14 the inverse trigonometric functions

The Inverse Trigonometric Functions

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


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)

v f(v)

u = v f(u) = f(v)


uf(u)=f(v)

v

u = v

a non-one-to-one function



Example A. a. g(x) = 2x + 1 is 1–1

u f(u)

v f(v)

u = v f(u) = f(v)


uf(u)=f(v)

v

u = v




u f(u)

v f(v)

u = v f(u) = f(v)


uf(u)=f(v)

v

u = v



Example A. a. g(x) = 2x + 1 is 1–1 because if u v, then 2u 2v so 2u + 1 2v + 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).

u f(u)

v f(v)

u = v f(u) = f(v)


uf(u)=f(v)

v

u = v




u f(u)

v f(v)

u = v f(u) = f(v)


uf(u)=f(v)

v

u = 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).b. f(x) = x2 is not 1–1


u f(u)

v f(v)

u = v f(u) = f(v)


uf(u)=f(v)

v

u = 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).b. f(x) = x2 is not 1–1 because 3 –3, but f(3) = f(–3) = 9.


u f(u)

v f(v)

u = v f(u) = f(v)


uf(u)=f(v)

v

u = 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).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.


u f(u)

v f(v)

u = v f(u) = f(v)


uf(u)=f(v)

v

u = v



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).



f(u)

v f(v)

u = v f(u) = f(v)

f(x) is a one-to-one function


f(x)

u


f(u)

v f(v)

u = v f(u) = f(v)



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)



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


f(u)

v f(v)

u = v f(u) = f(v)



u f(u)

v f(v)

u = v f(u) = f(v)


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


f(u)

v f(v)

u = v f(u) = f(v)



u f(u)

v f(v)

u = v f(u) = f(v)


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


f(u)

v f(v)

u = v f(u) = f(v)



u f(u)

v f(v)

u = v f(u) = f(v)


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


f(u)

v f(v)

u = v f(u) = f(v)



u f(u)

v f(v)

u = v f(u) = f(v)


f –1(x)f(x)

Furthermore f –1(f(x)) = x

x

A

f(x)

B

f

u



f(u)

v f(v)

u = v f(u) = f(v)



u f(u)

v f(v)

u = v f(u) = f(v)


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



f(u)

v f(v)

u = v f(u) = f(v)



u f(u)

v f(v)

u = v f(u) = f(v)


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



f(u)

v f(v)

u = v f(u) = f(v)



u f(u)

v f(v)

u = v f(u) = f(v)


f –1(x)f(x)

x

A

f(x)

B

f

f –1

f –1(f(x)) = x


f– 1(x)

A B

f –1

f –1 (x)

x

u



f(u)

v f(v)

u = v f(u) = f(v)



u f(u)

v f(v)

u = v f(u) = f(v)


f –1(x)f(x)

x

A

f(x)

B

f

f –1

f –1(f(x)) = x


A B

f

f –1

f(f –1 (x)) = x

xf– 1(x)

u


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


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.





Set y = f(x) = 2x + 1 and solve for x,




Set y = f(x) = 2x + 1 and solve for x, we getx = (y – 1)/2 = f–1(y)




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.





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.


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}.




Hence g–1 exists.



Hence g–1 exists. To find it, set y = g(x) = x2,



Hence g–1 exists. To find it, set y = g(x) = x2, solve for xwe’ve x = ±√y.



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)



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




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.


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)


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)


y = f(x)

y = f–1 (x)

y = x


(a, b)

(b, a)

This is true for all points on the graph.


y = f(x)

y = f–1 (x)

So the graphs of f and f–1 are symmetric diagonally.

y = x(a, b)

(b, a)


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]


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]


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]


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

π0

1

–1 a = cos()

a



Specifically, given [-1, 1]cos–1(a) = if cos() = a and [0, π].

π0

1

–1 a = cos()

a



Specifically, given [-1, 1]cos–1(a) = if cos() = a and [0, π].

Example C.

π0

1

–1 a = cos()

a


a. cos–1(½) = π/3

c. cos–1(cos(5π/3)) = π/3b. cos–1(–½) = 2π/3


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


0

1

–1

y

x


0

1

y = cos(x)

y = x

π


1

–1

y

x


1

y = cos(x)

y = x

π–1

01

π


π

π

–1

x

y = cos–1(x)

y = cos(x)

y = x

–10

1

1

y


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)


1

(0, 0) = 0 = π


(a, b)

1

(0, 0) = 0 = π


= 0 = π

(a, b)

b1

(0, 0) a=cos()


= 0 = π

(a, b)

b1

(0, 0)

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

a=cos()


(a, b)

b1

(0, 0)

= 0 = πa(0, 0)

= 0 = πa=cos()

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

1


(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


(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.


(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


(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.


1

a = cos–1(b)


Hence the opposite side is √1 – a2.

1

a

√1 – a2

= cos–1(b)



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

.


√1 – a2


b. cot(cos–1(2a))

a. sin(cos–1(–3/5))

–3

5

=cos–1(–3/5)

= cos–1(b)


1

a

.


√1 – a2


b. cot(cos–1(2a))

a. sin(cos–1(–3/5))

–3

54

=cos–1(–3/5)

= cos–1(b)


1

a

.


√1 – a2


b. cot(cos–1(2a))

a. sin(cos–1(–3/5))

= 4/5 –3

54

=cos–1(–3/5)

= cos–1(b)


1

a

.


√1 – a2


b. cot(cos–1(2a))1

2a

a. sin(cos–1(–3/5))

= 4/5 –3

54

=cos–1(–3/5)

= cos–1(b)


1

a

.


√1 – a2



2a

√1 – 4a2

a. sin(cos–1(–3/5))

= 4/5 –3

54

=cos–1(–3/5)

= cos–1(b)


1

a

.


√1 – a2



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


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


b

(0, 0)


= – π/2

(a, b)a



b

(0, 0)

sin–1(b) =


= – π/2

(a, b)a



b

(0, 0)

sin–1(b) =


= – π/2

(a, b)a

The right triangle representing sin–1(b) is shown here.

1b

= sin–1(b)



b

(0, 0)

sin–1(b) =


= – 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



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].


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

y = sin(x)1

–1


–π/2



π/2

1

–1

y = sin(x)

–π/2




–1

–π/2

y = sin–1(x)

1

–1

–π/2

π/2

1–1π/2

y = x

y = sin(x)




–π/2

π/2

1–1

y = sin–1(x)



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) .


= 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) .


= 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) .


= 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



= π/2

b

(0, 0)

1

= –π/2

a

1

t = tan–1(t)


= tan–1(b/a)

(a, b), t = b/a

The right triangle representing tan–1(t) is shown here.



= 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


= tan–1(b/a)

(a, b), t = b/a



–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


–π/2

y = tan(x)

–π/2



–π/2

π/2

y = tan(x)

–π/2

– π/2

y = tan–1(x)

reflect



– 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


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

14 the inverse trigonometric functions

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