The Inverse Trigonometric Functions-x

1. A function is one–to–one (1–1) if different inputs
produce different outputs.
The Inverse Trigonometric Functions

2. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
The Inverse Trigonometric Functions

3. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
The Inverse Trigonometric Functions

4. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
u
f(u)=f(v)v
u = v
a non–one–to–one function
The Inverse Trigonometric Functions

5. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
Example A.
a. g(x) = 2x + 1 is 1–1
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
u
f(u)=f(v)v
u = v
a non–one–to–one function
The Inverse Trigonometric Functions

6. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
u
f(u)=f(v)v
u = v
a non–one–to–one function
The Inverse Trigonometric Functions
Example A.
a. g(x) = 2x + 1 is 1–1 because if u  v, then 2u  2v
so 2u + 1  2v + 1

7. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
Example A.
a. g(x) = 2x + 1 is 1–1 because if u  v, then 2u  2v
so 2u + 1  2v + 1 or g(u)  g(v).
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
u
f(u)=f(v)v
u = v
a non–one–to–one function
The Inverse Trigonometric Functions

8. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
u
f(u)=f(v)v
u = v
a non–one–to–one function
The Inverse Trigonometric Functions
Example A.
a. g(x) = 2x + 1 is 1–1 because if u  v, then 2u  2v
so 2u + 1  2v + 1 or g(u)  g(v).
b. f(x) = x2 is not 1–1

9. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
u
f(u)=f(v)v
u = v
a non–one–to–one function
The Inverse Trigonometric Functions
Example A.
a. g(x) = 2x + 1 is 1–1 because if u  v, then 2u  2v
so 2u + 1  2v + 1 or g(u)  g(v).
b. f(x) = x2 is not 1–1 because 3  –3,
but f(3) = f(–3) = 9.

10. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
u
f(u)=f(v)v
u = v
a non–one–to–one function
The Inverse Trigonometric Functions
Example A.
a. g(x) = 2x + 1 is 1–1 because if u  v, then 2u  2v
so 2u + 1  2v + 1 or g(u)  g(v).
b. f(x) = x2 is not 1–1 because 3  –3,
but f(3) = f(–3) = 9.
c. sin(x) is not 1–1 since sin(0) = sin(π) = 0.

11. A function is one–to–one (1–1) if different inputs
produce different outputs. That is, f(x) is 1–1 iff
for every pair of inputs u  v it must be that f(u)  f(v).
u f(u)
v f(v)
u = v f(u) = f(v)
a one–to–one function
u
f(u)=f(v)v
u = v
a non–one–to–one function
The Inverse Trigonometric Functions
Trig–functions are not 1–1.
Example A.
a. g(x) = 2x + 1 is 1–1 because if u  v, then 2u  2v
so 2u + 1  2v + 1 or g(u)  g(v).
b. f(x) = x2 is not 1–1 because 3  –3,
but f(3) = f(–3) = 9.
c. sin(x) is not 1–1 since sin(0) = sin(π) = 0.

12. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
The Inverse Trigonometric Functions

13. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
f(x)
u

14. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f(x)
u

15. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
u

16. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
Specifically if A denotes the domain and B denotes
the range of f(x), then f –1(x) has B as the domain and
A as the range
u

17. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
Specifically if A denotes the domain and B denotes
the range of f(x), then f –1(x) has B as the domain and
A as the range and f –1(x) is 1–1 also.
u

18. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
Specifically if A denotes the domain and B denotes
the range of f(x), then f –1(x) has B as the domain and
A as the range and f –1(x) is 1–1 also.
Furthermore f –1(f(x)) = x
u

19. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
Furthermore f –1(f(x)) = x
x
A
f(x)
B
f
u
Specifically if A denotes the domain and B denotes
the range of f(x), then f –1(x) has B as the domain and
A as the range and f –1(x) is 1–1 also.

20. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
Furthermore f –1(f(x)) = x
x
A
f(x)
B
f
f –1
f –1(f(x)) = x
u
Specifically if A denotes the domain and B denotes
the range of f(x), then f –1(x) has B as the domain and
A as the range and f –1(x) is 1–1 also.

21. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
x
A
f(x)
B
f
f –1
f –1(f(x)) = x
Furthermore f –1(f(x)) = x and f (f –1(x)) = x
u
Specifically if A denotes the domain and B denotes
the range of f(x), then f –1(x) has B as the domain and
A as the range and f –1(x) is 1–1 also.

22. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
x
A
f(x)
B
f
f –1
f –1(f(x)) = x
Furthermore f –1(f(x)) = x and f (f –1(x)) = x
f– 1(x)
A B
f –1
f –1 (x)
x
u
Specifically if A denotes the domain and B denotes
the range of f(x), then f –1(x) has B as the domain and
A as the range and f –1(x) is 1–1 also.

23. The inverse relation of a 1–1 function f(x) is itself a
function and it’s called the inverse function of f(x).
It’s denoted as f –1(x).
f(u)
v f(v)
u = v f(u) = f(v)
f(x) is a one–to–one function
The Inverse Trigonometric Functions
u f(u)
v f(v)
u = v f(u) = f(v)
f –1(x) is a well defined function
f –1(x)f(x)
x
A
f(x)
B
f
f –1
f –1(f(x)) = x
Furthermore f –1(f(x)) = x and f (f –1(x)) = x
A B
f
f –1
f(f –1 (x)) = x
xf– 1(x)
u
Specifically if A denotes the domain and B denotes
the range of f(x), then f –1(x) has B as the domain and
A as the range and f –1(x) is 1–1 also.

24. Some inverse functions may be solved algebraically
from f(x).
The Inverse Trigonometric Functions

25. Some inverse functions may be solved algebraically
from f(x). To do this, solve the equation y = f(x) for the
variable x in terms of y, the solution is x = f–1(y).
The Inverse Trigonometric Functions

26. Some inverse functions may be solved algebraically
from f(x). To do this, solve the equation y = f(x) for the
variable x in terms of y, the solution is x = f–1(y).
However, in general it’s not possible to solve
algebraically for f–1(x).
The Inverse Trigonometric Functions

27. Some inverse functions may be solved algebraically
from f(x). To do this, solve the equation y = f(x) for the
variable x in terms of y, the solution is x = f–1(y).
However, in general it’s not possible to solve
algebraically for f–1(x).
Example B. Solve for f–1(x) given that f(x) = 2x + 1.
The Inverse Trigonometric Functions

28. Some inverse functions may be solved algebraically
from f(x). To do this, solve the equation y = f(x) for the
variable x in terms of y, the solution is x = f–1(y).
However, in general it’s not possible to solve
algebraically for f–1(x).
Example B. Solve for f–1(x) given that f(x) = 2x + 1.
The Inverse Trigonometric Functions
Set y = f(x) = 2x + 1 and solve for x,

29. Some inverse functions may be solved algebraically
from f(x). To do this, solve the equation y = f(x) for the
variable x in terms of y, the solution is x = f–1(y).
However, in general it’s not possible to solve
algebraically for f–1(x).
Example B. Solve for f–1(x) given that f(x) = 2x + 1.
The Inverse Trigonometric Functions
Set y = f(x) = 2x + 1 and solve for x, we get
x = (y – 1)/2 = f–1(y)

30. Some inverse functions may be solved algebraically
from f(x). To do this, solve the equation y = f(x) for the
variable x in terms of y, the solution is x = f–1(y).
However, in general it’s not possible to solve
algebraically for f–1(x).
Example B. Solve for f–1(x) given that f(x) = 2x + 1.
The Inverse Trigonometric Functions
Set y = f(x) = 2x + 1 and solve for x, we get
x = (y – 1)/2 = f–1(y)
Switching to the variable x we have f–1(x) = (x – 1)/2.

31. Some inverse functions may be solved algebraically
from f(x). To do this, solve the equation y = f(x) for the
variable x in terms of y, the solution is x = f–1(y).
However, in general it’s not possible to solve
algebraically for f–1(x).
Example B. Solve for f–1(x) given that f(x) = 2x + 1.
The Inverse Trigonometric Functions
Set y = f(x) = 2x + 1 and solve for x, we get
x = (y – 1)/2 = f–1(y)
If a function f with domain A is not 1–1,
we may downsize the domain A so that f is 1–1 in the
new domain.
Switching to the variable x we have f–1(x) = (x – 1)/2.

32. Some inverse functions may be solved algebraically
from f(x). To do this, solve the equation y = f(x) for the
variable x in terms of y, the solution is x = f–1(y).
However, in general it’s not possible to solve
algebraically for f–1(x).
Example B. Solve for f–1(x) given that f(x) = 2x + 1.
The Inverse Trigonometric Functions
Set y = f(x) = 2x + 1 and solve for x, we get
x = (y – 1)/2 = f–1(y)
If a function f with domain A is not 1–1,
we may downsize the domain A so that f is 1–1 in the
new domain. We then may talk about f and f–1 in
relation to this new domain.
Switching to the variable x we have f–1(x) = (x – 1)/2.

33. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R.
The Inverse Trigonometric Functions

34. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R. However, if we set the domain to be
A = {x ≥ 0} (non–negative numbers) then the function
g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}.
The Inverse Trigonometric Functions

35. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R. However, if we set the domain to be
A = {x ≥ 0} (non–negative numbers) then the function
g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}.
The Inverse Trigonometric Functions
Hence g–1 exists.

36. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R. However, if we set the domain to be
A = {x ≥ 0} (non–negative numbers) then the function
g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}.
The Inverse Trigonometric Functions
Hence g–1 exists. To find it,
set y = g(x) = x2,

37. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R. However, if we set the domain to be
A = {x ≥ 0} (non–negative numbers) then the function
g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}.
The Inverse Trigonometric Functions
Hence g–1 exists. To find it,
set y = g(x) = x2, solve for x
and get x = ±√y.

38. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R. However, if we set the domain to be
A = {x ≥ 0} (non–negative numbers) then the function
g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}.
The Inverse Trigonometric Functions
Hence g–1 exists. To find it,
set y = g(x) = x2, solve for x
and get x = ±√y.
Since x is non–negative we
must have
x = √y = g–1(y)

39. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R. However, if we set the domain to be
A = {x ≥ 0} (non–negative numbers) then the function
g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}.
The Inverse Trigonometric Functions
Hence g–1 exists. To find it,
set y = g(x) = x2, solve for x
and get x = ±√y.
Since x is non–negative we
must have
x = √y = g–1(y) or that
g–1(x) = √x.

40. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R. However, if we set the domain to be
A = {x ≥ 0} (non–negative numbers) then the function
g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}.
The Inverse Trigonometric Functions
Hence g–1 exists. To find it,
set y = g(x) = x2, solve for x
and get x = ±√y.
Since x is non–negative we
must have
x = √y = g–1(y) or that
g–1(x) = √x.
g(x) = x2
g–1(x) = √x
Here are their graphs.
y = x

41. There is no inverse for x2 with the set of all real
numbers R as the domain because x2 is not a 1–1
function over R. However, if we set the domain to be
A = {x ≥ 0} (non–negative numbers) then the function
g(x) = x2 is a 1–1 function with the range B = {y ≥ 0}.
The Inverse Trigonometric Functions
Hence g–1 exists. To find it,
set y = g(x) = x2, solve for x
and get x = ±√y.
Since x is non–negative we
must have
x = √y = g–1(y) or that
g–1(x) = √x.
g(x) = x2
g–1(x) = √x
Here are their graphs.
Note that they are symmetric about the line y = x.
y = x

42. The Inverse Trigonometric Functions
Let f and f–1 be a pair of inverse functions and that
f(a) = b

43. The Inverse Trigonometric Functions
Let f and f–1 be a pair of inverse functions and that
f(a) = b so that f–1(b) = a.

44. The Inverse Trigonometric Functions
y = f(x)
y = f–1 (x)
(a, b)
Let f and f–1 be a pair of inverse functions and that
f(a) = b so that f–1(b) = a. Therefore (a, b) is a point
on the graph of y = f(x) and (b, a) is a point on the
graph of y = f–1(x).
(b, a)

45. The Inverse Trigonometric Functions
y = f(x)
y = f–1 (x)
y = x
Let f and f–1 be a pair of inverse functions and that
f(a) = b so that f–1(b) = a. Therefore (a, b) is a point
on the graph of y = f(x) and (b, a) is a point on the
graph of y = f–1(x). The points (a, b) and (b, a) are
mirror images with respect to the line y = x.
(a, b)
(b, a)

46. The Inverse Trigonometric Functions
y = f(x)
y = f–1 (x)
y = x
Let f and f–1 be a pair of inverse functions and that
f(a) = b so that f–1(b) = a. Therefore (a, b) is a point
on the graph of y = f(x) and (b, a) is a point on the
graph of y = f–1(x). The points (a, b) and (b, a) are
mirror images with respect to the line y = x.
(a, b)
(b, a)
This is true for all points on the graph.

47. The Inverse Trigonometric Functions
y = f(x)
y = f–1 (x)
So the graphs of f and f–1 are symmetric diagonally.
y = x(a, b)
(b, a)
Let f and f–1 be a pair of inverse functions and that
f(a) = b so that f–1(b) = a. Therefore (a, b) is a point
on the graph of y = f(x) and (b, a) is a point on the
graph of y = f–1(x). The points (a, b) and (b, a) are
mirror images with respect to the line y = x.
This is true for all points on the graph.

48. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.

49. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.

50. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function.

51. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)

52. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)
cos–1(a): [–1, 1] [0, π]

53. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
π0
1
–1 a = cos()

a
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)
cos–1(a): [–1, 1] [0, π]

54. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
π0
1
–1 a = cos()

a
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a).
cos–1(a): [–1, 1] [0, π]

55. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
Specifically, given a in [–1, 1],
cos–1(a) =  if
cos() = a
π0
1
–1 a = cos()

a
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)
cos–1(a): [–1, 1] [0, π]

56. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
Specifically, given a in [–1, 1],
cos–1(a) =  if
cos() = a and θ is in [0, π].
π0
1
–1 a = cos()

a
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)
cos–1(a): [–1, 1] [0, π]

57. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
Specifically, given a in [–1, 1],
cos–1(a) =  if
cos() = a and θ is in [0, π].
Example C.
π0
1
–1 a = cos()

a
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)
cos–1(a): [–1, 1] [0, π]
a. cos–1(½) = π/3
c. cos–1(cos(5π/3)) = π/3
b. cos–1(–½) = 2π/3

58. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
Specifically, given a in [–1, 1],
cos–1(a) =  if
cos() = a and [0, π].
Example C.
π0
1
–1 a = cos()

a
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)
cos–1(a): [–1, 1] [0, π]
a. cos–1(½) = π/3
c. cos–1(cos(5π/3)) = π/3
b. cos–1(–½) = 2π/3

59. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
Specifically, given a in [–1, 1],
cos–1(a) =  if
cos() = a and [0, π].
Example C.
π0
1
–1 a = cos()

a
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)
cos–1(a): [–1, 1] [0, π]
a. cos–1(½) = π/3
c. cos–1(cos(5π/3)) = π/3
b. cos–1(–½) = 2π/3

60. The Inverse Trigonometric Functions
Trig–functions are not 1–1 due to their periodicity.
For example cos(0) = cos(2π) = cos(4π) = … = 1.
Specifically, given a in [–1, 1],
cos–1(a) =  if
cos() = a and [0, π].
Example C.
π0
1
–1 a = cos()

a
If we downsize the domain to [0, π], then
a = cos(): [0, π] [–1, 1]
is a 1–1 function. Then its inverse exists and it’s
denoted as cos–1(a)
cos–1(a): [–1, 1] [0, π]
a. cos–1(½) = π/3
c. cos–1(cos(5π/3)) = π/3
b. cos–1(–½) = 2π/3

61. The Inverse Trigonometric Functions
0
1
–1
y
x
y = cos(x)
If we use x as the input variable and y as the output
variable and plot both functions, we have
y = cos(x): [0, π] to [–1,1]
and its refection across y = x
y = cos–1(x): [–1,1] to [0, π] .
π

62. The Inverse Trigonometric Functions
0
1
–1
y
x
If we use x as the input variable and y as the output
variable and plot both functions, we have
y = cos(x): [0, π] to [–1,1]
and its refection across y = x
y = cos–1(x): [–1,1] to [0, π] .
0
1
y = cos(x)
y = x
π

63. The Inverse Trigonometric Functions
1
–1
y
x
If we use x as the input variable and y as the output
variable and plot both functions, we have
y = cos(x): [0, π] to [–1,1]
and its refection across y = x
y = cos–1(x): [–1,1] to [0, π] .
1
y = cos(x)
y = x
π
–1
0
1
π

64. The Inverse Trigonometric Functions
π
π
–1
x
y = cos–1(x)
y = cos(x)
y = x
–1
0
1
1
y
If we use x as the input variable and y as the output
variable and plot both functions, we have
y = cos(x): [0, π] to [–1,1]
and its refection across y = x
y = cos–1(x): [–1,1] to [0, π] .

65. The Inverse Trigonometric Functions
Using the variables  and a, view
cos() = a as a function about the unit circle.
We have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.
1
(0, 0)

66. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.

1

(0, 0)
 = 0 = π

67. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.

(a, b)
1

(0, 0)
 = 0 = π

68. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.  = 0 = π 
(a, b)
b
1

(0, 0) a=cos()

69. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.  = 0 = π 
(a, b)
b
1

(0, 0)
The cosine inverse
cos–1(a) =  (–1≤ a ≤ 1)
a=cos()

70. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.

(a, b)
b
1

(0, 0)
 = 0 = π
a(0, 0)
 = 0 = π
a=cos()
The cosine inverse
cos–1(a) =  (–1≤ a ≤ 1)
1

71. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.

(a, b)
b
1

(0, 0)
 = 0 = π
a=cos()
The cosine inverse
cos–1(a) =  (–1≤ a ≤ 1)
= length of the arc whose
end point has a as the
x–coordinate.  = 0 = π
a(0, 0)
1

72. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.

(a, b)
b
1

(0, 0)
 = 0 = π
(a, b)
a
b
1
(0, 0)
 = 0 = π
a=cos()
The cosine inverse
cos–1(a) =  (–1≤ a ≤ 1)
= length of the arc whose
end point has a as the
x–coordinate.

73. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.

(a, b)
b
1

(0, 0)
 = 0 = π

(a, b)
a
b
cos–1(a)
= 
(0, 0)
 = 0 = π
a=cos()
The cosine inverse
cos–1(a) =  (–1≤ a ≤ 1)
= length of the arc whose
end point has a as the
x–coordinate. (0 ≤  ≤ π)
1

74. The Inverse Trigonometric Functions
Using the variables  and a and view
cos() = a as a function about the unit circle,
we have the definition of
cos() = a = x–coordinate
of the end point (a, b) on
the unit circle defined by
the arc of length rad.

(a, b)
b
1

(0, 0)
 = 0 = π

(a, b)
a
b
(0, 0)
 = 0 = π
a=cos()
The cosine inverse
cos–1(a) =  (–1≤ a ≤ 1)
= length of the arc whose
end point has a as the
x–coordinate. (0 ≤  ≤ π)
cos–1(a)
= 
1
So cos–1(a) is also notated as arccos(a) or acos(a)
because cos–1(a) =  is the length of a circular arc.

75. The Inverse Trigonometric Functions
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.

76. The Inverse Trigonometric Functions
1
a
 = cos–1(a)
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.

77. The Inverse Trigonometric Functions
Hence the opposite side is √1 – a2.
1
a
√1 – a2
 = cos–1(a)
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.  = cos–1(a)

78. The Inverse Trigonometric Functions
Hence the opposite side is √1 – a2.
Example D. Draw and
find the trig–value.
b. cot(cos–1(2a))
a. sin(cos–1(–3/5))
1
a
√1 – a2
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.
 = cos–1(a)

79. The Inverse Trigonometric Functions
1
a
Hence the opposite side is √1 – a2.
√1 – a2
Example D. Draw and
find the trig–value.
b. cot(cos–1(2a))
a. sin(cos–1(–3/5))
–3
5
=cos–1(–3/5)
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.
 = cos–1(a)

80. The Inverse Trigonometric Functions
1
a
Hence the opposite side is √1 – a2.
√1 – a2
Example D. Draw and
find the trig–value.
b. cot(cos–1(2a))
a. sin(cos–1(–3/5))
–3
54
=cos–1(–3/5)
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.
 = cos–1(a)

81. The Inverse Trigonometric Functions
1
a
Hence the opposite side is √1 – a2.
√1 – a2
Example D. Draw and
find the trig–value.
b. cot(cos–1(2a))
a. sin(cos–1(–3/5))
= 4/5 –3
54
=cos–1(–3/5)
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.
 = cos–1(a)

82. The Inverse Trigonometric Functions
1
a
Hence the opposite side is √1 – a2.
√1 – a2
Example D. Draw and
find the trig–value.
b. cot(cos–1(2a))
1
2a

a. sin(cos–1(–3/5))
= 4/5 –3
54
=cos–1(–3/5)
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.
 = cos–1(a)

83. The Inverse Trigonometric Functions
1
a
Hence the opposite side is √1 – a2.
√1 – a2
Example D. Draw and
find the trig–value.
b. cot(cos–1(2a))
1
2a

√1 – 4a2
a. sin(cos–1(–3/5))
= 4/5 –3
54
=cos–1(–3/5)
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.
 = cos–1(a)

84. The Inverse Trigonometric Functions
1
a
Hence the opposite side is √1 – a2.
√1 – a2
Example D. Draw and
find the trig–value.
b. cot(cos–1(2a))
1
2a

√1 – 4a2
=
2a
√1 – 4a2
a. sin(cos–1(–3/5))
= 4/5 –3
54
=cos–1(–3/5)
With the variables  and a,
we have the following right
triangle to represent the
relation cos–1(a) = θ.
 = cos–1(a)

85. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.

86. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.
The sine inverse is defined for
–1≤ b ≤ 1.

87. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.
The sine inverse is defined for
–1≤ b ≤ 1.
sin–1(b) =  ɛ [–π/2, π/2]
= length of the arc whose
end point has b as the
y–coordinate

88. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.  = π/2
(0, 0)
The sine inverse is defined for
–1≤ b ≤ 1.
sin–1(b) =  ɛ [–π/2, π/2]
= length of the arc whose
end point has b as the
y–coordinate  = – π/2

89. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.  = π/2
b
(0, 0)
The sine inverse is defined for
–1≤ b ≤ 1.
 = – π/2
sin–1(b) =  ɛ [–π/2, π/2]
= length of the arc whose
end point has b as the
y–coordinate

90. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.  = π/2
b
(0, 0)
The sine inverse is defined for
–1≤ b ≤ 1.
 = – π/2
(a, b)a
sin–1(b) =  ɛ [–π/2, π/2]
= length of the arc whose
end point has b as the
y–coordinate

91. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.  = π/2

b
(0, 0)
sin–1(b) = 
The sine inverse is defined for
–1≤ b ≤ 1.
 = – π/2
(a, b)a
sin–1(b) =  ɛ [–π/2, π/2]
= length of the arc whose
end point has b as the
y–coordinate

92. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.  = π/2

b
(0, 0)
sin–1(b) = 
The sine inverse is defined for
–1≤ b ≤ 1.
 = – π/2
(a, b)a
The right triangle
representing sin–1(b) is
shown here.
1
b
 = sin–1(b)
sin–1(b) =  ɛ [–π/2, π/2]
= length of the arc whose
end point has b as the
y–coordinate

93. The Inverse Trigonometric Functions
The inverses of the other trig–functions are defined
similarly so that each trig–function has its own
domain and range.  = π/2

b
(0, 0)
sin–1(b) = 
The sine inverse is defined for
–1≤ b ≤ 1.
 = – π/2
(a, b)a
The right triangle
representing sin–1(b) is
shown here.
The adjacent of the triangle
is √1 – b2 .
1
b
 = sin–1(b)
√1 – b2
sin–1(b) =  ɛ [–π/2, π/2]
= length of the arc whose
end point has b as the
y–coordinate

94. The Inverse Trigonometric Functions
and the inverse of sine
y = sin–1(x) : [–1, 1] [–π/2, π/2].
In variables x and y
y = sin(x): [–π/2, π/2] [–1, 1].

95. The Inverse Trigonometric Functions
and the inverse of sine
y = sin–1(x) : [–1, 1] [–π/2, π/2].
We get the graph of y = sin–1(x) by reflecting y = sin(x)
across y = x.
In variables x and y
y = sin(x): [–π/2, π/2] [–1, 1].

96. The Inverse Trigonometric Functions
π/2
y = sin(x)
1
–1
and the inverse of sine
y = sin–1(x) : [–1, 1] [–π/2, π/2].
We get the graph of y = sin–1(x) by reflecting y = sin(x)
across y = x.
–π/2
In variables x and y
y = sin(x): [–π/2, π/2] [–1, 1].

97. The Inverse Trigonometric Functions
π/2
1
–1
y = sin(x)
–π/2
and the inverse of sine
y = sin–1(x) : [–1, 1] [–π/2, π/2].
We get the graph of y = sin–1(x) by reflecting y = sin(x)
across y = x.
In variables x and y
y = sin(x): [–π/2, π/2] [–1, 1].

98. The Inverse Trigonometric Functions
–1
–π/2
y = sin–1(x)
1
–1
–π/2
π/2
1–1
π/2
y = x
y = sin(x)
and the inverse of sine
y = sin–1(x) : [–1, 1] [–π/2, π/2].
We get the graph of y = sin–1(x) by reflecting y = sin(x)
across y = x.
In variables x and y
y = sin(x): [–π/2, π/2] [–1, 1].

99. The Inverse Trigonometric Functions
–π/2
π/2
1–1
y = sin–1(x)
and the inverse of sine
y = sin–1(x) : [–1, 1] [–π/2, π/2].
We get the graph of y = sin–1(x) by reflecting y = sin(x)
across y = x.
In variables x and y
y = sin(x): [–π/2, π/2] [–1, 1].

100. The Inverse Trigonometric Functions
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1.

101. The Inverse Trigonometric Functions
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1.
So the tangent inverse exists
and tan– 1(t) = , –∞< t < ∞,

102. The Inverse Trigonometric Functions
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1.
So the tangent inverse exists
and tan– 1(t) = , –∞< t < ∞
with tan() = t

103. The Inverse Trigonometric Functions
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1.
So the tangent inverse exists
and tan– 1(t) = , –∞< t < ∞
with tan() = t = b/a
where (a, b) is the end point
of the arc define by 
with  ɛ (–π/2, π/2) .

104. The Inverse Trigonometric Functions
 = π/2
(0, 0)
 = –π/2
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1.
So the tangent inverse exists
and tan– 1(t) = , –∞< t < ∞
with tan() = t = b/a
where (a, b) is the end point
of the arc define by 
with  ɛ (–π/2, π/2) .

105. The Inverse Trigonometric Functions
 = π/2
(0, 0)
 = –π/2
(a, b),
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1. t = b/a
So the tangent inverse exists
and tan– 1(t) = , –∞< t < ∞
with tan() = t = b/a
where (a, b) is the end point
of the arc define by 
with  ɛ (–π/2, π/2) .

106. The Inverse Trigonometric Functions
 = π/2

b
(0, 0)
1
 = –π/2
(a, b)
a
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1.
 = tan–1(b/a)
(a, b), t = b/a
So the tangent inverse exists
and tan– 1(t) = , –∞< t < ∞
with tan() = t = b/a
where (a, b) is the end point
of the arc define by 
with  ɛ (–π/2, π/2) .

107. The Inverse Trigonometric Functions
 = π/2

b
(0, 0)
1
 = –π/2
a
1
t
 = tan–1(t)
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1.
 = tan–1(b/a)
(a, b), t = b/a
The right triangle
representing tan–1(t) is
shown here.
So the tangent inverse exists
and tan– 1(t) = , –∞< t < ∞
with tan() = t = b/a
where (a, b) is the end point
of the arc define by 
with  ɛ (–π/2, π/2) .

108. The Inverse Trigonometric Functions
 = π/2

b
(0, 0)
1
 = –π/2
a
The right triangle
representing tan–1(t) is
shown here. We note that
the hypotenuse is 1 + t2.
1
t
 = tan–1(t)
1 + t2
Set the domain of tangent to be (–π/2, π/2) so
tan() = t: (–π/2, π/2) (–∞ , ∞)
is 1–1.
 = tan–1(b/a)
(a, b), t = b/a
So the tangent inverse exists
and tan– 1(t) = , –∞< t < ∞
with tan() = t = b/a
where (a, b) is the end point
of the arc define by 
with  ɛ (–π/2, π/2) .

109. The Inverse Trigonometric Functions
–π/2
y = tan(x)
Using x and y, we get the graph of y = tan–1(x) by
reflecting y = tan(x) across y = x.
–π/2

110. The Inverse Trigonometric Functions
–π/2
y = tan(x)
–π/2
Using x and y, we get the graph of y = tan–1(x) by
reflecting y = tan(x) across y = x.

111. The Inverse Trigonometric Functions
–π/2
π/2
y = tan(x)
–π/2
– π/2
y = tan–1(x)
reflect
Using x and y, we get the graph of y = tan–1(x) by
reflecting y = tan(x) across y = x.

112. The Inverse Trigonometric Functions
31,38,45,56,57,58,76,77,102,113,114.

113. The Inverse Trigonometric Functions
– π/2
y = tan–1(x)
We summarize the sin–1(x), cos–1(x), and tan–1(x) here.
π/2
y = cos–1(x)
y =sin–1(x)
Domain Range Rt– Δ
[–1, 1]
(–∞, ∞)
[0, π]
[–π/2, π/2] –1 1
–π/2
π/2
–1 1
π
x
1
x
1
x
1



√1+x2
√1–x2
√1–x2
Graph
[–1, 1]
(–π/2, π/2)
x
x
x

114. The Inverse Trigonometric Functions
y = cot–1(x)
We summarize the sec–1(x), csc–1(x), and cot–1(x) here.
π/2
y = sec–1(x)
y =csc–1(x)
Domain Range Rt– Δ
1 ≤ | x |
1 ≤ | x |
0≤  ≤ π
[–π/2, π/2]
(–π/2, π/2)
x
1
1
x
x
1



√1+x2
√x2–1
Graph
√x2–1
 = π/2
(1, π/2)
(–1, –π/2)
(1, 0)
(–1, π)
 = 0
(0, π/2)
(0,– π/2)
 = 0
(–∞, ∞)
x
x
x