ¡ 4
Example 7 (a) What is sin (cos¡1 x)?(b) What is sin¡1 x + cos¡1 x?
Solution: (a) Set µ = cos¡1 x so µ is in [0; ¼ ] where sin µ > 0: Thus
p
¡
¢
sin cos¡1 x = sin µ = 1 ¡ cos2 µ
p
= 1 ¡ x2 :
(b) Set ® = sin¡1 x: Then x = sin ®: Draw a right triangle whose one
angle is ®, hypotenuse = 1; and the opposite side is a = x: Then this side x
¼
is adjacent to the third angle in the triangle, i.e., ¯ = ¡®; and cos ¯ = x=1;
2
or ¯ = cos¡1 x: We conclude
sin¡1 x + cos¡1 x = ®+ ¯ =
Inverse function for Tangent Function
Since the graph of y = tan x is
and the period is ¼ : So one entire period
8
¼
2
passes the horizontal
and thus invertible. The inverse function of this
³ test
¼ ¼´
restricted piece in ¡ ;
is denoted as
2 2
y = tan¡1 x or y = arctan x:
Some basic facts:
y = tan¡1 x has the domain (¡1; 1) and range
³ ¼ ¼´
¡ ;
2 2
³ ¼ ¼´
tan (tan x) = x only for x in ¡ ;
2 2
¡
¢
tan tan¡1 x = x for all x:
¡1
p ¢
¡
Example 8 Find (a) tan¡1 (¡1) ; (b) tan tan¡1 5
9
³ ¼´
¼
Solution: (a) Since tan ¡
= ¡1; tan¡1 (¡1) = ¡
4
p ¢ p 4
¡
(b) tan tan¡1 5 = 5
Example 9 (a) Given tan µ = x=2; 0 < µ < ¼ =2: Express tan 2µ as a function of x:
(b) Simplify sec (tan¡1 x)
Solution: (a) Using double-angle formula,
³x´
2
2 tan µ
4x
tan 2µ =
=
³2x ´2 =
2
1 ¡ tan µ
4 ¡ x2
1¡
2
(b) Set µ = tan¡1 x: x = tan µ:
p
p
¡
¢
sec tan¡1 x = sec x = 1 + tan2 µ = 1 + x2
Homework:
In Exercise 1-10, evaluate each of the quantities that is de…ned. If a
quantity is unde…ned, explain it.
Ãp !
3
1. sin¡1
=
2
2. cos¡1 (¡1) =
p
3. tan¡1 3 =
µ
µ ¶¶
5
¡1
4. sin sin
¡
=
4
µ µ ¶¶
4¼
¡1
5. cos
cos
=
5
³
³ ¼ ´´
6. tan¡1 tan ¡
=
5
µ
¶
4¼
¡1
7. sin
sin
=
3
10
µ
µ ¶¶
3
8. cos cos
=
7
µ
µ ¶¶
3¼
¡1
9. sin sin
=
2
¡1
10. cos (cos¡1 2) =
In Exercise 11-15, …nd and simplify the exact value of each quantity.
µ
µ ¶¶
4
¡1
11. tan sin
5
µ
µ ¶¶
4
12. cos arcsin
9
13. sin (tan¡1 2)
µ
µ ¶¶
1
¡1
14. sin cos
4
µ
¶
¡1 4
15. cos tan
3
Simplify the following expressions.
16. tan (cos¡1 x)
17. sin (tan¡1 x)
¡
¢
18. cos sin¡1 x + cos¡1 x
11