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Chapter 13 1 Glencoe Algebra 2

Student-Built Glossary

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 13. As you study the chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Algebra Study Notebook to review vocabulary at the end of the chapter.

Vocabulary Term

Found on Page

Defi nition/Description/Example

cofunction identity

negative angle identity

Pythagorean identity

quotient identity

(continued on the next page)

001_020_ALG2_A_CRM_C13_CR_660516.indd 1001_020_ALG2_A_CRM_C13_CR_660516.indd 1 12/21/10 11:57 AM12/21/10 11:57 AM

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Student-Built Glossary

Vocabulary TermFound

on PageDefi nition/Description/Example

reciprocal identity

trigonometric equation

trigonometric identity


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13-1 Study Guide and InterventionTrigonometric Identities

Find Trigonometric Values A trigonometric identity is an equation involving trigonometric functions that is true for all values for which every expression in the equation is defined.

Basic

Trigonometric

Identities

Quotient Identities tan θ = sin θ −

cos θ cot θ = cos θ

−

sin θ

Reciprocal Identities csc θ = 1 −

sin θ sec θ = 1

−

cos θ cot θ = 1

−

tan θ

Pythagorean Identities cos2 θ + sin2 θ = 1 tan2 θ + 1 = sec2 θ cot2 θ+ 1 = csc2 θ

Find the exact value of cot θ if csc θ = - 11 −

5 and 180° < θ < 270°.

cot2 θ + 1 = csc2 θ Trigonometric identity

cot 2 θ + 1 = (-

11 −

5 )

2 Substitute - 11

−

5 for csc θ.

cot 2 θ + 1 = 121 −

25 Square - 11

−

5 .

cot 2 θ = 96 −

25 Subtract 1 from each side.

cot θ = ±

4 √ � 6

−

5

Take the square root of each side.

Since θ is in the third quadrant, cot θ is positive. Thus, cot θ = 4 √ � 6 −

5 .

ExercisesFind the exact value of each expression if 0° < θ < 90°.

1. If cot θ = 4, find tan θ. 2. If cos θ = √ � 3

−

2 , find csc θ.

3. If sin θ = 3 −

5 , find cos θ. 4. If sin θ = 1 −

3 , find sec θ.

5. If tan θ = 4 −

3 , find cos θ. 6. If sin θ = 3 −

7 , find tan θ.

Find the exact value of each expression if 90° < θ < 180°.

7. If cos θ = -

7 −

8 , find sec θ. 8. If csc θ = 12 −

5 , find cot θ.


9. If cos θ = 6 −

7 , find sin θ. 10. If csc θ = -

9 −

4 , find sin θ.

Example


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13-1 Study Guide and Intervention (continued)

Trigonometric Identities

Simplify Expressions The simplified form of a trigonometric expression is written as a numerical value or in terms of a single trigonometric function, if possible. Any of the trigonometric identities can be used to simplify expressions containing trigonometric functions.

Simplify (1 - cos2 θ) sec θ cot θ + tan θ sec θ cos2 θ.

(1 - cos2 θ) sec θ cot θ + tan θ sec θ cos2 θ = sin 2 θ · 1 −

cos θ · cos θ

−

sin θ + sin θ

−

cos θ · 1 −

cos θ · cos 2 θ

= sin θ + sin θ

= 2 sin θ

Simplify sec θ · cot θ −

1 - sin θ - csc θ

−

1 + sin θ .

sec θ · cot θ −

1 - sin θ - csc θ

−

1 + sin θ =

1 −

cos θ · cos θ

−

sin θ

−

1 + sin θ -

1 −

sin θ

−

1 + sin θ

= 1 −

sin θ (1 + sin θ) - 1 −

sin θ (1 - sin θ)

−−

(1 - sin θ)(1 + sin θ)

= 1 −

sin θ + 1 - 1 −

sin θ + 1 −−

1 - sin 2 θ

= 2 −

cos 2 θ or 2 sec2 θ

Exercises

Simplify each expression.

1. tan θ · csc θ −

sec θ 2. sin θ · cot θ

−

sec 2 θ - tan 2 θ

3. sin 2 θ - cot θ · tan θ −−

cot θ · sin θ 4. cos θ

−

sec θ - tan θ

5. tan θ · cos θ −

sin θ + cot θ · sin θ · tan θ · csc θ 6. csc 2 θ - cot 2 θ −

tan θ · cos θ

7. 3 tan θ · cot θ + 4 sin θ · csc θ + 2 cos θ · sec θ 8. 1 - cos 2 θ −

tan θ · sin θ

Example 1

Example 2


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13-1 Skills PracticeTrigonometric Identities


1. If tan θ = 1, find sec θ. 2. If tan θ = 1 −

2 , find cos θ.

3. If sec θ = 2, find cos θ. 4. If cos θ = 8 −

17 , find csc θ.


5. If cos θ = -

4 −

5 , find sin θ. 6. If cot θ = -

3 −

2 , find cos θ.


7. If tan θ = 1, find cos θ. 8. If sin θ = -

√ � 2 −

2 , find tan θ.

9. If csc θ = -2, find cos θ. 10. If cos θ = -

2 √ � 5 −

5 , find tan θ.

11. If csc θ = -2, find cot θ. 12. If sin θ = -

5 −

13 , find tan θ.


13. sin θ sec θ 14. csc θ sin θ

15. cot θ sec θ 16. cos θ −

sec θ

17. tan θ + cot θ 18. csc θ tan θ - tan θ sin θ

19. 1 - sin 2 θ −

sin θ + 1

20. csc θ + cot θ

21. sin 2 θ + cos 2 θ −

1 - cos 2 θ 22. 1 + tan 2 θ

−

1 + sec θ


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13-1 PracticeTrigonometric Identities


1. If cos θ = 5 −

13 , find sin θ. 2. If cot θ = 1 −

2 , find sin θ.

3. If tan θ = 4, find sec θ. 4. If tan θ = 2 −

5 , find cot θ.


5. If sin θ = -

15 −

17 , find sec θ. 6. If csc θ = -

3 −

2 , find cot θ.


7. If cos θ = 3 −

10 , find cot θ. 8. If csc θ = -8, find sec θ.

9. If tan θ = -

1 −

2 , find sin θ. 10. If cos θ = 1 −

3 , find cot θ.


11. csc θ tan θ 12. sin2 θ −

tan2 θ 13. sin2 θ cot2 θ

14. cot2 θ + 1 15. csc2 θ - cot2 θ −

1 - cos2 θ 16. csc θ - sin θ

−

cos θ

17. sin θ + cos θ cot θ 18. cos θ −

1 - sin θ -

cos θ −

1 + sin θ 19. sec2 θ cos2 θ + tan2 θ

20. AERIAL PHOTOGRAPHY The illustration shows a plane taking an aerial photograph of point A. Because the point is directly below the plane, there is no distortion in the image. For any point B not directly below the plane, however, the increase in distance creates distortion in the photograph. This is because as the distance from the camera to the point being photographed increases, the exposure of the film reduces by (sin θ)(csc θ - sin θ). Express (sin θ)(csc θ - sin θ) in terms of cos θ only.

21. WAVES The equation y = a sin θt represents the height of the waves passing a buoy at a time t in seconds. Express a in terms of csc θt.

A B

θ


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13-1 Word Problem PracticeTrigonometric Identities

1. MAPS The figure below shows a map of some buildings in Aiden’s hometown. The sine of the angle θ formed by the high school, the middle school, and Aiden’s house is 3

−

7 .

Aiden’sHouse

Library High SchoolMiddle School

θ

a. What is the cosine of the angle?

b. What is the tangent of the angle?

c. What are the sine, cosine, and tangent of the angle formed by the library, the middle school, and Aiden’s house?

2. GEOMETRY When a line is drawn in the coordinate plane, the tangent of the angle that the line makes with the horizontal axis is equal to the slope of the line (for non-vertical lines).

θ

The cosine of the angle θ the line in the

figure makes with the horizontal is 1−3

.

a. Explain two ways to determine the slope of the line. Draw a representative triangle with the length of the side adjacent to angle θequal to 1 and the hypotenuse equal to 3.

b. Compute the sine and tangent of the angle.

c. What is the slope of the line?


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13-1 Enrichment

Planetary Orbits

The orbit of a planet around the Sun is an ellipse with the Sun at one focus. Let the pole of a polar coordinate system be that focus and the polar axis be toward the other focus. The polar equation of an ellipse is

r = 2ep −

1 - e cos θ . Since 2p = b 2 − c and b2 = a2 - c2,

2p = a 2 - c 2 − c = a 2 − c (1 - c 2 −

a 2 ) . Because e = c − a ,

2p = a ( a − c ) (1 - ( c − a ) 2 ) = a ( 1 − e ) (1 - e2).

Therefore, 2ep = a(1 - e2). Substituting into the polar equation of an ellipse yields an equation that is useful for finding distances from the planet to the Sun.

r = a(1 - e 2 ) −

1 - e cos θ

Note that e is the eccentricity of the orbit and a is the length of the semi-major axis of the ellipse. Also, a is the mean distance of the planet from the Sun.

The mean distance of Venus from the Sun is 67.24 million miles and the eccentricity of its orbit is 0.006788. Find the minimum and maximum distances of Venus from the Sun.

The minimum distance occurs when θ = π.

r = 67.24 × 10 6 (1 - 0.00678 8 2 ) −−

1 - 0.006788 cos π = 66.78 × 10 6 or 66.78 million miles

The maximum distance occurs when θ = 0.

r = 67.24 × 10 6 (1 - 0.00678 8 2 ) −−

1 - 0.006788 cos 0 = 67.70 × 10 6 or 67.70 million miles

Complete each of the following.

1. The mean distance of Mars from the Sun is 141.64 million miles and the eccentricity of its orbit is 0.093382. Find the minimum and maximum distances of Mars from the Sun.

2. The minimum distance of Earth from the Sun is 91.445 million miles, and the eccentricity of its orbit is 0.016734. Find the mean and maximum distances of Earth from the Sun.

Example

r

Polar Axis

θ


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Study Guide and InterventionVerifying Trigonometric Identities

Transform One Side of an Equation Use the basic trigonometric identities along with the definitions of the trigonometric functions to verify trigonometric identities. Often it is easier to begin with the more complicated side of the equation and transform that expression into the form of the simpler side.

Verify that each equation is an identity.

Exercises


1. 1 + csc2 θ · cos2 θ = csc2 θ 2. sin θ −

1 - cos θ - cot θ

−

1 + cos θ = 1 - cos 3 θ

−

sin 3 θ

Example

a. sin θ −

cot θ - sec θ = -cos θ

Transform the left side.

sin θ −

cot θ - sec θ � -cos θ

sin θ −

cos θ −

sin θ

- 1 −

cos θ � -cos θ

sin 2 θ −

cos θ - 1 −

cos θ � -cos θ

sin 2 - 1 −

cos θ � -cos θ

- cos 2 θ −

cos θ � -cos θ

-cos θ = -cos θ �

b. tan θ −

csc θ + cos θ = sec θ

Transform the left side.

tan θ −

csc θ + cos θ � sec θ

sin θ

−

cos θ

−

1 −

sin θ

+ cos θ � sec θ

sin 2 θ −

cos θ + cos θ � sec θ

sin 2 θ + cos 2 θ −

cos θ � sec θ

1 −

cos θ � sec θ

sec θ = sec θ �


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Study Guide and Intervention (continued)

Verifying Trigonometric Identities

Transform Each Side of an Equation The following techniques can be helpful in verifying trigonometric identities.• Substitute one or more basic identities to simplify an expression.• Factor or multiply to simplify an expression.• Multiply both numerator and denominator by the same trigonometric expression.• Write each side of the identity in terms of sine and cosine only. Then simplify each side.

Verify that tan 2 θ + 1 −−

sin θ · tan θ · sec θ + 1 = sec 2 θ - tan 2 θ is an identity.

tan 2 θ + 1 −−

sin θ · tan θ · sec θ + 1 � sec 2 θ - tan 2 θ

sec 2 θ −−

sin θ · sin θ −

cos θ · 1 −

cos θ + 1

� 1 −

cos 2 θ - sin 2 θ

−

cos 2 θ

1 −

cos 2 θ

−

sin 2 θ −

cos 2 θ + 1

� 1 - sin 2 θ −

cos 2 θ

1 −

cos 2 θ

−

sin 2 θ + cos 2 θ −

cos 2 θ

� cos 2 θ −

cos 2 θ

1 −

sin 2 θ + cos 2 θ � 1

1 = 1 �

Exercises


1. csc θ · sec θ = cot θ + tan θ 2. tan 2 θ −

1 - cos 2 θ = sec θ

−

cos θ

3. cos θ · cot θ −

sin θ = csc θ

−

sin θ · sec 2 θ 4. csc 2 θ - cot 2 θ

−

sec 2 θ = cot 2 θ(1 - cos 2 θ)

Example


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1. tan θ cos θ = sin θ 2. cot θ tan θ = 1

3. csc θ cos θ = cot θ 4. 1 – sin

2

θ −

cos θ= cos θ

5. (tan θ)(1 - sin 2 θ) = sin θ cos θ 6. csc θ−

sec θ= cot θ

7. sin2 θ −

1 - sin2 θ = tan2 θ 8. c os 2 θ

−

1 – sin θ = 1 + sin θ


Skills PraticeVerifying Trigonometric Identities



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PracticeVerifying Trigonometric Identities


1. sin 2 θ + cos 2 θ −

cos 2 θ = sec 2 θ 2. cos 2 θ

−

1 - sin 2 θ = 1

3. (1 + sin θ)(1 - sin θ) = cos2 θ 4. tan4 θ + 2 tan2 θ + 1 = sec4 θ

5. cos2 θ cot2 θ = cot2 θ - cos2 θ 6. (sin2 θ)(csc2 θ + sec2 θ) = sec2 θ

7. PROJECTILES The square of the initial velocity of an object launched from the ground is

v 2 = 2gh −

sin 2 θ , where θ is the angle between the ground and the initial path h is the

maximum height reached, and g is the acceleration due to gravity. Verify the identity

2gh −

sin 2 θ =

2gh sec 2 θ −

sec 2 θ - 1 .

8. LIGHT The intensity of a light source measured in candles is given by I = ER2 sec θ, where E is the illuminance in foot candles on a surface, R is the distance in feet from the light source, and θ is the angle between the light beam and a line perpendicular to the surface. Verify the identity ER2(1 + tan2 θ) cos θ = ER2 sec θ.


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Word Problem PracticeVerifying Trigonometric Identities

1. EXPERIMENTS Kyle is performing an experiment for his physics class. He uses a motion detector connected to a computer to collect data. The first time he performs the experiment, the computer produces the graph below. The graph is of the function

y = cos2 x −

1 - sin x .

y

x

3

2

1

-1

-2

-3

-90 90 180 270 360 450

The second time he performs the experiment, the computer produces the graph below. The graph is of the y = sin x.

y

x

3

2

1

-1

-2

-3

-90 90 180 270 360 450

Use the graphs to write an identity involving cos2 x −

1 - sin x and sin x.

2. GRAPHING FUNCTIONS Brandi is doing her trigonometry homework and needs to graph the function

y =

sin2 x - tan2 x−

sec2 x . She thinks that

it would be easier to graph the function if she could rewrite it in a simpler way, either without a denominator or as an expression containing only one trigonometric function. After some work, Brandi decides that she can graph y = -sin4 x instead of the given function.

a. Is it possible for Brandi to simplify the function in the way she claims?

b. If Brandi graphs the given function and her simpler function on the same set of axes, what will she find?

c. What does this mean aboutsin2 x - tan2 x−

sec2 x and -sin4 x?


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Enrichment

Trigonometric IdentitiesAs you continue to study mathematics, you will find that one of the most important applications of trigonometric identities is to allow you to integrate non-trigonometric functions. Integration is used to analyze and apply real-world functions.

Pythagorean Identity sin2 θ + cos2 θ = 1

Quotient Identity tan θ = sin θ −

cos θ

Reciprocal Identities csc θ = 1 −

sin θ sec θ = 1

−

cos θ cot θ = 1

−

tan θ

You can use known identities such as the Pythagorean and Quotient Identities to write a trigonometric identity relating any of the trigonometric functions to any other.

Use the identities above to complete the table below.

Function

in terms

of:

sin θ cos θ tan θ csc θ sec θ cot θ

sin θ = sin θ ± √ �� 1 - cos2 θ ± tan θ −

√ �� 1 + tan2 θ ±

√ �� sec2 θ - 1 −

secθ

cos θ = cos θ ± 1 −

√ �� 1 + tan2 θ

1 −

sec θ

tan θ = ± sin θ −

√ �� 1 - sin2 θ

tan θ

1 −

cot θ

csc θ = 1 −

sin θ csc θ ± √ �� 1 + cot2 θ

sec θ = 1 −

cos θ ± √ �� 1 + tan2 θ sec θ

cot θ = 1 −

tan θ ± √ �� csc2 θ - 1 cot θ


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Study Guide and InterventionSum and Difference of Angles Identities

Sum and Difference Identities The following formulas are useful for evaluating an expression like sin 15º from the known values of sine and cosine of 60º and 45º.

Sum and

Difference

of Angles

The following identities hold true for all values of α and β.

cos (α ± β) = cos α · cos β ∓ sin α · sin β

sin (α ± β) = sin α · cos β ± cos α · sin β

Find the exact value of each expression.

a. cos 345º cos 345º = cos (300º + 45º) = cos 300º · cos 45º - sin 300º · sin 45º

= 1 −

2 ·

√ � 2 −

2 - (-

√ � 3 −

2 ) ·

√ � 2 −

2

= √ � 2 + √ � 6

−

4

b. sin (-105º) sin (-105º) = sin (45º - 150º) = sin 45º · cos 150º - cos 45º · sin 150º

= √ � 2

−

2 · (-

√ � 3 −

2 ) -

√ � 2 −

2 ·

1 −

2

= - √ � 2 + √ � 6

−

4

Exercises


1. sin 105º 2. cos 285º 3. cos (-75º)

4. cos (-165º) 5. sin 195º 6. cos 420º

7. sin (-75º) 8. cos 135º 9. cos (-15º)

10. sin 345º 11. cos (-105º) 12. sin 495º

Example


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Sum and Difference of Angles Identities

Verify Trigonometric Identities You can also use the sum and difference of angles formulas to verify identities.

Verify that cos (θ + 3π

−

2 ) = sin θ is an identity.

cos (θ + 3π

−

2 ) � sin θ Original equation

cos θ · cos 3π

−

2 - sin θ · sin 3π

−

2 � sin θ Sum of Angles Formula

cos θ · 0 - sin θ · (-1) � sin θ Evaluate each expression.

sin θ = sin θ � Simplify.

Verify that sin (

θ - π −

2 ) + cos (θ + π) = -2 cos θ is an identity.

sin (θ - π −

2 ) + cos (θ + π) � -2 cos θ Original equation

sin θ · cos π −

2 - cos θ · sin π −

2 + cos θ · cos π - sin θ · sin π � -2 cos θ Sum and Difference of

Angles Formulas

sin θ · 0 - cos θ · 1 + cos θ · (-1) - sin θ · 0 � -2 cos θ Evaluate each expression.

-2 cos θ = -2 cos θ � Simplify.

ExercisesVerify that each equation is an identity.

1. sin (90° + θ) = cos θ

2. cos (270° + θ) = sin θ

3. sin ( 2π

−

3 - θ) + cos (θ - 5π

−

6 ) = sin θ

4. cos ( 3π

−

4 + θ) -

(sin θ - π −

4 ) = - √ � 2 sin θ

Example 1

Example 2


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1. sin 330° 2. cos (-165°) 3. sin (-225°)

4. cos 135° 5. sin (- 45)° 6. cos 210°

7. cos (-135°) 8. sin 75° 9. sin (-195°)

Verify that each equation is an identity. 10. sin (90° + θ) = cos θ

11. sin (180° + θ) = -sin θ

12. cos (270° - θ) = -sin θ

13. cos (θ - 90°) = sin θ

14. sin (θ - π −

2 ) = - cos θ

15. cos (π + θ) = - cos θ


13-3 Skills PracticeSum and Difference of Angles Identities


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13-3 PracticeSum and Difference of Angles Identities


1. cos 75° 2. cos 375° 3. sin (-165°)

4. sin (-105°) 5. sin 150° 6. cos 240°

7. sin 225° 8. sin (-75°) 9. sin 195°


10. cos (180° - θ) = -cos θ

11. sin (360° + θ) = sin θ

12. sin (45° + θ) - sin (45° - θ) = √ � 2 sin θ

13. cos (x - π −

6 ) + sin

(x - π −

3 ) = sin x

14. SOLAR ENERGY On March 21, the maximum amount of solar energy that falls on a square foot of ground at a certain location is given by E sin (90° - ϕ), where ϕ is the latitude of the location and E is a constant. Use the difference of angles formula to find the amount of solar energy, in terms of cos ϕ, for a location that has a latitude of ϕ.

15. ELECTRICITY In a certain circuit carrying alternating current, the formula c = 2 sin (120t) can be used to find the current c in amperes after t seconds.

a. Rewrite the formula using the sum of two angles.

b. Use the sum of angles formula to find the exact current at t = 1 second.


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Word Problem PracticeSum and Difference of Angles Identities

1. ART As part of a mosaic that an artist is making, she places two right triangular tiles together to make a new triangular piece. One tile has lengths of 3 inches, 4 inches, and 5 inches. The other tile has lengths 4 inches, 4 √�3 inches, and 8 inches. The pieces are placed with the sides of 4 inches against each other, as shown in the figure below.

A

85

3

4

4√3

a. What is the exact value of the sine of angle A?

b. What is the exact value of the cosine of angle A?

c. What is the measure of angle A?

d. Is the new triangle formed from the two triangles also a right triangle?

2. MANUFACTURING A robotic arm performs two operations in the course of manufacturing each car door that comes down the assembly line. As the door arrives in front of the robotic arm, the arm is parallel to the assembly line. Once the door is in place, the arm rotates counterclockwise 72° to perform the first operation. After performing the first operation, the arm rotates another 33° counterclockwise to perform the second operation. After the second operation is completed, the robotic arm rotates clockwise back to its starting position.

72°33°

a. After completing the second operation, through what angle does the robotic arm rotate to return to its starting position?

b. What is the exact value of the sine of the angle through which the arm rotates to return to its starting position?

c. What is the exact value of the cosine of the angle through which the arm rotates to return to its starting position?


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Enrichment

Identities for the Products of Sines and Cosines

By adding the identities for the sines of the sum and difference of the measures of two angles, a new identity is obtained.

sin (α + β) = sin α cos β + cos α sin β sin (α - β) = sin α cos β - cos α sin β(i) sin (α + β) + sin (α - β) = 2 sin α cos β

This new identity is useful for expressing certain products as sums.

Write sin 3θ cos θ as a sum.

In the identity let α = 3θ and β = θ so that 2 sin 3θ cos θ = sin (3θ + θ) + sin (3θ - θ). Thus, sin 3θ cos θ = 1 −

2 sin 4θ + 1 −

2 sin 2θ.

By subtracting the identities for sin (α + β) and sin (α - β), a similar identity for expressing a product as a difference is obtained.

(ii) sin (α + β) - sin (α - β) = 2 cos α sin β

1. Use the identities for cos (α + β) and cos (α - β) to find identities for expressing the products 2 cos α cos β and 2 sin α sin β as a sum or difference.

2. Find the value of sin 105º cos 75º without using tables.

3. Express cos θ sin θ −

2 as a difference.

Example


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Study Guide and InterventionDouble-Angle and Half-Angle Identities

Double-Angle Identities

Double-Angle

Identities

The following identities hold true for all values of θ.

sin 2θ = 2 sin θ · cos θ cos 2θ = cos2 θ - sin2 θ

cos 2θ = 1 - 2 sin2 θ

cos 2θ = 2 cos2 θ - 1

Find the exact values of sin 2θ, and cos 2θ if sin θ = -

9 −

10

and θ is between 180º and 270º.First, find value of cos θ. cos2 θ = 1 - sin2 θ cos2 θ + sin2 θ = 1

cos2 θ = 1 - (-

9 −

10 )

2 sin θ = -

9 −

10

cos2 θ = 19 −

100

cos θ = ± √ �� 19

−

10

Since θ is in the third quadrant, cos θ is negative. Thus cos θ = - √ �� 19

−

10 .

To find sin 2θ, use the identity sin 2θ = 2 sin θ · cos θ.

sin 2θ = 2 sin θ · cos θ.

= 2 (-

9 −

10 ) (-

√ �� 19 −

10 )

= 9 √ �� 19 −

50

The value of sin 2θ is 9 √ �� 19 −

50 .

To find cos 2θ, use the identity cos 2θ = 1 - 2 sin2 θ. cos 2θ = 1 - 2 sin2 θ

= 1 - 2 (-

9 −

10 )

2

= -

31 −

50

The value of cos 2θ is = -

31 −

50 .

ExercisesFind the exact values of sin 2θ and cos 2θ.

1. sin θ = 1 −

4 , 0º < θ < 90º 2. sin θ = -

1 −

8 , 270º < θ < 360º

3. cos θ = -

3 −

5 , 180º < θ < 270º 4. cos θ = -

4 −

5 , 90º < θ < 180º

5. sin θ = -

3 −

5 , 270º < θ < 360º 6. cos θ = -

2 −

3 , 90º < θ < 180º

Example


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Double-Angle and Half-Angle Identities

Half-Angle Identities

Half-Angle

Identities

The following identities hold true for all values of a.

sin α −

2 = ±

√ ��

1 - cos α −

2 cos α −

2 = ±

√ ��

1 + cos α −

2

Find the exact value of sin α −

2 if sin α = 2 −

3 and α is between

90º and 180º.First find cos α. cos2 α = 1 - sin2 α cos2 α + sin2 α = 1

cos2 α = 1 - ( 2 −

3 )

2 sin α = 2 −

3

cos2 α = 5 −

9 Simplify.

cos α = ± √ � 5

−

3 Take the square root of each side.

Since α is in the second quadrant, cos α = - √ � 5

−

3 .

sin α −

2 = ±

√ ��

1 - cos α −

2 Half-Angle formula

= ± √

��

1 - (-

√ � 5 −

3 ) −

2 cos α = -

√ � 5 −

3

= ± √

��

3 + √ � 5 −

6 Simplify.

= ± √

��

18 + 6 √ � 5 −

6 Rationalize.

Since α is between 90º and 180º, α −

2 is between 45º and 90º. Thus sin α −

2 is positive and

equals √

��

18 + 6 √ � 5 −

6 .

ExercisesFind the exact values of sin α −

2 and cos α −

2 .

1. cos α = -

3 −

5 , 180º < α < 270º 2. cos α = -

4 −

5 , 90º < α < 180º

3. sin α = -

3 −

5 , 270º < α < 360º 4. cos α = -

2 −

3 , 90º < α < 180º


5. cos 22 1 −

2 º 6. sin 67.5º 7. cos 7π

−

8

Example


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Skills PracticeDouble-Angle and Half-Angle Identities

Find the exact values of sin 2θ, cos 2θ, sin θ−2

, and cos θ−2

.

1. cos θ =

7−

25 , 0° < θ < 90° 2. sin θ = -

4−

5 , 180° < θ < 270°

3. sin θ =

40−

41 , 90° < θ < 180° 4. cos θ =

3−

7 , 270° < θ < 360°

5. cos θ = -

3−

5 , 90° < θ < 180° 6. sin θ =

5−

13 , 0° < θ < 90°


7. cos 22.5° 8. sin 165°

9. cos 105° 10. sin π−8

11. sin 15π

−

8 12. cos 75°


13. sin 2θ = 2 tan θ −

1 + tan 2 θ 14. tan θ + cot θ = 2 csc 2θ


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PracticeDouble-Angle and Half-Angle Identities

Find the exact values of sin 2θ, cos 2θ, sin θ−2

, and cos θ−2

for each of the following.

1. cos θ = 5 −

13 , 0º < θ < 90º 2. sin θ = 8 −

17 , 90º < θ < 180º

3. cos θ = 1 −

4 , 270º < θ < 360º 4. sin θ = -

2 −

3 , 180º < θ < 270º


5. tan 105º 6. tan 15º 7. cos 67.5º 8. sin (-

π

−

8 )


9. sin2 θ −

2 = tan θ - sin θ

−

2 tan θ

10. sin 4θ = 4 cos 2θ sin θ cos θ

11. AERIAL PHOTOGRAPHY In aerial photography, there is a reduction in film exposure for any point X not directly below the camera. The reduction E

θ is given by

Eθ = E0 cos4 θ, where θ is the angle between the perpendicular line from the camera to

the ground and the line from the camera to point X, and E0 is the exposure for the point directly below the camera. Using the identity 2 sin2 θ = 1 - cos 2θ, verify that

E0 cos4 θ = E0 ( 1 −

2 + cos 2θ

−

2

) 2.

12. IMAGING A scanner takes thermal images from altitudes of 300 to 12,000 meters. The width W of the swath covered by the image is given by W = 2H′ tan θ, where H′ is the

height and θ is half the scanner’s field of view. Verify that 2H′ sin 2θ −

1 + cos 2θ = 2H′ tan θ.


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Word Problem PracticeDouble-Angle and Half-Angle Identities

1. GEOMETRY The large triangle shown in the figure below is an isosceles right triangle. The small triangle inside the large triangle was formed by bisecting each of the isosceles angles of the right triangle.

a. What is the exact value of the sine of either of the congruent angles of the small triangle?

b. What is the exact value of the cosine of either of the congruent angles of the small triangle?

c. What is the exact value of the sine of the obtuse angle of the small triangle?

d. What is the exact value of the cosine of the obtuse angle of the small triangle?

2. RAMPS A ramp for loading goods onto a truck was mistakenly built with the dimensions shown in the figure below. The degree measure of the angle the ramp makes with the ground should have been twice the degree measure of the angle shown in the figure.

18

2θ

a. Find the exact values of the sine and cosine of the angle the ramp should have made with the ground.

b. If the ramp had been built properly, what would the degree measures of the two acute angles have been?


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Enrichment

Alternating Current

The figure at the right represents an alternating current generator. A rectangular coil of wire is suspended between the poles of a magnet. As the coil of wire is rotated, it passes through the magnetic field and generates current.

As point X on the coil passes through the points A and C, its motion is along the direction of the magnetic field between the poles. Therefore, no current is generated. However, through points B and D, the motion of X is perpendicular to the magnetic field. The maximum current may have a positive This induces maximum current in the coil. Between A or negative value. and B, B and C, C and D, and D and A, the current in the coil will have an intermediate value. Thus, the graph of the current of an alternating current generator is closely related to the sine curve.

The actual current i in a household current is given by i = IM sin(120πt + α) where IM is the maximum value of the current t is the elapsed time in seconds, and α is the angle determined by the position of the coil at time tn.

If α = π −

2 , find a value of t for which i = 0.

If i = 0, then IM sin (120πt + α) = 0. i = IM sin (120πt + α)

Since IM ≠ 0, sin (120πt + α) = 0. If ab = 0 and a ≠ 0, then b = 0.

Let 120πt + α = s. Thus, sin s = 0.s = π because sin π = 0. 120πt + α = π Substitute 120πt + α for s.

120πt + π −

2 = π Substitute

π

−

2 for α.

= 1 −

240 Solve for t.

This solution is the least positive value of t that satisfies the problem.

Using the equation for the actual current in a household circuit, i = IM sin(120πt + α), solve each problem. For each problem, find the first positive value of t.

1. If α = 0, find a value of t for 2. If α = 0, find a value of t for whichwhich i = 0. i = +IM .

3. If α = π −

2 , find a value of t for which 4. If α = π −

4 , find a value of t for which

i = -IM . i = 0.

Example

XA

B D

C

O

A

B

C

D

i(amperes)

t(seconds)


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Study Guide and InterventionSolving Trigonometric Equations

Solve Trigonometric Equations You can use trigonometric identities to solve trigonometric equations, which are true for only certain values of the variable.

Solve 4 sin2 θ - 1 = 0 if 0° < θ < 360°.

4 sin2 θ - 1 = 0 4 sin2 θ = 1 sin2 θ = 1 −

4

sin θ = ±

1 −

2

θ = 30º, 150º, 210º, 330º

Solve sin 2θ + cos θ = 0 for all values of θ. Give your answer in both radians and degrees.

sin 2θ + cos θ = 0 2 sin θ cos θ + cos θ = 0 cos θ (2 sin θ + 1) = 0cos θ = 0 or 2 sin θ + 1 = 0 sin θ = -

1 −

2

θ = 90º + k · 180º; θ = 210º + k · 360º,

θ = π −

2 + k · π 330º + k · 360º;

θ = 7π

−

6 + k · 2π,

11π

−

6 + k · 2π

Exercises

Solve each equation for the given interval.

1. 2 cos2 θ + cos θ = 1, 0 ≤ θ < 2π 2. sin2 θ cos2 θ = 0, 0 ≤ θ < 2π

3. cos 2θ = √ � 3

−

2 , 0º ≤ θ < 360º 4. 2 sin θ - √ � 3 = 0, 0 ≤ θ < 2π

Solve each equation for all values of θ if θ is measured in radians.

5. 4 sin2 θ - 3 = 0 6. 2 cos θ sin θ + cos θ = 0

Solve each equation for all values of θ if θ is measured in degrees.

7. cos 2θ + sin2 θ = 1 −

2 8. tan 2θ = -1

Example 1 Example 2


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Solving Trigonometric Equations

Extraneous Solutions Some trigonometric equations have no solutions. For example, the equation sin θ = 3 has no solution because all values of sin θ are between -1 and 1.

Solve 2 cos2 x + 3 cos x - 2 = 0 if 0 ≤ θ ≤ 2π.

2 cos2 θ + cos θ - 1 = 0 Original equation

(cos θ + 2)(2 cos θ - 1) = 0 Factor.

cos θ + 2 = 0 or 2 cos θ - 1 = 0

cos θ = -2 2 cos θ = 1

cos θ = 1 −

2

θ =

π

−

3 ; 5π

−

3

There is no solution to cos θ = -2 since all values of cos θ are between -1 and 1, inclusive.The solutions are π −

3 and 5π

−

3 .

Exercises

Solve each equation if 0 ≤ θ ≤ 2π.

1. sin2 θ +

7 −

2 sin θ +

3 −

2 = 0 2. 2 tan4 θ = sec2 θ

3. 8 cos θ = 4 cos2 θ + 3 4. 2 csc2 θ = -(3 csc θ + 1)

5. 2 sin2 θ = 6 - 5 √ � 2 sin θ 6. 2 cos4 θ + 9 sin2 θ = 5

Example


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Skills PracticeSolving Trigonometric Equations


1. sin θ = √ � 2

−

2 , 0° ≤ θ ≤ 360° 2. 2 cos θ = - √ � 3 , 90° < θ < 180°

3. tan 2 θ = 1, 180° < θ < 360° 4. 2 sin θ = 1, 0 ≤ θ < π

5. sin 2 θ + sin θ = 0, π ≤ θ < 2π 6. 2 cos 2 θ + cos θ = 0, 0 ≤ θ < π


7. 2 cos 2 θ - cos θ = 1 8. sin 2 θ - 2 sin θ + 1 = 0

9. sin θ + sin θ cos θ = 0 10. sin 2 θ = 1

11. 4 cos θ = -1 + 2 cos θ 12. tan θ cos θ = 1 −

2


13. 2 sin θ + 1 = 0 14. 2 cos θ + √ � 3 = 0

15. √ � 2 sin θ + 1 = 0 16. 2 cos 2 θ = 1

17. 4 sin 2 θ = 3 18. cos 2θ = -1

Solve each equation.

19. 3 cos 2 θ - sin 2 θ = 0 20. sin θ + sin 2θ = 0

21. 2 sin2 θ = sin θ + 1 22. cos θ + sec θ = 2


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PracticeSolving Trigonometric Equations


1. sin 2θ = cos θ, 90º ≤ θ < 180º 2. √ � 2 cos θ = sin 2θ , 0º ≤ θ , 360º

3. cos 4θ = cos 2θ, 180º ≤ θ < 360º 4. cos θ + cos (90 - θ) = 0, 0 ≤ θ < 2π

5. 2 + cos θ = 2 sin2 θ, π ≤ θ ≤ 3π

−

2 6. tan2 θ + sec θ = 1, π −

2 ≤ θ < π


7. cos2 θ = sin2 θ 8. cot θ = cot3 θ

9. √ � 2 sin3 θ = sin2 θ 10. cos2 θ sin θ = sin θ

11. 2 cos 2θ = 1 - 2 sin2 θ 12. sec2 θ = 2


13. sin2 θ cos θ = cos θ 14. csc2 θ - 3 csc θ + 2 = 0

15. 3 −

1 + cos θ = 4(1 - cos θ) 16. √ � 2 cos2 θ = cos2 θ

Solve each equation.

17. 4 sin2 θ = 3 18. 4 sin2 θ - 1 = 0

19. 2 sin2 θ - 3 sin θ = -1 20. cos 2θ + sin θ - 1 = 0

21. WAVES Waves are causing a buoy to float in a regular pattern in the water. The vertical position of the buoy can be described by the equation h = 2 sin x. Write an expression that describes the position of the buoy when its height is at its midline.

22. ELECTRICITY The electric current in a certain circuit with an alternating current can be described by the formula i = 3 sin 240t, where i is the current in amperes and t is the time in seconds. Write an expression that describes the times at which there is no current.


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Word Problem PracticeSolving Trigonometric Equations

1. SANDCASTLES The water level on Sunset Beach can be modeled by the function y = 7 + 7 sin

(

π

−

6t), where y is the

distance in feet of the shoreline above the low tide mark and t is the number of hours past 6 A.M. At 2 P.M., Victoria built her sandcastle 10 feet above the low tide mark. At what time will the shoreline reach Victoria’s sandcastle?

2. BATTERY The amount of light emitted from a battery indicator bulb pulses while the battery is charging. This can be modeled by the equation y = 60 + 60 sin

(

π

−

4t), where y is the

lumens emitted from the bulb and tis the number of seconds since the beginning of a pulse. At what time will the amount of light emitted be equal to 110 lumens?

3. BUILDINGS In downtown Centerville, the length of the shadows of the hotel and the bank depends on the angle of inclination of the Sun θ.

Bank

Hotel

θθ

82 m

40 m

50 m

a. Express the length of the shadow of each building as a function of the angle of inclination.

b. What is the greatest angle of inclination of the Sun such that the base of the bank is touched by the shadow of the hotel?

c. At what angle of inclination of the Sun will the bank’s shadow be equal to the height of the hotel?


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Enrichment

Mechanical EngineeringMechanical engineering technicians help build machines. Devices such as watches, railroad-car suspension systems, and space-shuttle cameras, require the use of springs. Technicians who work with these types of machines must study the stretch and compression of springs made of various metals.

Suppose a mechanical engineering technician pulls a weight down 4 centimeters from its equilibrium position and releases it. If other forces are neglected, the weight will bob up and down as shown in the diagram below.

An equation for the displacement d (in centimeters) of the weight over time t (in seconds) is given by the equation below.

D = -4 cos ( π t −

3 )

After how many seconds will the weight first be 1 centimeter above the equilibrium point?

Sketch the graph at the right. From the graph, t ≈ 1.8.

The weight will be 1 centimeter above the equilibrium point for the first time after about 1.8 seconds.

1. From the graph, find the displacement of the weight after 3 seconds. Does your answer agree with the diagram above?

2. From the graph, estimate the first time that the displacement will be 2 centimeters below equilibrium.

3. Find the second time that the weight is at equilibrium.

-2

-4

2

4

1 2 3 4 5 6

d d = -4 cos

t

3πt

compressed

equillbrium

stretchedt = 0 t = 3 t = 4.5 t = 6t = 1.5

4 cm

4 cm


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