1.derivatives of trigonometric functions

7/28/2019 1.Derivatives of Trigonometric Functions

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DERIVATIVES OF

TRIGONOMETRIC FUNCTIONS

The derivative of sin x

The derivative of cos x

The derivative of tan x

The derivative of cot x

The derivative of sec x

The derivative of csc x

THE DERIVATIVE of sin x is cos x. To prove that, we will use thefollowing identity:

sinA sinB = 2 cos (A + B) sin (AB).

(Topic 20of Trigonometry.)

Problem 1. Use that identity to show:

sin (x+ h) sinx=

To see the proof, pass your mouse over the colored area.To cover the answer again, click "Refresh" ("Reload").Do the problem yourself first!

sin (x+ h) sinx= 2 cos (x+ h+ x) sin (x+ hx)

= 2 cos (2x+ h) sin h

=

Before going on to the derivative of sin x, however, we must provealemma; which is a preliminary, subsidiary theorem needed to prove aprinciple theorem. That lemma requires the following identity:
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Problem 2. Show that tan divided by sin is equal to1

cos :

tan sin

=1

cos .

(SeeTopic 20of Trigonometry.)

tan sin

= tan 1

sin = sin

cos

1sin

= 1cos

The lemma we have to prove is discussed inTopic 14of Trigonometry.(Take a look at it.) Here it is:

LEMMA. When is measured in radians, then

Proof. It is not possible to prove that by applying the usual theorems onlimits (Lesson 2). We have to go to geometry, and to the meanings ofsin and radian measure.

Let O be the center of a unit circle, that is, a circle of radius 1;and let be the first quadrant central angle BOA, measured in radians.

Then, since arc length s= r, and r= 1, arc BA is equal to . (Topic14of Trigonometry.)Draw angle B'OA equal to angle , thus making arcAB'equal to arcBA;

draw the straight line BB', cuttingAO at P;and draw the straight lines BC, B'Ctangent to the circle.

Then
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BB'< arc BAB'< BC+ CB'.

Now, in that unit circle, BP= PB'= sin , (Topic 17of Trigonometry),so that BB'= 2 sin ;

and BC= CB'= tan . (For, tan =BCOB

=B 1

= BC.)

The continued inequalityabovetherefore becomes:

2 sin < 2< 2 tan .

On dividing each term by 2 sin :

1 cos .

(Lesson 11 of Algebra,Theorem 5.)

On changing the signs, the sense changes again :

1 < sin

< cos,

(Lesson 11 of Algebra,Theorem 4),and if we add 1 to each term:

0 < 1 sin

< 1 cos.

Now, as becomes very close to 0 ( 0), cos becomes very close to1; therefore, 1 cos becomes very close to 0. The expression in themiddle, beinglessthan 1 cos, becomes even closer to 0 (and on the left isbounded by 0), therefore the expression in the middle will definitely approach0. This means:
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Which is what we wanted to prove.

The student should keep in mind that for a variable to "approach" 0 orany limit (Definition 2.1), does not mean that the variable ever equalsthat limit.

The derivative of sin x

ddx

sin x= cos x

To prove that, we will apply the definition of the derivative (Lesson 5).

First, we will calculate the difference quotient.

sin (x+ h) sinxh

= , Problem 1,

=, on dividing numeratorand denominator by 2,

=

We will now take the limit as h 0. But the limit of a product is equalto the product of the limits. (Lesson 2.) And the factor on the right has theform sin /. Therefore, according to theLemma, its limit is 1. Therefore,

ddxsin x= cos x.

We have established the formula.

The derivative of cos x

d cos x= sinx
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dx

To establish that, we will use the following identity:

cos x= sin (2

x).

A function of any angle is equal to the cofunction of its complement.(Topic 3of Trigonometry).

Therefore, on applying thechain rule:


The derivative of tan x

d

dx tan x= secx

Now, tan x=sin xcos x

. (Topic 20of Trigonometry.)

Therefore according to thequotient rule:d

dxtan x=

ddx

sin xcos x

=cos x cos x sinx(sinx)

cosx

=cosx+ sinx

cosx

=1

cosx

= secx.

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Problem 3. The derivative of cot x. Prove:ddx

cot x= cscx

d

dxcot x = d

dx

cos x

sin x= sin x(sinx) cosx cosx

sinx

= (sinx+ cosx)sinx

= 1

sinx

= cscx.

The derivative of sec x

ddx

sec x = sec xtan x

Since sec x=1

cos x= (cos x)1 , then, on using thechain ruleand

the generalpower rule:


Problem 4. The derivative of csc x. Prove:d csc x= cscxcot x
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dx

ddx

csc x=ddx

1sin x

=

=

=

=

=

Example. Calculate the derivative of sin ax.

Solution. On applying thechain rule,dd

sin ax = cos axddx

ax = cos ax 2ax = 2axcos ax.

Problem 5. Calculate these derivatives.

a)ddxsin 5x = 5 cos 5

b)d

dx sinx = sin xcos

c)dd

2 cos 3x = 6 sin 3x

d)d

dxxcos x = cos xxsin

e)dd

sin 2xcos x = 2 cos 2xcos x sin 2xsin

f)ddx

tan (3x) = 18xsec (3x)
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g)d

dx2 cot

2= csc2

h)d

dxsec 4x = 4 sec 4x tan 4

i)d

dxacsc bx = abcsc bxcot b

j) =

Problem 6. ABC is a right angle, and the straight line AD is rotatingso that the angle is increasing in the positivedirection. At what rate -- how many radians persecond -- is it increasing if BC is constant at 3 cm,and AB (call it x) is decreasing at the rate of3 cm/sec, and its length is 6 cm?

Therefore,


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1.derivatives of trigonometric functions

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