Inverse TrigonometricFunctions

September 11, 2019

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Restricted Sine Function.

The trigonometric function sinx is not a one-to-one function.

H�6, 1�2L H5�6, 1�2L

-Π -Π

2

Π

2Π

-1.0

-0.5

0.5

1.0y= sin x

We still want an inverse, so what to do? We must restrict the domainof the sin function.We want to choose some interval (as large as we can find) so thatsin(x) is one-to-one on that interval. From the graph we see that[−π

2,π

2

]looks likes a good choice. To be super cautious note

d sin(x)

dx= cos(x) > 0 on

[−π

2,π

2

]and 0 only at the end points so

sin(x) is increasing on[−π

2,π

2

].

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The restricted sine function

Define the restricted sine function by

f(x) =

sinx −π

26 x 6

π

2

undefined otherwise

We have Domain(f) =[−π

2,π

2

]and Range(f) = [−1, 1] and graph

-Π

2-

Π

4

Π

4

Π

2

-1.0

-0.5

0.5

1.0y= fHxL

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Inverse Sine Function (arcsinx = sin−1 x).

The restricted sine function is one-to-one and hence has an inverse,shown in red in the diagram below.

H�2, 1L

H-Π

4,

-1

2

L

H1, Π�2L

H-1

2

,-Π

4L

-Π

2-

Π

4

Π

4

Π

2

-1.5

-1.0

-0.5

0.5

1.0

1.5

This inverse function, f−1(x), is denoted by

f−1(x) = sin−1 x or arcsinx.

sin−1 is terrible notation since you will confuse it with the cosecantfunction.

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Properties of arcsin(x).

Domain(arcsin) = [−1, 1] and Range(arcsin) = [−π2 ,π2 ].

Since f−1(x) = y if and only if f(y) = x, we have:

arcsinx = y if and only if sin(y) = x and − π

26 y 6

π

2.

Since f(f−1(x)) = x f−1(f(x)) = x we have:

sin(arcsin(x)) = x for x ∈ [−1, 1]

arcsin(sin(x)) = x for x ∈[−π

2,π

2

].

From the graph: arcsin(x) is an odd function and soarcsin(−x) = − arcsin(x).

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Evaluating arcsin(x).

Example Evaluate arcsin(−1√2

)using the marked circle above.

I We see that the point(−1√2, −π

4

)is on the graph of y = arcsin(x).

I Therefore arcsin(−1√2

)= −π

4.

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Example Evaluate arcsin

(√3

2

)and arcsin

(−√

3

2

).

I arcsin

(√3

2

)= y is the same statement as:

y is an angle between −π2

andπ

2with sin y =

√3

2.

I Consulting our unit circle, we see that y =π

3.

I arcsin

(−√3

2

)= − arcsin

(√3

2

)= −π

3

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More Examples For arcsin(x)

Example Evaluate arcsin(sinπ). It is tempting to write π until you

realize the answer is between −π2

andπ

2.

I We have sinπ = 0, hence arcsin(sinπ) = arcsin(0) = 0.

Example Evaluate cos(arcsin(

√3/2)

).

I We saw above that arcsin

(√3

2

)=π

3.

I Therefore cos

(arcsin

(√3

2

))= cos

(π3

)=

1

2.

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Preparation for the method of TrigonometricSubstitution

Example Give a formula in terms of x for tan(arcsin(x)

).

I We draw a right angled triangle with θ = arcsin(x).

1 - x2

θ

x1

I From this we see that tan(arcsin(x)

)= tan(θ) =

x√1− x2

.

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Derivative of arcsin(x).

d

dxarcsin(x) =

1√1− x2

, −1 < x < 1.

I Proof We have arcsin(x) = y if and only if sin(y) = x and−π/2 6 y 6 π/2. Using implicit differentiation, we get

cos(y)dy

dx= 1 or

dy

dx=

1

cos(y).

I Now we know that cos2(y) + sin2(y) = 1, hence we have thatcos2(y) + x2 = 1 and

cos(y) = ±√

1− x2

I For y between −π2

andπ

2, cos(y) > 0 so cos(y) =

√1− x2 and

d

dxarcsin(x) =

1√1− x2

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Derivative of arcsin(x): Example

Example Find the derivatived

dxarcsin

(√cos(x)

).

I We haved

dxarcsin

(√cos(x)

)=

1√1− (

√cos(x) )2

d

dx

√cosx

I

=1√

1− cosx· − sinx

2√

cosx=

− sinx

2√

cosx√

1− cosx

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Inverse Cosine Function

Inverse Cosine Function We can define the functioncos−1 x = arccos(x) similarly. The details are given at the end ofPilkington’s lecture notes.

Domain(arccos) = [−1, 1] and Range(arccos) = [0, π].

arccos(x) = y if and only if cos(y) = x and 0 6 y 6 π.

cos(arccos(x)) = x for x ∈ [−1, 1] arccos(cos(x)) = x for x ∈[0, π].

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Derivative of Inverse Cosine Function

Inverse Cosine Function Also by implicit differentiation, we canshow that (see end of Pilkington’s lecture notes)

d

dxarccos(x) = − d

dxarcsin(x) =

−1√1− x2

Note that since both have the same derivative, we must have

arccos(x) = − arcsin(x) + C

for some constant C. Letting x = 0, we get that

arcsin(x) + arccos(x) =π

2.

This means that the arccos is not an odd function. In fact

arccos(−x) = π − arccos(x).

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Restricted Tangent Function

The tangent function is not a one to one function.

The restricted tangent function is given by

h(x) =

tanx −π

2< x <

π

2

undefined otherwise

We see from the graph of the restricted tangent function (or from itsderivative) that the function is one-to-one and hence has an inverse,which we denote by

h−1(x) = tan−1 x or arctanx.

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H�4, 1L

-Π

2-

Π

4

Π

4

Π

2

-6

-4

-2

2

4

6

y= hHxL

H1, Π�4L

-5 -4 -3 -2 -1 1 2 3 4 5

-Π

2

-Π

4

Π

4

Π

2

y= arctanHxL

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Properties of arctan(x).

Domain(arctan) = (−∞,∞) and Range(arctan) =(−π

2,π

2

).

Since h−1(x) = y if and only if h(y) = x, we have:

arctan(x) = y if and only if tan(y) = x and − π

2< y <

π

2.

Since h(h−1(x)) = x and h−1(h(x)) = x, we have:

tan(arctan(x)) = x for x ∈ (−∞,∞)

arctan(tan(x)) = x for x ∈(−π

2,π

2

)

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From the graph, we have: arctan(−x) = − arctan(x).

Also, since limx→(π

2 )−tanx =∞ and lim

x→(−π2 )+

tanx = −∞,

we have limx→∞

arctan(x) =π

2and lim

x→−∞arctan(x) = −π

2

H1, Π�4L

-5 -4 -3 -2 -1 1 2 3 4 5

-Π

2

-Π

4

Π

4

Π

2

y= arctanHxL

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Evaluating arctan(x)

Example Find arctan(1) and arctan

(1√3

).

I arctan(1) is the unique angle, θ, between −π2and π

2with

tan(θ) =sin(θ)

cos(θ)= 1. By inspecting the unit circle, we see that θ =

π

4.

I arctan

(1√3

)is the unique angle, θ, between −π

2and π

2with

tan(θ) =sin(θ)

cos(θ)=

1√3. By inspecting the unit circle, we see that

θ =π

6.

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Example Find cos(

arctan(

1√3

)).

I cos(arctan

(1√3

))= cos

(π6

)=

√3

2.

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Derivative of arctan(x).

Using implicit differentiation, we get

d

dxarctan(x) =

1

x2 + 1, −∞ < x <∞.

tan(y) = x; y′ sec2(y) = 1 so y′(1 + tan2(y)

)= 1 so y′ =

1

1 + x2.

We can use the chain rule in conjunction with the above derivative.

Example Find the domain and derivative of arctan(ln(x)

)I Domain = Domain of lnx = (0,∞)

Id

dxarctan

(ln(x)

)=

1

1 + (lnx)21

x=

1

x(1 + (lnx)2).

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Integration Formulas

Reversing the derivative formulas above, we get∫1√

1− x2dx = arcsin(x) + C,

∫1

1 + x2dx = arctan(x) + C,

Example ∫ 1/2

0

1

1 + 4x2dx

I We use substitution. Let u = 2x, then du = 2dx, u(0) = 0,

u(1/2) = 1.

I ∫ 1/2

0

1

1 + 4x2dx =

1

2

∫ 1

0

1

1 + u2du =

I1

2arctan(u)

∣∣∣10=

1

2

(arctan(1)− arctan(0)

)=

1

2

(π4− 0)=π

8.

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Integration

Example ∫1√

9− x2dx

I ∫1√

9− x2dx =

∫1

3√

1− x2

9

dx =1

3

∫1√

1− x2

9

dx

I Let u = x3, then dx = 3du

I ∫1√

9− x2dx =

1

3

∫3√

1− u2du = arcsin(u) + C = arcsin

(x3

)+ C

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arcsec(x)

Here is a graph of the secant function. There are vertical asymptotes

atπ

2+ kπ, k any integer.

There is disagreement in the literature as to how to restrict it. The

book uses[0,π

2

)∪[π,

3π

2

)so we will too, but some authors prefer[

0,π

2

)∪(π

2, π].

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We are using the red curves, but some authors prefer the red in thefirst quadrant union the blue in the fourth. Our domain is[0,π

2

)∪[π,

3π

2

)and the range is (−∞,−1] ∪ [1,∞). (The other

choice has the same range.)

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The red curve is the graph y = arcsec(x). The domain

is(−∞,−1] ∪ [1,∞) and the range is[0,π

2

)∪[π,

3π

2

). There are two

limits (or horizontal asymptotes)

limx→−∞

arcsec(x) =3π

2and lim

x→∞arcsec(x) =

π

2

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d

dxarcsec(x) =

1

x√x2 − 1

y = arcsec(x); sec y = x; y′ sec(y) tan(y) = 1. Nowsec2(y) = 1 + tan2(y) so tan2(y) = sec2 y − 1 and hence

tan(y) = ±√

sec2 y − 1 = ±√x2 − 1. Hence y′ = ± 1

x√x2 − 1

.

From the graph, when x < 0 the curve is decreasing so the derivativeis negative and the sign is +. Similarly, when x > 0 the curve isincreasing so the derivative is positive and the sign is +.

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The “other” choice for the arcsec gives a function which is always

increasing but has derivative1

|x|√x2 − 1

. Since we will only use the

arcsec to integrate, we prefer the definition we chose.

There are also inverse trig functions arccot and arccsc. These do notallow us to integrate anything new so we will not discuss them.

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Basic formulas

I arcsin:I domain:[−1, 1], range:

[−π

2, π2

]I lim

x→1−1arcsin(x) =

π

2; limx→−1+

arcsin(x) = −π2

Id arcsin(x)

dx=

1√1− x2

;

∫dx√1− x2

= arcsin(x) + C

I arctan:I domain:(−∞,∞), range:

(−π

2, π2

)I lim

x→∞arctan(x) =

π

2; limx→−∞

arctan(x) = −π2

Id arctan(x)

dx=

1

1 + x2;

∫dx

1 + x2= arctan(x) + C

I arcsec:I domain:(−∞,−1] ∪ [1,∞), range:

[0, π

2

)∪[π, 3π

2

)I lim

x→∞arcsec(x) =

π

2; limx→−∞

arcsec(x) = −3π

2I lim

x→1+arcsec(x) = 0; lim

x→−1−arcsec(x) = π

Id arcsec(x)

dx=

1

x√x2 − 1

;

∫dx

x√x2 − 1

= arcsec(x) + C

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