Inverse Trig Functions &Derivatives of Trig Functions

Cole ZmurchokMath 102 Section 106

November 28, 2016

Today...

1. More Trig Function Derivatives

2. Inverse Trig Functions

3. Related Rates with Trig Functions

I Course Evaluations: do these for all of yourcourses! Let me know if you need the website.

I WW 12 due Thursday

I Final Exam: Dec 12, 8:30 am

Warm-up

Q1. What is the period, T , of h(t) = 8− 6 sin(4t)?

A. T = 4

B. T = 1/4

C. T = 1

D. T = 2π

E. T = π/2

Warm-up

Q2. What is the amplitude of h(t) = 8− 6 sin(4t)?

A. 14

B. 8

C. 12

D. 6

E. -6

Warm-up

Q3. What is the derivative of tan(ex2

)?

A. sec2(ex2

)

B. sec2(ex2

)ex2

C. sec2(ex2

)ex2

x2

D. sec2(ex2

)ex2

2x

E. sec2(ex2

)ex2

x22x

Derivatives of Trigonometric Functions

I We already know thatI d

dxsinx = cosx

I ddx

cosx = − sinx

I We used the quotient rule to find

d

dxtanx =

d

dx

(sinx

cosx

)= sec2 x

Derivatives of Trigonometric Functions

I What is the derivative of cot(x)?

d

dxcotx =

d

dx

(cosxsinx

)=− sinx sinx− cosx cosx

sin2 x

=−1

sin2 x= − csc2 x

Inverse Trigonometric FunctionsI f(x) = sin x

I f−1(x) = arcsin(x) ”the angle whose sine is x”14.3. Inverse Trigonometric functions 283

θ

1

1−x2

x

Figure 14.10. This triangle has been constructed so that θ is an angle whose sineis x/1 = x. This means that θ = arcsin(x)

The functions sin(x) and arcsin(x), reverse (or “invert”) each other’s effect, that is:

arcsin(sin(x)) = x for − π/2 ≤ x ≤ π/2,

sin(arcsin(x)) = x for − 1 ≤ x ≤ 1.

There is a subtle point that the allowable values of x that can be “plugged in” are not exactlythe same for the two cases. In the first case, x is an angle whose sine we compute first, andthen reverse the procedure. In the second case, x is a number whose arc-sine is an angle.

We can evaluate arcsin(sin(x)) for any value of x, but the result may not agree withthe original value of x unless we restrict attention to the interval −π/2 ≤ x ≤ π/2. Forexample, if x = π, then sin(x) = 0 and arcsin(sin(x)) = arcsin(0) = 0, which is not thesame as x = π. For the other case, i.e. for sin(arcsin(x)), we cannot plug in any value ofx outside of −1 ≤ x ≤ 1, since arcsin(x) is simply not define at all, outside this interval.This demonstrates that care must be taken in handling the inverse trigonometric functions.

Inverse cosine

We cannot use the same interval to restrict the cosine function, because it has the same yvalues to the right and left of the origin. If we pick the interval 0 ≤ x ≤ π, this difficultyis avoided, since we arrive at a one-to-one function. We will call the restricted-domainversion of cosine by the name y = Cos(x) = cos(x) for 0 ≤ x ≤ π. (See red curve inFigure 14.11(a). On the interval 0 ≤ x ≤ π, we have 1 ≥ Cos(x) ≥ −1 and we define thecorresponding inverse function

y = arccos(x) − 1 ≤ x ≤ 1

as shown in blue in Figure 14.11(b).We understand the meaning of the expression y = arccos(x) as “the angle (in radi-

ans) whose cosine is x”. For example, arccos(0.5) = π/3 because π/3 is an angle whosecosine is 1/2=0.5. In Figure 14.12, we show a triangle constructed specifically so thatθ = arccos(x). Again, this follows from the fact that cos(θ) is adjacent over hypotenuse.The length of the third side of the triangle is obtained using the Pythagorian theorem.

The inverse relationship between the functions mean that

arccos(cos(x)) = x for 0 ≤ x ≤ π,

I The above triangle is constructed so thatsin θ = x

I This means that θ = arcsinx

Inverse Trigonometric Functions

Q4. Simplify the expression y = tan(arcsin(x))

A.√1− x2

B. x√1−x2

C. 1x

D. x

E.√1−x2

x

Inverse Trigonometric Functions

y = tan(arcsin(x))14.3. Inverse Trigonometric functions 283

θ

1

1−x2

x

Figure 14.10. This triangle has been constructed so that θ is an angle whose sineis x/1 = x. This means that θ = arcsin(x)

The functions sin(x) and arcsin(x), reverse (or “invert”) each other’s effect, that is:

arcsin(sin(x)) = x for − π/2 ≤ x ≤ π/2,

sin(arcsin(x)) = x for − 1 ≤ x ≤ 1.

There is a subtle point that the allowable values of x that can be “plugged in” are not exactlythe same for the two cases. In the first case, x is an angle whose sine we compute first, andthen reverse the procedure. In the second case, x is a number whose arc-sine is an angle.

We can evaluate arcsin(sin(x)) for any value of x, but the result may not agree withthe original value of x unless we restrict attention to the interval −π/2 ≤ x ≤ π/2. Forexample, if x = π, then sin(x) = 0 and arcsin(sin(x)) = arcsin(0) = 0, which is not thesame as x = π. For the other case, i.e. for sin(arcsin(x)), we cannot plug in any value ofx outside of −1 ≤ x ≤ 1, since arcsin(x) is simply not define at all, outside this interval.This demonstrates that care must be taken in handling the inverse trigonometric functions.

Inverse cosine

We cannot use the same interval to restrict the cosine function, because it has the same yvalues to the right and left of the origin. If we pick the interval 0 ≤ x ≤ π, this difficultyis avoided, since we arrive at a one-to-one function. We will call the restricted-domainversion of cosine by the name y = Cos(x) = cos(x) for 0 ≤ x ≤ π. (See red curve inFigure 14.11(a). On the interval 0 ≤ x ≤ π, we have 1 ≥ Cos(x) ≥ −1 and we define thecorresponding inverse function

y = arccos(x) − 1 ≤ x ≤ 1

as shown in blue in Figure 14.11(b).We understand the meaning of the expression y = arccos(x) as “the angle (in radi-

ans) whose cosine is x”. For example, arccos(0.5) = π/3 because π/3 is an angle whosecosine is 1/2=0.5. In Figure 14.12, we show a triangle constructed specifically so thatθ = arccos(x). Again, this follows from the fact that cos(θ) is adjacent over hypotenuse.The length of the third side of the triangle is obtained using the Pythagorian theorem.

The inverse relationship between the functions mean that

arccos(cos(x)) = x for 0 ≤ x ≤ π,

I θ = arcsin(x)

I y = tan θ = x√1−x2

Inversefunctions

Inversefunctionsonrestricteddomains

Onrestricteddomains

f(g(x))=xandg(f(x))=x

Inverse Trigonometric Functions

I f(x) = sin x

I f−1(x) = arcsin(x) ”the angle whose sine is x”

I Because sinx is periodic, we need to restrictthe domain to define an inverse function.

I https://www.desmos.com/calculator/

zx6c0htith

(2)DomainsThefunctionssin(x)andarcsin(x)areinversefunctionsonthefollowingdomains:(A)–π≤x≤πand-1≤x≤1(B) –π/2≤x≤π/2and-1≤x≤1(C)-1≤x≤1and–π/2≤x≤π/2(D)-1≤x≤1and–π≤x≤π(E)–π/2≤x≤π/2and–π/2≤x≤π/2

Derivatives of Inverse TrigonometricFunctions

Q6. If y = arcsinx, with −1 ≤ x ≤ 1, what is dydx?

A. arccosx

B. arctanx

C.√1− x2

D. 1√1−x2

E. 11+x2

Derivatives of Inverse TrigonometricFunctions

Q6. If y = arcsinx, with −1 ≤ x ≤ 1, what is dydx?

A. arccosx

B. arctanx

C.√1− x2

D. 1√1−x2

E. 11+x2

Derivatives of Inverse TrigonometricFunctions

1. y = arcsinx

2. Inverse: x = sin y

3. Implicit Differentiation: 1 = cos y dydx

4. Use identity or triangle to write

cos y =

√1− sin2 y =

√1− x2

5. Solve for dydx :

dy

dx=

1√1− x2

Extra Practice

1. What are the domains of f(x) = tan x andf−1(x) = arctan x?

2. Show that the derivative of arctanx is 11+x2

Derivatives of Inverse Trig Functions

Subtle points!

I Inverse trigonometric functions are onlyinverses of the trigonometric functions on theirrestricted domains!

I Note that

arcsin

(sin

(5π

6

))=π

6

Sin(5π/6)=1/2

Arcsin(1/2)=π/6

arcsin(sin(x))≠xInthisexample,wefoundthat:

arcsin(sin(5π/6))=π/6≠5π/6

However,ontherestricteddomains,–π/2≤x≤π/2and-1≤x≤1

arcsin(sin(x))=x andsin(arcsin(x))=x.

Trig and related rates

1. If the height of an isosceles triangle with base2m changes at a rate of 3 m/s, how quickly isthe angle opposite the base changing when theheight is

√3 m?

2. A ladder of length L is leaning against a wallso that its point of contact with the ground isa distance x from the wall, and its point ofcontact with the wall is at height y. The ladderslips away at a constant rate C. How fast doesthe angle between the ladder and wall change?

Summary

I Using differentiation rules, we can find thederivatives of other trigonometric functions

I Inverse trigonometric functions also havederivatives

I Triangles and identities are useful whensimplifying inverse trigonometric relationships

I Caution required: when thinking about thedomains of inverse trig functions

I Trigonometric functions are useful in relatedrates problems

Answers

1. E

2. D

3. D

4. B

5. B

Related Exam Problems

1. Use Newton’s method to find the smallestpositive critical point of the function

g(x) = e−x sin(10x)