MATH 161 Calculus I - Millersville University of …banach.millersville.edu/~bob/math161/derivtrig/main.pdfLimits and Trigonometry We can establish formally the limits of the trigonometric

Derivatives of the Trigonometric FunctionsMATH 161 Calculus I

J. Robert Buchanan

Department of Mathematics

Summer 2018

Background

We can establish formally the limits of the trigonometricfunctions using circles, angles, and geometry.

Recall:I The length s of the arc of a circle of radius r which

subtends a central angle θ is s = r θ.I The area A of a sector of a circle of radius r subtended by

a central angle θ is A = 12 r2θ.

I The coordinates of the point on the circle of radius rcentered at the origin at the intersection of a ray making anangle θ with the positive x-axis are

(x , y) = (r cos θ, r sin θ).

Limits and TrigonometryWe can establish formally the limits of the trigonometricfunctions using the unit circle and geometry.

θ

θ

sin θ

(cos θ,sin θ)

x

y

Basic Limits (1 of 2)

Lemmalimθ→0

sin θ = 0.

Proof.If 0 < θ < π

2 then

0 ≤ sin θ ≤ θlimθ→0

0 ≤ limθ→0

sin θ ≤ limθ→0

θ

0 ≤ limθ→0

sin θ ≤ 0

and use the Squeeze Theorem.

Basic Limits (2 of 2)

Lemmalimθ→0

cos θ = 1.

Proof.If 0 < θ < π

2 then

limθ→0

cos θ = limθ→0

√1− sin2 θ

=

√1−

(limθ→0

sin θ

)2

=√

1− 02 = 1.

Justification of a Common Limit (1 of 3)

θ

R(1,0)O

P(cos θ,sin θ)

Q(1,tan θ)

x

y

4OPR =12

(1) sin θ

4OQR =12

(1) tan θ

OPR =12

(12)θ

Justification of a Common Limit (2 of 3)

Lemma

limθ→0

sin θ

θ= 1

Justification of a Common Limit (3 of 3)

Proof.

0 < area ∆OPR < area OPR < area ∆OQR

0 <12

sin θ <θ

2<

12

tan θ

0 < sin θ < θ < tan θ

0 < 1 <θ

sin θ<

tan θ

sin θ=

1cos θ

cos θ <sin θ

θ< 1

limθ→0

cos θ < limθ→0

sin θ

θ< lim

θ→01

1 ≤ limθ→0

sin θ

θ≤ 1

and use the Squeeze Theorem.

Limit Involving cos θ

Lemma

limθ→0

1− cos θ

θ= 0

Proof.

limθ→0

1− cos θ

θ= lim

θ→0

(1− cos θ)

θ

(1 + cos θ)

(1 + cos θ)

= limθ→0

1− cos2 θ

θ(1 + cos θ)

= limθ→0

sin2 θ

θ(1 + cos θ)

= limθ→0

sin θ

θ

sin θ

1 + cos θ= (1)(0) = 0

Limit Involving cos θ

Lemma

limθ→0

1− cos θ

θ= 0

Proof.

limθ→0

1− cos θ

θ= lim

θ→0

(1− cos θ)

θ

(1 + cos θ)

(1 + cos θ)

= limθ→0

1− cos2 θ

θ(1 + cos θ)

= limθ→0

sin2 θ

θ(1 + cos θ)

= limθ→0

sin θ

θ

sin θ

1 + cos θ= (1)(0) = 0

Sum of Angles Formulas

We will need the following trigonometric identities known as thesum of angles formulas:

sin(A + B) = sin A cos B + cos A sin Bcos(A + B) = cos A cos B − sin A sin B

Basic Derivatives (1 of 2)

Theorem

ddx

[sin x ] = cos x

Proof

ddx

[sin x ] = limh→0

sin(x + h)− sin xh

= limh→0

sin x cos h + cos x sin h − sin xh

= limh→0

(sin x cos h − sin x) + cos x sin hh

= limh→0

sin x(cos h − 1) + cos x sin hh

= (sin x) limh→0

cos h − 1h

+ (cos x) limh→0

sin hh

= (sin x)(0) + (cos x)(1)

= cos x

Basic Derivatives (2 of 2)

Theorem

ddx

[cos x ] = − sin x

Proof

ddx

[cos x ] = limh→0

cos(x + h)− cos xh

= limh→0

cos x cos h − sin x sin h − cos xh

= limh→0

(cos h − 1) cos x − sin x sin hh

= limh→0

cos xcos h − 1

h− lim

h→0sin x

sin hh

= cos x(0)− sin x(1)

= − sin x

Derivatives of the Remaining Trigonometric Functions

ddx

[tan x ] = sec2 x

ddx

[cot x ] = − csc2 x

ddx

[sec x ] = sec x tan x

ddx

[csc x ] = − csc x cot x

Justification

We can find the derivative of the function f (x) = tan x using thequotient rule:

ddx

[tan x ] =ddx

[sin xcos x

]

=cos x cos x − sin x(− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x

=1

cos2 x= sec2 x .

Justification

We can find the derivative of the function f (x) = tan x using thequotient rule:

ddx

[tan x ] =ddx

[sin xcos x

]=

cos x cos x − sin x(− sin x)

cos2 x

=cos2 x + sin2 x

cos2 x

=1

cos2 x= sec2 x .

Examples

Find the derivatives of the following functions.I y = x sin xI y = x2 + sin2 xI y = 4x3 − cos(x2)

I y =√

tan x + 1I y = x3 sec2(4x)

I y = sin(cos x)

Solutions (1 of 3)

y = x sin xy ′ = (1) sin x + x cos x

= sin x + x cos x

y = x2 + sin2 xy ′ = 2x + 2 sin x(sin x)′

= 2x + 2 sin x cos x

Solutions (1 of 3)

y = x sin xy ′ = (1) sin x + x cos x

= sin x + x cos x

y = x2 + sin2 xy ′ = 2x + 2 sin x(sin x)′

= 2x + 2 sin x cos x

Solutions (2 of 3)

y = 4x3 − cos(x2)

y ′ = 12x2 − (− sin(x2))(x2)′

= 12x2 + 2x sin(x2)

y =√

tan x + 1

y ′ =12

(tan x + 1)−1/2(tan x + 1)′

=(tan x + 1)′

2(tan x + 1)1/2

=sec2 x

2√

tan x + 1

Solutions (2 of 3)

y = 4x3 − cos(x2)

y ′ = 12x2 − (− sin(x2))(x2)′

= 12x2 + 2x sin(x2)

y =√

tan x + 1

y ′ =12

(tan x + 1)−1/2(tan x + 1)′

=(tan x + 1)′

2(tan x + 1)1/2

=sec2 x

2√

tan x + 1

Solutions (3 of 3)

y = x3 sec2(4x)

y ′ = 3x2 sec2(4x) + x3(2 sec(4x))(sec(4x))′

= 3x2 sec2(4x) + 2x3 sec(4x)(sec(4x))′

= 3x2 sec2(4x) + 2x3 sec(4x)(sec(4x) tan(4x))(4x)′

= 3x2 sec2(4x) + 8x3 sec2(4x) tan(4x)

y = sin(cos x)

y ′ = cos(cos x)(cos x)′

= cos(cos x)(− sin x)

= − sin x cos(cos x)

Solutions (3 of 3)

y = x3 sec2(4x)

y ′ = 3x2 sec2(4x) + x3(2 sec(4x))(sec(4x))′

= 3x2 sec2(4x) + 2x3 sec(4x)(sec(4x))′

= 3x2 sec2(4x) + 2x3 sec(4x)(sec(4x) tan(4x))(4x)′

= 3x2 sec2(4x) + 8x3 sec2(4x) tan(4x)

y = sin(cos x)

y ′ = cos(cos x)(cos x)′

= cos(cos x)(− sin x)

= − sin x cos(cos x)

Applications

The trigonometric functions arise naturally in many mechanicaland electrical systems.

Suppose s(t) = a sin(ωt) + b cos(ωt) then

v(t) = aω cos(ωt)− bω sin(ωt)a(t) = −aω2 sin(ωt)− bω2 cos(ωt)

anda(t) + ω2s(t) = s′′(t) + ω2s(t) = 0.

Homework

I Read Section 2.6I Exercises: 1–35 odd