MATH 201 - Week 11

MATH 201 - Week 11

Ferenc Balogh

Concordia University

2008 Winter

Based on the textbook

J. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson

MATH 201 - Week 11

Overview

1 Addition and Subtraction Formulas - Section 7.2FormulasApplicationsA sin x + B cos x

2 Double-Angle, Half-Angle and Product-Sum Formulas - Section7.3

Squares of Trigonometric FunctionsHalf-Angle FormulasProduct-Sum Formulas

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Formulas

Addition Formulas

sin(s + t) = sin s cos t + cos s sin t

cos(s + t) = cos s cos t − sin s sin t

tan(s + t) =tan s + tan t

1− tan s tan t

Subtraction Formulas

sin(s − t) = sin s cos t − cos s sin t

cos(s − t) = cos s cos t + sin s sin t

tan(s − t) =tan s − tan t

1 + tan s tan t

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Formulas

Addition Formulas

sin(s + t) = sin s cos t + cos s sin t

cos(s + t) = cos s cos t − sin s sin t

tan(s + t) =tan s + tan t

1− tan s tan t

Subtraction Formulas

sin(s − t) = sin s cos t − cos s sin t

cos(s − t) = cos s cos t + sin s sin t

tan(s − t) =tan s − tan t

1 + tan s tan t

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Formulas

Example. Prove the addition formula for the tangent using theaddition formulas for the sine and cosine functions.

Solution.

tan(s + t) =sin(s + t)

cos(s + t)

=sin s cos t + cos s sin t

cos s cos t − sin s sin t

=cos s cos t

cos s cos t

tan s + tan t

1− tan s tan t

=tan s + tan t

1− tan s tan t.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Formulas

Example. Prove the addition formula for the tangent using theaddition formulas for the sine and cosine functions.Solution.

tan(s + t) =sin(s + t)

cos(s + t)

=sin s cos t + cos s sin t

cos s cos t − sin s sin t

=cos s cos t

cos s cos t

tan s + tan t

1− tan s tan t

=tan s + tan t

1− tan s tan t.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Formulas

Example. Prove the addition formula for the tangent using theaddition formulas for the sine and cosine functions.Solution.

tan(s + t) =sin(s + t)

cos(s + t)

=sin s cos t + cos s sin t

cos s cos t − sin s sin t

=cos s cos t

cos s cos t

tan s + tan t

1− tan s tan t

=tan s + tan t

1− tan s tan t.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Formulas

Example. Prove the addition formula for the tangent using theaddition formulas for the sine and cosine functions.Solution.

tan(s + t) =sin(s + t)

cos(s + t)

=sin s cos t + cos s sin t

cos s cos t − sin s sin t

=cos s cos t

cos s cos t

tan s + tan t

1− tan s tan t

=tan s + tan t

1− tan s tan t.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Formulas

Example. Prove the addition formula for the tangent using theaddition formulas for the sine and cosine functions.Solution.

tan(s + t) =sin(s + t)

cos(s + t)

=sin s cos t + cos s sin t

cos s cos t − sin s sin t

=cos s cos t

cos s cos t

tan s + tan t

1− tan s tan t

=tan s + tan t

1− tan s tan t.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Find the exact value of sin 5π12 and tan π

12 .

Solution. Since 5π12 = 3π

12 + 2π12 = π

4 + π6 we have

sin5π

12= sin

(π4

+π

6

)= sin

(π4

)cos(π

6

)+ cos

(π4

)sin(π

6

)=

1√2

√3

2+

1√2

1

2=

√3 + 1

2√

2.

Similarly, using π12 = π

3 −π4 we get

tan( π

12

)=

tan π3 − tan π

4

1 + tan π3 tan π

4

=

√3− 1

1 +√

3 · 1=

√3− 1

1 +√

3.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Find the exact value of sin 5π12 and tan π

12 .

Solution. Since 5π12 = 3π

12 + 2π12 = π

4 + π6 we have

sin5π

12= sin

(π4

+π

6

)= sin

(π4

)cos(π

6

)+ cos

(π4

)sin(π

6

)=

1√2

√3

2+

1√2

1

2=

√3 + 1

2√

2.

Similarly, using π12 = π

3 −π4 we get

tan( π

12

)=

tan π3 − tan π

4

1 + tan π3 tan π

4

=

√3− 1

1 +√

3 · 1=

√3− 1

1 +√

3.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Find the exact value of sin 5π12 and tan π

12 .

Solution. Since 5π12 = 3π

12 + 2π12 = π

4 + π6 we have

sin5π

12= sin

(π4

+π

6

)= sin

(π4

)cos(π

6

)+ cos

(π4

)sin(π

6

)=

1√2

√3

2+

1√2

1

2=

√3 + 1

2√

2.

Similarly, using π12 = π

3 −π4 we get

tan( π

12

)=

tan π3 − tan π

4

1 + tan π3 tan π

4

=

√3− 1

1 +√

3 · 1=

√3− 1

1 +√

3.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Find the exact value of

sin 20◦ cos 70◦ + cos 20◦ sin 70◦.

Solution.

sin 20◦ cos 70◦ + cos 20◦ sin 70◦ = sin(20◦ + 70◦)

= sin 90◦ = 1.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Find the exact value of

sin 20◦ cos 70◦ + cos 20◦ sin 70◦.

Solution.

sin 20◦ cos 70◦ + cos 20◦ sin 70◦ = sin(20◦ + 70◦)

= sin 90◦ = 1.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Prove the cofunction identity

sin(π

2− x)

= cos x .

Solution.

LHS = sin(π

2− x)

= sinπ

2cos x − cos

π

2sin x

= 1 · cos x − 0 · sin x

= cos x = RHS .

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Prove the cofunction identity

sin(π

2− x)

= cos x .

Solution.

LHS = sin(π

2− x)

= sinπ

2cos x − cos

π

2sin x

= 1 · cos x − 0 · sin x

= cos x = RHS .

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Prove the cofunction identity

sin(π

2− x)

= cos x .

Solution.

LHS = sin(π

2− x)

= sinπ

2cos x − cos

π

2sin x

= 1 · cos x − 0 · sin x

= cos x = RHS .

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Prove the cofunction identity

sin(π

2− x)

= cos x .

Solution.

LHS = sin(π

2− x)

= sinπ

2cos x − cos

π

2sin x

= 1 · cos x − 0 · sin x

= cos x = RHS .

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Prove the identity

sin(x + y)− sin(x − y) = 2 cos x sin y .

Solution.

LHS = sin(x + y)− sin(x − y)

= (sin x cos y + cos x sin y)− (sin x cos y − cos x sin y)

= sin x cos y + cos x sin y − sin x cos y + cos x sin y

= 2 cos x sin y = RHS .

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Prove the identity

sin(x + y)− sin(x − y) = 2 cos x sin y .

Solution.

LHS = sin(x + y)− sin(x − y)

= (sin x cos y + cos x sin y)− (sin x cos y − cos x sin y)

= sin x cos y + cos x sin y − sin x cos y + cos x sin y

= 2 cos x sin y = RHS .

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Prove the identity

sin(x + y)− sin(x − y) = 2 cos x sin y .

Solution.

LHS = sin(x + y)− sin(x − y)

= (sin x cos y + cos x sin y)− (sin x cos y − cos x sin y)

= sin x cos y + cos x sin y − sin x cos y + cos x sin y

= 2 cos x sin y = RHS .

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

Applications

Example. Prove the identity

sin(x + y)− sin(x − y) = 2 cos x sin y .

Solution.

LHS = sin(x + y)− sin(x − y)

= (sin x cos y + cos x sin y)− (sin x cos y − cos x sin y)

= sin x cos y + cos x sin y − sin x cos y + cos x sin y

= 2 cos x sin y = RHS .

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

Example. Write the expression

√3

2sin x +

1

2cos x

in terms of sine only.

Solution. Notice that

cosπ

6=

√3

2, sin

π

6=

1

2.

Therefore√

3

2sin x +

1

2cos x = cos

π

6sin x + sin

π

6cos x

= sin(x +

π

6

).

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

Example. Write the expression

√3

2sin x +

1

2cos x

in terms of sine only.

Solution. Notice that

cosπ

6=

√3

2, sin

π

6=

1

2.

Therefore√

3

2sin x +

1

2cos x = cos

π

6sin x + sin

π

6cos x

= sin(x +

π

6

).

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

Example. Write the expression

√3

2sin x +

1

2cos x

in terms of sine only.

Solution. Notice that

cosπ

6=

√3

2, sin

π

6=

1

2.

Therefore√

3

2sin x +

1

2cos x = cos

π

6sin x + sin

π

6cos x

= sin(x +

π

6

).

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

In general, an expression of the form

A sin x + B cos x ,

where A and B are real numbers and not both of them are zero, can beexpressed in terms of sine only.

To apply the trick above, we have to find an angle φ such that

cosφ =A√

A2 + B2sinφ =

B√A2 + B2

.

Then

A sin x + B cos x =√

A2 + B2

(A√

A2 + B2sin x +

B√A2 + B2

cos x

)=

√A2 + B2 (cosφ sin x + sinφ cos x)

=√

A2 + B2 sin(x + φ).

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

In general, an expression of the form

A sin x + B cos x ,

where A and B are real numbers and not both of them are zero, can beexpressed in terms of sine only.To apply the trick above, we have to find an angle φ such that

cosφ =A√

A2 + B2sinφ =

B√A2 + B2

.

Then

A sin x + B cos x =√

A2 + B2

(A√

A2 + B2sin x +

B√A2 + B2

cos x

)=

√A2 + B2 (cosφ sin x + sinφ cos x)

=√

A2 + B2 sin(x + φ).

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

In general, an expression of the form

A sin x + B cos x ,

where A and B are real numbers and not both of them are zero, can beexpressed in terms of sine only.To apply the trick above, we have to find an angle φ such that

cosφ =A√

A2 + B2sinφ =

B√A2 + B2

.

Then

A sin x + B cos x =√

A2 + B2

(A√

A2 + B2sin x +

B√A2 + B2

cos x

)=

√A2 + B2 (cosφ sin x + sinφ cos x)

=√

A2 + B2 sin(x + φ).

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

A sin x + B cos x

A sin x + B cos x = k sin(x + φ),

wherek =

√A2 + B2

and

cosφ =A√

A2 + B2, sinφ =

B√A2 + B2

.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

Example. Write the expression

12 sin x + 5 cos x

in terms of sine only. What are the period, amplitude and phase shift?

Solution. In this example, A = 12, B = 5 and hence

k =√

A2 + B2 =√

122 + 52 =√

144 + 25 =√

169 = 13.

12 sin x + 5 cos x = 13

(12

13sin x +

5

13cos x

)= 13 sin(x + φ),

where

cosφ =12

13, sinφ =

5

13.

So the amplitude is 13, the period is 2π and the phase shift is given by

φ = arccos12

13≈ 22.62◦.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

Example. Write the expression

12 sin x + 5 cos x

in terms of sine only. What are the period, amplitude and phase shift?

Solution. In this example, A = 12, B = 5 and hence

k =√

A2 + B2 =√

122 + 52 =√

144 + 25 =√

169 = 13.

12 sin x + 5 cos x = 13

(12

13sin x +

5

13cos x

)= 13 sin(x + φ),

where

cosφ =12

13, sinφ =

5

13.

So the amplitude is 13, the period is 2π and the phase shift is given by

φ = arccos12

13≈ 22.62◦.

MATH 201 - Week 11

Addition and Subtraction Formulas - Section 7.2

A sin x + B cos x

Example. Write the expression

12 sin x + 5 cos x

in terms of sine only. What are the period, amplitude and phase shift?

Solution. In this example, A = 12, B = 5 and hence

k =√

A2 + B2 =√

122 + 52 =√

144 + 25 =√

169 = 13.

12 sin x + 5 cos x = 13

(12

13sin x +

5

13cos x

)= 13 sin(x + φ),

where

cosφ =12

13, sinφ =

5

13.

So the amplitude is 13, the period is 2π and the phase shift is given by

φ = arccos12

13≈ 22.62◦.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Double-Angle Formulas

sin 2x = 2 sin x cos x

cos 2x = cos2 x − sin2 x

= 1− 2 sin2 x

= 2 cos2 x − 1

tan 2x =2 tan x

1− tan2 x

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Proofs.We use the addition formulas:

sin 2x = sin(x + x)

= sin x cos x + cos x sin x

= 2 sin x cos x .

cos 2x = cos(x + x)

= cos x cos x − sin x sin x

= cos2 x − sin2 x .

tan 2x = tan(x + x)

=tan x + tan x

1− tan x tan x

=2 tan x

1− tan2 x.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Proofs.We use the addition formulas:

sin 2x = sin(x + x)

= sin x cos x + cos x sin x

= 2 sin x cos x .

cos 2x = cos(x + x)

= cos x cos x − sin x sin x

= cos2 x − sin2 x .

tan 2x = tan(x + x)

=tan x + tan x

1− tan x tan x

=2 tan x

1− tan2 x.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Proofs.We use the addition formulas:

sin 2x = sin(x + x)

= sin x cos x + cos x sin x

= 2 sin x cos x .

cos 2x = cos(x + x)

= cos x cos x − sin x sin x

= cos2 x − sin2 x .

tan 2x = tan(x + x)

=tan x + tan x

1− tan x tan x

=2 tan x

1− tan2 x.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Proofs.We use the addition formulas:

sin 2x = sin(x + x)

= sin x cos x + cos x sin x

= 2 sin x cos x .

cos 2x = cos(x + x)

= cos x cos x − sin x sin x

= cos2 x − sin2 x .

tan 2x = tan(x + x)

=tan x + tan x

1− tan x tan x

=2 tan x

1− tan2 x.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Example. Find sin 2x , cos 2x and tan 2x if

sin x = −3

5

and x belongs to the third quadrant.

Solution. We have to find the value of cos x first:

cos x = ±√

1− sin2 x = ±√

1− 9

25= ±

√16

25= ±4

5.

The sign must be negative because x belongs to the third quadrant.Therefore tan x = 3

4 and

sin 2x = 2 sin x cos x = 2 ·(−3

5

)·(−4

5

)=

24

25

cos 2x = cos2 x − sin2 x =16

25− 9

25=

7

25

tan 2x =2 tan x

1− tan2 x=

2 · 34

1− 916

=64716

=6

4· 16

7=

24

7.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Example. Find sin 2x , cos 2x and tan 2x if

sin x = −3

5

and x belongs to the third quadrant.

Solution. We have to find the value of cos x first:

cos x = ±√

1− sin2 x = ±√

1− 9

25= ±

√16

25= ±4

5.

The sign must be negative because x belongs to the third quadrant.Therefore tan x = 3

4 and

sin 2x = 2 sin x cos x = 2 ·(−3

5

)·(−4

5

)=

24

25

cos 2x = cos2 x − sin2 x =16

25− 9

25=

7

25

tan 2x =2 tan x

1− tan2 x=

2 · 34

1− 916

=64716

=6

4· 16

7=

24

7.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Example. Find sin 2x , cos 2x and tan 2x if

sin x = −3

5

and x belongs to the third quadrant.

Solution. We have to find the value of cos x first:

cos x = ±√

1− sin2 x = ±√

1− 9

25= ±

√16

25= ±4

5.

The sign must be negative because x belongs to the third quadrant.Therefore tan x = 3

4 and

sin 2x = 2 sin x cos x = 2 ·(−3

5

)·(−4

5

)=

24

25

cos 2x = cos2 x − sin2 x =16

25− 9

25=

7

25

tan 2x =2 tan x

1− tan2 x=

2 · 34

1− 916

=64716

=6

4· 16

7=

24

7.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Example. Find sin 2x , cos 2x and tan 2x if

sin x = −3

5

and x belongs to the third quadrant.

Solution. We have to find the value of cos x first:

cos x = ±√

1− sin2 x = ±√

1− 9

25= ±

√16

25= ±4

5.

The sign must be negative because x belongs to the third quadrant.Therefore tan x = 3

4 and

sin 2x = 2 sin x cos x = 2 ·(−3

5

)·(−4

5

)=

24

25

cos 2x = cos2 x − sin2 x =16

25− 9

25=

7

25

tan 2x =2 tan x

1− tan2 x=

2 · 34

1− 916

=64716

=6

4· 16

7=

24

7.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

Squares of Trigonometric Functions

sin2 x =1− cos 2x

2

cos2 x =1 + cos 2x

2

tan2 x =1− cos 2x

1 + cos 2x.

These are used to lower the powers of the trigonometric functionsin simplifying certain expressions.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

Example. Lower the powers of the trigonometric functions in thefollowing expression:

sin4 x cos2 x .

Solution.

sin4 x cos2 x =(sin2 x

)2cos2 x

=

(1− cos 2x

2

)21 + cos 2x

2

=

(1− 2 cos 2x + cos2 2x

4

)1 + cos 2x

2

=1

8

(1− 2 cos 2x +

1 + cos 4x

2

)(1 + cos 2x)

=1

16(3− 4 cos 2x + cos 4x) (1 + cos 2x)

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

Example. Lower the powers of the trigonometric functions in thefollowing expression:

sin4 x cos2 x .

Solution.

sin4 x cos2 x =(sin2 x

)2cos2 x

=

(1− cos 2x

2

)21 + cos 2x

2

=

(1− 2 cos 2x + cos2 2x

4

)1 + cos 2x

2

=1

8

(1− 2 cos 2x +

1 + cos 4x

2

)(1 + cos 2x)

=1

16(3− 4 cos 2x + cos 4x) (1 + cos 2x)

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

Example. Lower the powers of the trigonometric functions in thefollowing expression:

sin4 x cos2 x .

Solution.

sin4 x cos2 x =(sin2 x

)2cos2 x

=

(1− cos 2x

2

)21 + cos 2x

2

=

(1− 2 cos 2x + cos2 2x

4

)1 + cos 2x

2

=1

8

(1− 2 cos 2x +

1 + cos 4x

2

)(1 + cos 2x)

=1

16(3− 4 cos 2x + cos 4x) (1 + cos 2x)

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

Example. Lower the powers of the trigonometric functions in thefollowing expression:

sin4 x cos2 x .

Solution.

sin4 x cos2 x =(sin2 x

)2cos2 x

=

(1− cos 2x

2

)21 + cos 2x

2

=

(1− 2 cos 2x + cos2 2x

4

)1 + cos 2x

2

=1

8

(1− 2 cos 2x +

1 + cos 4x

2

)(1 + cos 2x)

=1

16(3− 4 cos 2x + cos 4x) (1 + cos 2x)

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

Example. Lower the powers of the trigonometric functions in thefollowing expression:

sin4 x cos2 x .

Solution.

sin4 x cos2 x =(sin2 x

)2cos2 x

=

(1− cos 2x

2

)21 + cos 2x

2

=

(1− 2 cos 2x + cos2 2x

4

)1 + cos 2x

2

=1

8

(1− 2 cos 2x +

1 + cos 4x

2

)(1 + cos 2x)

=1

16(3− 4 cos 2x + cos 4x) (1 + cos 2x)

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

Example. Lower the powers of the trigonometric functions in thefollowing expression:

sin4 x cos2 x .

Solution.

sin4 x cos2 x =(sin2 x

)2cos2 x

=

(1− cos 2x

2

)21 + cos 2x

2

=

(1− 2 cos 2x + cos2 2x

4

)1 + cos 2x

2

=1

8

(1− 2 cos 2x +

1 + cos 4x

2

)(1 + cos 2x)

=1

16(3− 4 cos 2x + cos 4x) (1 + cos 2x)

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

=1

16

(3− 4 cos 2x + cos 4x + 3 cos 2x − 4 cos2 2x + cos 4x cos 2x

)

=1

16(3− cos 2x + cos 4x − 2(1 + cos 4x) + cos 4x cos 2x)

=1

16(1− cos 2x − cos 4x + cos 4x cos 2x) .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

=1

16

(3− 4 cos 2x + cos 4x + 3 cos 2x − 4 cos2 2x + cos 4x cos 2x

)=

1

16(3− cos 2x + cos 4x − 2(1 + cos 4x) + cos 4x cos 2x)

=1

16(1− cos 2x − cos 4x + cos 4x cos 2x) .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Squares of Trigonometric Functions

=1

16

(3− 4 cos 2x + cos 4x + 3 cos 2x − 4 cos2 2x + cos 4x cos 2x

)=

1

16(3− cos 2x + cos 4x − 2(1 + cos 4x) + cos 4x cos 2x)

=1

16(1− cos 2x − cos 4x + cos 4x cos 2x) .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Half-Angle Formulas

Squares of Trigonometric Functions

sinu

2= ±

√1− cos u

2

cosu

2= ±

√1 + cos u

2

tanu

2=

1− cos u

sin u

=sin u

1 + cos u.

The signs are determined by the quadrant in which the terminalpoint of u

2 lies.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Half-Angle Formulas

Example. Find the exact value of sin 9π8 .

Solution.

sin9π

8= ±

√1− cos 9π

4

2.

The angle 9π8 is in the third quadrant so its sine is negative and

cos9π

4= cos

(π4

+ 2π)

= cosπ

4=

1√2.

Hence

sin9π

8= −

√1− cos 9π

4

2

= −

√1− 1√

2

2= −

√√2− 1

2√

2

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Half-Angle Formulas

Example. Find the exact value of sin 9π8 .

Solution.

sin9π

8= ±

√1− cos 9π

4

2.

The angle 9π8 is in the third quadrant so its sine is negative and

cos9π

4= cos

(π4

+ 2π)

= cosπ

4=

1√2.

Hence

sin9π

8= −

√1− cos 9π

4

2

= −

√1− 1√

2

2= −

√√2− 1

2√

2

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Half-Angle Formulas

Example. Find the exact value of sin 9π8 .

Solution.

sin9π

8= ±

√1− cos 9π

4

2.

The angle 9π8 is in the third quadrant so its sine is negative and

cos9π

4= cos

(π4

+ 2π)

= cosπ

4=

1√2.

Hence

sin9π

8= −

√1− cos 9π

4

2

= −

√1− 1√

2

2= −

√√2− 1

2√

2

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Product-to-Sum Formulas

sin u cos v =1

2

(sin(u + v) + sin(u − v)

)cos u sin v =

1

2

(sin(u + v)− sin(u − v)

)cos u cos v =

1

2

(cos(u + v) + cos(u − v)

)sin u sin v =

1

2

(cos(u + v)− cos(u − v)

).

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Prove the product-to-sum formula

cos u sin v =1

2

(sin(u + v)− sin(u − v)

).

Solution. Using the addition formulas we get

RHS =1

2

(sin(u + v)− sin(u − v)

)=

1

2

((sin u cos v + cos u sin v)− (sin u cos v − cos u sin v)

)=

1

2

(sin u cos v + cos u sin v − sin u cos v + cos u sin v

)=

1

2

(2 cos u sin v

)= cos u sin v = LHS .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Prove the product-to-sum formula

cos u sin v =1

2

(sin(u + v)− sin(u − v)

).

Solution. Using the addition formulas we get

RHS =1

2

(sin(u + v)− sin(u − v)

)

=1

2

((sin u cos v + cos u sin v)− (sin u cos v − cos u sin v)

)=

1

2

(sin u cos v + cos u sin v − sin u cos v + cos u sin v

)=

1

2

(2 cos u sin v

)= cos u sin v = LHS .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Prove the product-to-sum formula

cos u sin v =1

2

(sin(u + v)− sin(u − v)

).

Solution. Using the addition formulas we get

RHS =1

2

(sin(u + v)− sin(u − v)

)=

1

2

((sin u cos v + cos u sin v)− (sin u cos v − cos u sin v)

)

=1

2

(sin u cos v + cos u sin v − sin u cos v + cos u sin v

)=

1

2

(2 cos u sin v

)= cos u sin v = LHS .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Prove the product-to-sum formula

cos u sin v =1

2

(sin(u + v)− sin(u − v)

).

Solution. Using the addition formulas we get

RHS =1

2

(sin(u + v)− sin(u − v)

)=

1

2

((sin u cos v + cos u sin v)− (sin u cos v − cos u sin v)

)=

1

2

(sin u cos v + cos u sin v − sin u cos v + cos u sin v

)

=1

2

(2 cos u sin v

)= cos u sin v = LHS .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Prove the product-to-sum formula

cos u sin v =1

2

(sin(u + v)− sin(u − v)

).

Solution. Using the addition formulas we get

RHS =1

2

(sin(u + v)− sin(u − v)

)=

1

2

((sin u cos v + cos u sin v)− (sin u cos v − cos u sin v)

)=

1

2

(sin u cos v + cos u sin v − sin u cos v + cos u sin v

)=

1

2

(2 cos u sin v

)

= cos u sin v = LHS .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Prove the product-to-sum formula

cos u sin v =1

2

(sin(u + v)− sin(u − v)

).

Solution. Using the addition formulas we get

RHS =1

2

(sin(u + v)− sin(u − v)

)=

1

2

((sin u cos v + cos u sin v)− (sin u cos v − cos u sin v)

)=

1

2

(sin u cos v + cos u sin v − sin u cos v + cos u sin v

)=

1

2

(2 cos u sin v

)= cos u sin v = LHS .

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Write sin 2x cos 3x as a sum of trigonometric functions.

Solution. Using the first product-to-sum formula with u = 2x ,v = 3x we have

sin 2x cos 3x =1

2

(sin(2x + 3x) + sin(2x − 3x)

)=

1

2

(sin 5x + sin(−x)

)=

1

2

(sin 5x − sin x

).

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Write sin 2x cos 3x as a sum of trigonometric functions.

Solution. Using the first product-to-sum formula with u = 2x ,v = 3x we have

sin 2x cos 3x =1

2

(sin(2x + 3x) + sin(2x − 3x)

)

=1

2

(sin 5x + sin(−x)

)=

1

2

(sin 5x − sin x

).

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Write sin 2x cos 3x as a sum of trigonometric functions.

Solution. Using the first product-to-sum formula with u = 2x ,v = 3x we have

sin 2x cos 3x =1

2

(sin(2x + 3x) + sin(2x − 3x)

)=

1

2

(sin 5x + sin(−x)

)

=1

2

(sin 5x − sin x

).

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Write sin 2x cos 3x as a sum of trigonometric functions.

Solution. Using the first product-to-sum formula with u = 2x ,v = 3x we have

sin 2x cos 3x =1

2

(sin(2x + 3x) + sin(2x − 3x)

)=

1

2

(sin 5x + sin(−x)

)=

1

2

(sin 5x − sin x

).

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Product-to-Sum Formulas

sin u + sin v = 2 sinu + v

2cos

u − v

2

sin u − sin v = 2 cosu + v

2sin

u − v

2

cos u + cos v = 2 cosu + v

2cos

u − v

2

cos u − cos v = −2 sinu + v

2sin

u − v

2.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Write sin 2x − sin 7x as a product of trigonometricfunctions.

Solution. Using the second sum-to-product formula with u = 2x ,v = 7x we have

sin 2x − sin 7x = 2 cos2x + 7x

2sin

2x − 7x

2

= 2 cos9x

2sin−5x

2

= −2 cos9x

2sin

5x

2.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Write sin 2x − sin 7x as a product of trigonometricfunctions.

Solution. Using the second sum-to-product formula with u = 2x ,v = 7x we have

sin 2x − sin 7x = 2 cos2x + 7x

2sin

2x − 7x

2

= 2 cos9x

2sin−5x

2

= −2 cos9x

2sin

5x

2.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Write sin 2x − sin 7x as a product of trigonometricfunctions.

Solution. Using the second sum-to-product formula with u = 2x ,v = 7x we have

sin 2x − sin 7x = 2 cos2x + 7x

2sin

2x − 7x

2

= 2 cos9x

2sin−5x

2

= −2 cos9x

2sin

5x

2.

MATH 201 - Week 11

Double-Angle, Half-Angle and Product-Sum Formulas - Section 7.3

Product-Sum Formulas

Example. Write sin 2x − sin 7x as a product of trigonometricfunctions.

Solution. Using the second sum-to-product formula with u = 2x ,v = 7x we have

sin 2x − sin 7x = 2 cos2x + 7x

2sin

2x − 7x

2

= 2 cos9x

2sin−5x

2

= −2 cos9x

2sin

5x

2.