14 the inverse trigonometric functions

The Inverse Trigonometric Functions

A function is one-to-one (1–1) if different inputs produce different outputs.

The Inverse Trigonometric Functions

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

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

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

uf(u)=f(v)

v

u = v

a non-one-to-one function

The Inverse Trigonometric Functions

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

uf(u)=f(v)

v

u = v

a non-one-to-one function

The Inverse Trigonometric Functions

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

uf(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

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

uf(u)=f(v)

v

u = v

a non-one-to-one function

The Inverse Trigonometric Functions

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

uf(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

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

uf(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.

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

uf(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.

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

uf(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.

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

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

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

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

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

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

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

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.

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.

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.

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.

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.

Some inverse functions may be solved algebraically from f(x).

The Inverse Trigonometric Functions

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

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

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

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,

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 getx = (y – 1)/2 = f–1(y)

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 getx = (y – 1)/2 = f–1(y)Switch to the variable x, we’ve f–1(x) = (x – 1)/2.

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 getx = (y – 1)/2 = f–1(y)Switch to the variable x, we’ve f–1(x) = (x – 1)/2.

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.

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 getx = (y – 1)/2 = f–1(y)Switch to the variable x, we’ve f–1(x) = (x – 1)/2.

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.

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

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

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.

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,

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 xwe’ve x = ±√y.

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 xwe’ve x = ±√y. Since x is non–negative we must have x = √y = g–1(y)

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 xwe’ve x = ±√y. Since x is non–negative we must have x = √y = g–1(y) or that g–1(x) = √x.

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 xwe’ve 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

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 xwe’ve 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

The Inverse Trigonometric FunctionsLet f and f–1 be a pair of inverse functions and that

f(a) = b

The Inverse Trigonometric FunctionsLet f and f–1 be a pair of inverse functions and that

f(a) = b so that f–1(b) = a.

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 that (b, a) is a point on

the graph of y = f–1(x).

(b, a)

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 that (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)

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 that (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.

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 that (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.

The Inverse Trigonometric FunctionsTrig–functions are not 1–1 due to their periodicity.

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

The Inverse Trigonometric FunctionsTrig–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.

The Inverse Trigonometric FunctionsTrig–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)

The Inverse Trigonometric FunctionsTrig–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]

The Inverse Trigonometric FunctionsTrig–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]

The Inverse Trigonometric FunctionsTrig–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]

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

Specifically, given [-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]

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

Specifically, given [-1, 1]cos–1(a) = if cos() = a and [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]

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

Specifically, given [-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/3b. cos–1(–½) = 2π/3

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

Specifically, given [-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/3b. cos–1(–½) = 2π/3

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

Specifically, given [-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/3b. cos–1(–½) = 2π/3

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

Specifically, given [-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/3b. cos–1(–½) = 2π/3

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 = xy = cos–1(x): [–1,1] to [0, π] .

π

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 = xy = cos–1(x): [–1,1] to [0, π] .

0

1

y = cos(x)

y = x

π

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 = xy = cos–1(x): [–1,1] to [0, π] .

1

y = cos(x)

y = x

π–1

01

π

The Inverse Trigonometric Functions

π

π

–1

x

y = cos–1(x)

y = cos(x)

y = x

–10

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 = xy = cos–1(x): [–1,1] to [0, π] .

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad.

1

(0, 0)

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = 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

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = 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

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad.

= 0 = π

(a, b)

b1

(0, 0) a=cos()

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad.

= 0 = π

(a, b)

b1

(0, 0)

The cosine inversecos–1(a) = (–1≤ a ≤ 1)

a=cos()

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad.

(a, b)

b1

(0, 0)

= 0 = πa(0, 0)

= 0 = πa=cos()

The cosine inversecos–1(a) = (–1≤ a ≤ 1)

1

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad.

(a, b)

b1

(0, 0) = 0 = π

a=cos()

The cosine inversecos–1(a) = (–1≤ a ≤ 1)= length of the arc whose end point has a as the x–coordinate. = 0 = π

a(0, 0)

1

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad.

(a, b)

b1

(0, 0)

= 0 = π

(a, b)

a

b1

(0, 0)

= 0 = πa=cos()

The cosine inversecos–1(a) = (–1≤ a ≤ 1)= length of the arc whose end point has a as the x–coordinate.

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad.

(a, b)

b1

(0, 0)

= 0 = π

(a, b)

a

bcos–1(a)=

(0, 0)

= 0 = πa=cos()

The cosine inversecos–1(a) = (–1≤ a ≤ 1)= length of the arc whose end point has a as the x–coordinate. (0 ≤ ≤ π)

1

The Inverse Trigonometric FunctionsUsing the variables and a and view cos() = a as a function about the unit circle, we have the definition ofcos() = a = x–coordinate of the end point (a, b) on the unit circle defined by the arc of length rad.

(a, b)

b1

(0, 0)

= 0 = π

(a, b)

a

b

(0, 0)

= 0 = πa=cos()

The cosine inversecos–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.

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a.

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a.

1

a = cos–1(b)

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a.

Hence the opposite side is √1 – a2.

1

a

√1 – a2

= cos–1(b)

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a.

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

= cos–1(b)

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a

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)

= cos–1(b)

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a

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)

= cos–1(b)

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a

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)

= cos–1(b)

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a

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)

= cos–1(b)

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a

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)

= cos–1(b)

The Inverse Trigonometric FunctionsWith the variables , a, we’ve the following right triangle represent the relation cos–1() = a

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)

= cos–1(b)

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

The Inverse Trigonometric FunctionsThe 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.

The Inverse Trigonometric FunctionsThe 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

The Inverse Trigonometric FunctionsThe 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

The Inverse Trigonometric FunctionsThe 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

The Inverse Trigonometric FunctionsThe 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

The Inverse Trigonometric FunctionsThe 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

The Inverse Trigonometric FunctionsThe 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.

1b

= sin–1(b)

sin–1(b) = ɛ [–π/2, π/2]= length of the arc whose end point has b as the y–coordinate

The Inverse Trigonometric FunctionsThe 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 .

1b

= 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

The Inverse Trigonometric Functions

and the inverse of siney = sin–1(x) : [-1, 1] [-π/2, π/2].

In variables x and yy = sin(x): [-π/2, π/2] [-1, 1].

The Inverse Trigonometric Functions

and the inverse of siney = 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 yy = sin(x): [-π/2, π/2] [-1, 1].

The Inverse Trigonometric Functions

π/2

y = sin(x)1

–1

and the inverse of siney = 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 yy = sin(x): [-π/2, π/2] [-1, 1].

The Inverse Trigonometric Functions

π/2

1

–1

y = sin(x)

–π/2

and the inverse of siney = 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 yy = sin(x): [-π/2, π/2] [-1, 1].

The Inverse Trigonometric Functions

–1

–π/2

y = sin–1(x)

1

–1

–π/2

π/2

1–1π/2

y = x

y = sin(x)

and the inverse of siney = 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 yy = sin(x): [-π/2, π/2] [-1, 1].

The Inverse Trigonometric Functions

–π/2

π/2

1–1

y = sin–1(x)

and the inverse of siney = 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 yy = sin(x): [-π/2, π/2] [-1, 1].

The Inverse Trigonometric FunctionsSet the domain of tangent to be (–π/2, π/2) sotan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1.

The Inverse Trigonometric FunctionsSet the domain of tangent to be (–π/2, π/2) sotan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. So the tangent inverse existstan– 1(t) = , –∞< t < ∞

The Inverse Trigonometric FunctionsSet the domain of tangent to be (–π/2, π/2) sotan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. So the tangent inverse existstan– 1(t) = , –∞< t < ∞with tan() = t

The Inverse Trigonometric FunctionsSet the domain of tangent to be (–π/2, π/2) sotan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. So the tangent inverse existstan– 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) .

The Inverse Trigonometric Functions

= π/2

(0, 0)

= –π/2

Set the domain of tangent to be (–π/2, π/2) sotan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. So the tangent inverse existstan– 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) .

The Inverse Trigonometric Functions

= π/2

(0, 0)

= –π/2

(a, b),

Set the domain of tangent to be (–π/2, π/2) sotan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1. t = b/a

So the tangent inverse existstan– 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) .

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) sotan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1.

= tan–1(b/a)

(a, b), t = b/a

So the tangent inverse existstan– 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) .

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) sotan() = 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 existstan– 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) .

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 thatthe hypotenuse is 1 + t2.

1

t = tan–1(t)

1 + t2

Set the domain of tangent to be (–π/2, π/2) sotan() = t: (–π/2, π/2) (–∞ , ∞) is 1–1.

= tan–1(b/a)

(a, b), t = b/a

So the tangent inverse existstan– 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) .

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

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.

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.

The Inverse Trigonometric Functions

– π/2

y = tan–1(x)

We summarize the sin–1(x), cos–1(x), and sin–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

x1

x

1

√1+x2

√1–x2

√1–x2

Graph

[–1, 1]

(–π/2, π/2)

x

x

x

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

1x

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