Lecture 9 : Derivatives of Trigonometric Functions ...

[Pages:9]Lecture 9 : Derivatives of Trigonometric Functions (Please review Trigonometry under Algebra/Precalculus Review on the class webpage.) In this section we will look at the derivatives of the trigonometric functions

sin x, cos x, tan x , sec x, csc x, cot x.

Here the units used are radians and sin x = sin(x radians). Recall that sin x and cos x are defined and continuous everywhere and

sin x

1

1

cos x

tan x = , sec x = , csc x = , cot x = ,

cos x

cos x

sin x

sin x

are continuous on their domains (all values of x where the denominator is non-zero). The graphs of the above functions are shown at the end of this lecture to help refresh your memory: Before we calculate the derivatives of these functions, we will calculate two very important limits.

First Important Limit

sin lim = 1. 0

See the end of this lecture for a geometric proof of the inequality,

shown in the picture below for > 0,

?1

? 0.5

From this we can easily derive that

sin < < tan .

1.6

1.4

1.2

1

0.8

D

0.6

B

0.4

0.2

O

!

? 0.2

1

|AD| = tan !

E

sin ! !

0.5

C 1A

1.5

? 0.4

? 0.6

? 0.8

?1

? 1.2

? 1.4

? 1.6

sin

cos < < 1

and we can use the squeeze theorem to prove that the limit shown above is 1.

1

Another Important Limit From the above limit, we can derive that :

cos - 1

lim

=0

0

Example Calculate the limits:

sin 5x

lim

,

x0 sin 3x

sin(x3)

lim

.

x0 x

Derivatives of Trigonometric Functions d

1. From our trigonometric identities, we can show that sin x = cos x : dx

d

sin(x + h) - sin(x)

sin(x) cos(h) + cos(x) sin(h) - sin(x)

sin x = lim

= lim

=

dx

h0

h

h0

h

sin(x)[cos(h) - 1] + cos(x) sin(h)

[cos(h) - 1]

sin(h)

lim

= lim sin(x)

+ lim cos(x)

h0

h

h0

h

h0

h

[cos(h) - 1]

sin(h)

= sin(x) lim

+ cos(x) lim

= cos(x).

h0

h

h0 h

d 2. We can also show that cos x = - sin(x) :

dx

d

cos(x + h) - cos(x)

cos(x) cos(h) - sin(x) sin(h) - cos(x)

cos x = lim

= lim

=

dx

h0

h

h0

h

2

cos(x)[cos(h) - 1]

sin(x) sin(h)

= lim

- lim

h0

h

h0

h

[cos(h) - 1]

sin(h)

= cos(x) lim

- sin(x) lim

= - sin(x).

h0

h

h0 h

3. Using the derivatives of sin(x) and cos(x) and the quotient rule, we can deduce that d tan x = sec2(x) : dx

Example Find the derivative of the following function: 1 + cos x

g(x) = x + sin x

Higher Derivatives We see that the higher derivatives of sin x and cos x form a pattern in that they repeat with a cycle of four. For example, if f (x) = sin x, then

f (x) = cos x, f (x) = - sin x, f (3)(x) = - cos x, f (4)(x) = sin x, f (5)(x) = cos x, . . .

(Note the derivatives follow a similar pattern for cos(x). )

Example Let f (x) = sin x. What is

f (20)(x)?

3

A mass on a spring released at some point other than its equilibrium position will follow a pattern of simple harmonic motion (x(t) = A sin(Cx + D) or equivalently x(t) = A cos(Cx + D) ), when there is no friction or other forces to dampen the effect. The values of A, C and D depend on the elasticity of the spring, the mass and the point at which the mass is released. You will be able to prove this easily

later when you learn about differential equations.

Example An object at the end of a vertical spring is stretched 5cm beyond its rest position and released at time t = 0. Its position at time t is given by x(t) with the positive direction as shown in a downward direction, where

x(t) = 5 cos(t).

(a) Find the velocity and acceleration at time t.

(b)

Find the position,

velocity and acceleration

of the mass at time t =

4

.

In which direction is it

moving at that time?

4

The following is a summary of the derivatives of the trigonometric functions. You should be able to verify all of the formulas easily.

d sin x = cos x,

dx

d cos x = - sin x,

dx

d tan x = sec2 x dx

d csc x = - csc x cot x,

dx

d sec x = sec x tan x,

dx

d cot x = - csc2 x dx

Example The graph below shows the variations in day length for various degrees of Lattitude.

At 60o North, at what times of the year is the length of the day changing most rapidly?

Extras

Example (Preparation for Related Rates) A police car is parked 40 feet from the road at the

point P in the diagram below. Your vehicle is approaching on the road as in the diagram below and the

police are pointing a radar gun at your car. Let x denote the distance from your car to the police car

and let be the angle between the line of sight of the radar gun and the road. How fast is x changing

with respect to when =

4

?

(Please attempt this problem before looking at the solution on the

following page.)

x !

P 40ft

5

Solution We have that the variables x and are related in the following way:

Therefore

40 = sin().

x 40 =x

sin()

and

When

=

4

,

dx

- cos()

= 40 d

sin2()

.

dx d

=

4

= 40

-

cos(

4

)

sin2(

4

)

-1/ 2

= 40

= -40 2

1/2

feet per radian.

18

16

14

18

18

12

16

Graphs of Trigon16 ometric functions

10

? 4! ? 4!

? 3! ? 3!

? 2! ? 2!

14

12

10

8

6

4

18 2

?! ?!

16

14 ? 2

12 ?4

10 ?6

8 ?8

6 ? 10 4 ? 12 2 ? 14

? 16 ?2

? 18 ?4

?6

f(x) = sin(x)

!

2!

3!

4!

? 4!

? 3!

1 s(x) =

sin(x)

!

2!

3!

? 4!

? 3!

? 2! ? 2!

14

12

10

8

6

4

182

16 ?!

1?42 12 ?4 10 ?6

8 ?8 6 ? 10 4 ? 12 2 ? 14 ?! ??126

??148

?6

8 6 4

h(x) = tan(x)

2

? 4!

? 3!

g(x) = cos(x)

!

2!

3!

4!

1 r(x) = cos(x)

? 4!

? 3!

!

2!

3!

4!

? 2! ? 2!

?!

18

?2 16

?4 14

?6 12

?8 10

8? 10

6? 12

4? 14

2? 16

? 18 ?!

?2

?4

?6

!

2!

3!

4!

1 t(x) =

tan(x)

!

2!

3!

4!

?8

?8

?8

? 10

? 10

? 10

? 12

? 12

? 12

? 14 ? 16 ? 18

? 14 ? 14

? 16 ? 16

? 18 ? 18

6

Inequality

Let

be

an

angle

close

to

0,

and

between

0

and

2

.

Note that since sin = - sin(-), we have

sin

=

sin(-) -

and

lim0+

sin

=

lim0-

sin

.

Because

of

this,

we

need

only

consider

the

right

hand

limit,

lim0+

sin

with

>

0.

In the picture below, we see that , which is the length of the arc of the unit circle from A to B in

larger than the length of the line segment from A to B. The line segment from A to B is larger than

sin since it is the hypotenuse of a right triangle1.4with a side of length sin .

1.2

1

0.8

B

0.6

0.4 1

sin ! 0.2

O

!

?1

? 0.5

0.5

1A

? 0.2

? 0.4

? 0.6

? 0.8

?1

? 1.2

From this we can conclude that sin < or ?1.4

sin < 1.

Now consider the picture below. We can see intuitively that the length of the arc of the unit circle from A to B is smaller than the sum of the lengths of the line segments |AE| + |EB|. Because the line segment EB is a side of a right triangle with hypotenuse ED, we see that |EB| < |ED|. Thus we have

< |AE| + |EB| < |AE| + |ED| = |AD|

Note

now

that

|AD| |OA|

=

tan

and

|AD|

=

|OA| tan1.4

=

tan .

1.2

1

0.8

D

0.6

B

0.4

1

E

sin ! ! 0.2

O

!

?1

? 0.5

? 0.2

0.5

C 1A

? 0.4

? 0.6

? 0.8

?1

We now have that

? 1.2

? 1.4

sin

< tan =

giving

cos

sin cos <

since cos > 0 (when we multiply by positive numbers, inequalities are preserved).

Putting both inequalities together we get

sin cos < < 1

7

1. Calculate

Extra Problems

sin(x3)

lim

.

xo x

2. Calculate

lim 7x cot(3x).

x0

3. If g(x) = cos(x), what is g(42)(x)? 4. Find f (x) if f (x) = x2 cos(x) sin(x).

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