2.8 derivatives of the trigonometric functionsmath.buffalostate.edu/dcunningham/161sec2-8.pdf2.8...

2.8 Derivatives of the Trigonometric FunctionsWe need to investigate the differentiability of sine and cosine. We will temporarily denotethe angle under consideration by t, which is also the length of the subtended arc (see figure)on the unit circle, measured from the starting point (1, 0). For t strictly between 0 and ⇡

2

,there are equalities and inequalities that we obtain from considering following unit circle.

h

tsin(t)

0 < sin(t) < t <

sin(t)cos(t)

limt!0

sin(t) = 0

limt!0

cos(t) = 1

sin(t)t

< 1 cos(t) <

sin(t)t

< 1

limt!0

sin(t)t

= 1

0 < sin(t) < t

length of DB = sin(t)

0 < length of DB < t < h

h =

sin(t)

cos(t)

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Since the above circle is the unit circle, the angle t is the length of the corresponding arc,and (N) sin(t) = DB

1

= DB. Similarly, cos(t) = 0D

1

= 0D. Thus, the slope of the line segment0B is equal to DB

0D

=

sin(t)

cos(t)

. The slope of this line is also equal to h

1

= h. Hence, (?) sin(t)

cos(t)

= h.As t is the length of the subtended arc (see figure), we conclude that

0 < DB < t < h.

Thus, from (N) and (?) we obtain

0 < sin(t) < t <

sin(t)

cos(t)

.

Because 0 < sin(t) < t and lim

t!0

t = 0, the Pinching Theorem 1.3.15 implies that limt!0

sin(t) = 0.Since sin

2

(t) + cos

2

(t) = 1, it follows that (†) lim

t!0

cos(t) = 1. Since 0 < sin(t) < t, we see

that sin(t)

t

< 1. As cos(t) <

sin(t)

t

< 1, (†) and the Pinching Theorem 1.3.15 imply thatlim

t!0

sin(t)

t

= 1. This discussion implies our next theorem.

79

Theorem 2.8.1. The following equations hold:

lim

✓!0

sin(✓) = 0, lim

✓!0

cos(✓) = 1, lim

✓!0

sin(✓)

✓

= 1;

and therefore,lim

✓!0

sin(✓) = sin(0) and lim

✓!0

cos(✓) = cos(0).

Theorem 2.8.2. The following equation holds:

lim

✓!0

cos(✓) � 1

✓

= 0.

Proof. We shall derive the desired limit as follows:

lim

✓!0

cos(✓) � 1

✓

= lim

✓!0

✓cos(✓) � 1

✓

· cos(✓) + 1

cos(✓) + 1

◆

= lim

✓!0

cos

2

(✓) � 1

✓(cos(✓) + 1)

by algebra

= lim

✓!0

� sin

2

(✓)

✓(cos(✓) + 1)

because sin

2

(✓) + cos

2

(✓) = 1

= � lim

✓!0

sin

2

(✓)

✓(cos(✓) + 1)

by limit law

= � lim

✓!0

sin(✓)

✓

· sin(✓)

cos(✓) + 1

by algebra

= � lim

✓!0

sin(✓)

✓

· lim✓!0

sin(✓)

cos(✓) + 1

by limit law

= �1 · lim✓!0

sin(✓)

cos(✓) + 1

by above Theorem 2.8.1

= �1 · sin(0)

cos(0) + 1

by Theorem 2.8.1

= �1 · 0

1 + 1

= 0 because sin(0) = 0 and cos(0) = 1

and this completes the proof.

Theorem 2.8.3. All of the trigonometric functions are differentiable on their domains withthe following derivatives:

(1) sin

0(x) = cos(x)

(2) cos

0(x) = � sin(x)

(3) tan

0(x) = sec

2

(x)

(4) cot

0(x) = � csc

2

(x)

(5) sec

0(x) = sec(x) tan(x)

(6) csc

0(x) = � csc(x) cot(x)

80

Proof. We shall show that all of the trigonometric functions are differentiable on their do-mains and compute their derivatives as follows:

(1) Let f(x) = sin(x). Then

f

0(x) = lim

h!0

f(x+ h) � f(x)

h

by definition of derivative

= lim

h!0

sin(x+ h) � sin(x)

h

by definition of f

= lim

h!0

sin(x) cos(h) + cos(x) sin(h) � sin(x)

h

by trig. identity (Sct. 2.7.5)

= lim

h!0

sin(x)(cos(h) � 1) + cos(x) sin(h)

h

by algebra

= lim

h!0

✓sin(x)(cos(h) � 1)

h

+

cos(x) sin(h)

h

◆by algebra

= lim

h!0

✓sin(x)(cos(h) � 1)

h

◆+ lim

h!0

✓cos(x) sin(h)

h

◆by limit law

= sin(x) lim

h!0

✓cos(h) � 1

h

◆+ cos(x) lim

h!0

✓sin(h)

h

◆by limit law

= sin(x) · 0 + cos(x) · 1 = cos(x) by Theorems 2.8.2 and 2.8.1.

(2) Let f(x) = cos(x). By the method used in (1), one can show that f 0(x) = � sin(x).

(3) Let f(x) = tan(x). Then

f

0(x) = tan

0(x)

=

✓sin(x)

cos(x)

◆0

by trig. identity (Sct. 2.7.5)

=

cos(x) sin

0(x) � sin(x) cos

0(x)

cos

2

(x)

by quotient rule

=

cos(x) cos(x) � sin(x)(� sin(x))

cos

2

(x)

by (1) and (2) above

=

cos

2

(x) + sin

2

(x)

cos

2

(x)

by algebra

=

1

cos

2

(x)

by trig. identity (Sct. 2.7.5)

= sec

2

(x) by trig. identity (Sct. 2.7.5)

(4) Let f(x) = cot(x). By the method used in (3), one can show that f 0(x) = � csc

2

(x).

81

(5) Let f(x) = sec(x). Then

f

0(x) = sec

0(x)

=

✓1

cos(x)

◆0

by trig. identity (Sct. 2.7.5)

= �� sin(x)

cos

2

(x)

by reciprocal rule

=

sin(x)

cos

2

(x)

by algebra

=

1

cos(x)

sin(x)

cos(x)

by algebra

= sec(x) tan(x) by trig. identities (Sct. 2.7.5).

(6) Let f(x) = csc(x). One can similarly show that f 0(x) = � csc(x) cot(x).

This completes the proof.Problem 19. Let f(x) = x sin(5x) and g(x) = e

cos(x). Evaluate f

0(x) and g

0(x).

Solution. First we evaluate f

0(x) using the product rule:

f

0(x) = x cos(5x)5 + sin(5x) · 1 = 5x cos(5x) + sin(5x).

Now we evaluate g

0(x) by using the chain rule:

g

0(x) = e

cos(x)

sin(x).

Problem 20. Evaluate⇣

sin(x)

x

⌘0.

Problem 21. Evaluate (ln(cos(x)))

0 and⇣5 cos(3x) +

sin(x)

x

⌘0.

Solution.

(ln(cos(x)))

0=

� sin(x)

cos(x)

= � sin(x)

cos(x)

.

✓5 cos(3x) +

sin(x)

x

◆0

= 5(� sin(3x)3) +

x cos(x) � sin(x)

x

2

= �15 sin(3x) +

x cos(x) � sin(x)

x

2

.

Problem 22. Evaluate⇣

cot(x)

x

⌘0.

Problem 23. Evaluate (x tan(x))

0, (7 csc(3x))0, and⇣sec(x) +

cot(x)

x

⌘0.

Solution.

(x tan(x))

0= x tan

0(x) + tan(x)(x)

0= x sec

2

(x) + tan(x).

(7 csc(3x))

0= 7(� csc(3x) cot(3x))3 = �21 csc(3x) cot(3x).

✓sec(x) +

cot(x)

x

◆0

= sec(x) tan(x) +

x(� csc

2

(x)) � cot(x)

x

2

= sec(x) tan(x) +

�x csc

2

(x) � cot(x)

x

2

.

82