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2.4. TRIGONOMETRIC FUNCTIONS (3.4)

83

2.4

Trigonometric functions (3.4)

In this section, we study di¤erentiation of the trigonometric functions sin x, cos x; tan x, cot x, sec x, csc x. Before we derive the formulae for the derivative of these functions, we must ...rst revisit limits.

2.4.1

Fundamental Trigonometric Limits

When deriving the formula for the derivative of sin x and cos x, two limits play sin x cos x 1 an important role. They are lim and lim . These are not limits x!0 x x!0 x we can evaluate by using the rules we have developed earlier. To ...rst get an idea of what the value of these limits might be, we look at a table of value. x :1 :01 :001 :1 :01 :001 sin x 0:998 334 0:999 983 0:999 999 0:998 334 0:999 983 0:999 999 x sin x = 1. We are going to prove this result by It therefore appears that lim x!0 x using the squeeze theorem. The goal, when using the squeeze theorem, is to ...nd

Figure 2.1: Trigonometric Function in a Unit Circle sin x x

two functions f and g such that f (x)

g (x), such that lim f (x) =

x!0

84

CHAPTER 2. RULES OF DIFFERENTIATION

sin x = L. We x ...nd these functions using geometry. From ...gure 2.1, one can see that

x!0

lim g (x) = L. By the squeeze theorem, it will follow that lim

x!0

Area of triangle OAP

Area of sector OAP

Area of triangle OAQ (2.1) base height, and that we

1 But, remembering that the area of a triangle is 2 have a unit circle, that is OA = 1, we see that Area of triangle OAP = 1 1 2 sin x = 2

sin x

(2.2)

and Area of triangle OAQ = 1 1 tan x 2 tan x = 2 sin x = 2 cos x (2.3)

Also remembering (see your calculus book, on the ...rst page) that the area of 1 the sector of angle in a circle of radius r is r2 , since we are in a unit circle, 2 we see that 1 Area of sector OAP = x (2.4) 2 Using equations 2.2, 2.3 and 2.4 in equation 2.1, we have sin x 2 Multiplying each side by 2 gives sin x Dividing each side by sin x gives 1 x sin x 1 cos x x sin x cos x 1 x 2 sin x 2 cos x

Since lim cos x = 1, it follows that lim

x!0

theorem,

x!0

1 = 1. Hence, by the squeeze x!0 cos x

lim

x =1 sin x

2.4. TRIGONOMETRIC FUNCTIONS (3.4)

85

This is not the limit we wanted. We can now get the limit we wanted using our limit rules. sin x 1 lim = lim x x!0 x x!0 sin x 1 = x lim x!0 sin x 1 = 1 =1 We are now ready to ...nd lim lim cos x x 1

x!0

cos x x

x!0

1

.

x!0

= lim

(cos x 1) (cos x + 1) x (cos x + 1) 2 cos x 1 = lim x!0 x (cos x + 1) sin2 x x!0 x (cos x + 1) sin x sin x = lim x!0 x cos x + 1 sin x sin x = lim lim x!0 cos x + 1 x x!0 0 = ( 1) 1+1 = lim

=0 Therefore, we have proven the following theorem: Theorem 130 (Fundamental Trigonometric Limits) sin x =1 x cos x 1 lim =0 x!0 x

x!0

lim

(2.5)

2.4.2

Di¤erentiation of Sine

We are now ready to ...nd the derivative of sin x, using the de...nition. Let us ...rst remind the reader of some trigonometric identities which will be useful sin (a + b) = sin a cos b + sin b cos a sin (a b) = sin a cos b sin b cos a cos (a + b) = cos a cos b sin a sin b cos (a b) = cos a cos b + sin a sin b (2.6)

86

CHAPTER 2. RULES OF DIFFERENTIATION

sin (x) = lim

0

sin (x + h) sin (x) h!0 h sin (x) cos (h) + sin (h) cos (x) sin (x) = lim (using equation 2.6) h!0 h sin (x) (cos (h) 1) + sin (h) cos (x) = lim h!0 h sin (x) (cos (h) 1) sin (h) cos (x) = lim + lim h!0 h!0 h h

Since we are taking the limit as h ! 0, both sin x and cos x are constants with respect to this limit. Therefore, sin (x) = sin x lim

0

cos (h) 1 sin (h) + cos x lim h!0 h h = sin x (0) + cos x (1) (by equation 2.5) = cos x

h!0

2.4.3

Di¤erentiation of Cosine

0

The derivation of the formula for (cos x) is almost identical to that of sin x and is left as an exercise. We have: cos (x) = You will notice the minus sign.

0

sin x

2.4.4

Di¤erentiation of Tangent

sin x . Therefore, cos x sin x cos x

0 0

By de...nition, tan x = (tan x) = =

0

(sin x) cos x sin x (cos x) (quotient rule) cos2 x cos2 x sin2 x = cos2 x 1 = cos2 x = sec2 x

0

2.4.5

Di¤erentiation of the Other Trigonometric Functions

Since the remaining trigonometric functions can be expressed as ratios of sin x and cos x, we can use the rules we have to derive the formula for their derivatives.

2.4. TRIGONOMETRIC FUNCTIONS (3.4)

87

The derivation of these formulae is left as an exercise. You will recall that cos x 1 1 1 cot x = = , sec x = , csc x = . We have: sin x tan x cos x sin x (cot x) =

0 0 0

csc2 x

(sec x) = sec x tan x (csc x) = csc x cot x

2.4.6

Summary for this Subsection

0

Theorem 131 We have proven and you must know the following rules: d sin x = cos x dx d cos x 0 2. (cos x) = sin x or = sin x dx d tan x 0 3. (tan x) = sec2 x or = sec2 x dx d cot x 0 4. (cot x) = csc2 x or = csc2 x dx d sec x 0 5. (sec x) = sec x tan x or = sec x tan x dx d csc x 0 6. (csc x) = csc x cot x or = csc x cot x dx Remark As in the formulas studied before, it is important to understand that x should not be taken literally. What matters is that the variable of di¤erentiation be the same as the variable in the function. In other words, d sin x d sin u d sin we have = cos x but we also have = cos u or = cos , dx du d d sin u and so on. On the other hand, cannot be done using this rule dx because the variable of the function is not the variable of di¤erentiation. If u is some function of x, then we have to wait until the next section before we can do it. This means that we cannot ...nd derivatives such as d sin x2 d cos (ex ) d tan (2x) or or even . On the other hand, as noticed dx dx dx d sin u = 0. several times above, if u is not a function of x, then dx 1. (sin x) = cos x or Examples Example 132 Find f 0 (x) for f (x) = x sin x. f 0 (x) = (x sin x)

0 0 0

= (x) sin x + x (sin x) (product rule) = sin x + x cos x

88

CHAPTER 2. RULES OF DIFFERENTIATION

Example 133 Find f 0 (x) for f (x) = sin x cos x f 0 (x) = (sin x cos x)

0 0 0

= (sin x) cos x + sin x (cos x) (product rule) = cos2 x sin2 x 1 sec x tan x

0

Example 134 Find f 0 (x) for f (x) = f 0 (x) = = = = = (1

0

sec x) tan x (1 sec x) (tan x) tan2 x (0 sec x tan x) tan x (1 sec x) sec2 x tan2 x 2 sec x tan x sec2 x + sec3 x tan2 x 2 sec x sec x tan2 x sec2 x tan2 x sec x (1 sec x) (recall that 1 + tan2 x = sec2 x) tan2 x

Example 135 Find the slope of the tangent to f (x) = cos x at x = . 3 This is a classical problem. We must ...nd a point on the tangent and the slope of the tangent. When x = ,y=f 3 on the tangent. = cos = 1 . Therefore, 2 is f 0 1 ; 3 2 is a point

3

3

The slope of the tangent to f (x) at x = f 0 (x) = sin x and therefore f 0 3 =

3

. Since f (x) = cos x, 3 p 3 sin = . 3 2

It follows that the equation of the tangent is p 1 3 y = x 2 2 3

2.4.7

Problems

1. Be able to do problems such as # 1, 3, 5, 7, 9, 17, 25, 27, 35, 36 on pages 218, 219. 2. Find lim

x!0

sin 5x (Answer: 1) 5x sin 5x (Answer: 5) x

3. Find lim

x!0

2.4. TRIGONOMETRIC FUNCTIONS (3.4) 4. lim sin x 1 (Answer: ) 5x 5

0

89

x!0

5. Prove that (cos x) = 6. Prove that (cot x) =

0 0 0

sin x csc2 x

7. Prove that (sec x) = sec x tan x 8. Prove that (csc x) = csc x cot x

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