Five-Minute Check (over Lesson 14–4)

Then/Now

New Vocabulary

Example 1:Solve Equations for a Given Interval

Example 2:Infinitely Many Solutions

Example 3:Real-World Example: Solve Trigonometric Equations

Example 4:Determine Whether a Solution Exists

Example 5:Solve Trigonometric Equations by Using Identities

Over Lesson 14–4

A. A

B. B

C. C

D. D0% 0%0%0%

Find the exact value of sin 2 when

cos = – and 180° < < 270°.__7

8

A.

B.

C.

D.

Over Lesson 14–4

A. A

B. B

C. C

D. D0% 0%0%0%

A.

B.

C.

D.

Over Lesson 14–4

A. A

B. B

C. C

D. D0% 0%0%0%

Find the exact value of sin 67.5° by using half-angle formulas.

A.

B.

C.

D.

Over Lesson 14–4

A. A

B. B

C. C

D. D0% 0%0%0%

Find the exact value of cos 22.5° by using double-angle formulas.

A.

B.

C.

D.

Over Lesson 14–4

A. A

B. B

C. C

D. D0% 0%0%0%

Find the exact value of tan by using double-angle formulas.

A.

B.

C.

D.

Over Lesson 14–4

A. A

B. B

C. C

D. D0% 0%0%0%

A. sin2 x

B. cos2 x

C. sec2 x

D. csc2 x

Simplify the expression tan x (cot x + tan x).

You verified trigonometric identities. (Lessons 14–2 through 14–4)

• Solve trigonometric equations.

• Find extraneous solutions from trigonometric equations.

• trigonometric equations

Solve Equations for a Given Interval

Solve 2cos2 – 1 = sin if 0 ≤ 180.

2cos2 – 1

=

sin

Original equation

2(1 – sin2 ) – 1 – sin

=

0

Subtract sin from each side.

2 – 2 sin2 – 1 – sin

=

0

Distributive Property

–2 sin2 – sin + 1

=

0

Simplify.

2 sin2 + sin – 1

=

0

Divide each side by –1.

(2 sin – 1)(sin + 1)

=

0

Factor.

Solve Equations for a Given Interval

2 sin – 1 = 0 sin + 1 = 0

Answer: Since 0° ≤ ≤ 180°, the solutions are 30°, and 150°.

2 sin = 1 sin = –1

= 30° or 150°

Now use the Zero Product Property.

= 270°

A. A

B. B

C. C

D. D0% 0%0%0%

A. 0°, 90°, 180°

B. 0°, 180°, 270°

C. 90°, 180°, 270°

D. 0°, 90°, 270°

Find all solutions of sin2 + cos 2 – cos = 0 for the interval 0 ≤ ≤ 360.

Infinitely Many Solutions

Look at the graph of y = cos – sin2

to find solutions of cos – sin2 = –

A. Solve cos + = sin2 for all values of if is

measured in degrees.

Infinitely Many Solutions

Answer: 60° + k ● 360° and 300° + k ● 360°, where k is measured in degrees.

The solutions are 60°, 300°, and so on, and –60°, –300°, and so on. The period of the function is 360°. So the solutions can be written as 60° + k ● 360° and 300° + k ● 360°, where k is measured in degrees.

Infinitely Many Solutions

2cos = –1

B. Solve 2cos = –1 for all values of if is measured in radians.

Infinitely Many Solutions

Answer: , where k is any

integer.

The solutions are , and so on, and ,

and so on. The period of the cosine function is

2 radians. So the solutions can be written as

, where k is any integer.

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. Solve cos 2 sin + 1 = 0 for all values of if is measured in degrees.

A. 0° + k ● 360° and 45° + k ● 360° where k is any integer

B. 45° + k ● 360° where k is any integer

C. 0° + k ● 360° and 90° + k ● 360° where k is any integer

D. 90° + k ● 360° where k is any integer

A. A

B. B

C. C

D. D0% 0%0%0%

B. Solve 2 sin = –2 for all values of if is measured in radians.

A.

B.

C.

D.

Solve Trigonometric Equations

AMUSEMENT PARKS When you ride a Ferris

wheel that has a diameter of 40 meters and turns

at a rate of 1.5 revolutions per minute, the height

above the ground, in meters, of your seat after

t minutes can be modeled by the equation

h = 21 – 20 cos 3t. How long after the Ferris wheel

starts will your seat first be meters

above the ground?

Solve Trigonometric Equations

Original equation

Replace h with

Subtract 21 fromeach side.

Divide each side by –20.

Take the Arccosine.

Solve Trigonometric Equations

Divide each side by 3.

The Arccosine of

Answer:

A. A

B. B

C. C

D. D0% 0%0%0%

A. about 7 seconds

B. about 10 seconds

C. about 13 seconds

D. about 16 seconds

AMUSEMENT PARKS When you ride a Ferris wheel thathas a diameter of 40 meters and turns at a rate of 1.5 revolutions per minute, the height above the ground,in meters, of your seat after t minutes can be modeled by the equation h = 21 – 20 cos 3t. How long after theFerris wheel starts will your seat first be 11 meters above the ground?

Determine Whether a Solution Exists

A. Solve the equation sin cos = cos2 if 0 ≤ ≤ 2.

sin cos

= cos2

Original equation

sin cos – cos2

= 0

Subtract cos2 from each side.

cos (sin – cos )

= 0

Factor.

cos = 0 or sin – cos = 0 Zero Product Property

sin = cos Divide each side by cos

Determine Whether a Solution Exists

Divide each side by cos .

tan = 1

Determine Whether a Solution Exists

Check

?

?

?

?

Determine Whether a Solution Exists

?

?

?

?

Answer:

Determine Whether a Solution Exists

B. Solve the equation cos = 1 – sin if 0° ≤ < 360°.

cos = 1 – sin Original equation

cos2 = (1 – sin)2 Square each side.

1 – sin2 = (1 – 2 sin + sin2 ) cos2 = 1 – sin20 = 2 sin2 – 2 sin Simplify.

0 = sin (2 sin – 2) Factor.

sin = 0 or 2 sin – 2 = 0 Zero Product Property

= 0 or 180 sin = 1 Solve for sin

= 90

Determine Whether a Solution Exists

Check

cos = 1 – sin cos = 1 – sin ? ?

cos (0°) = 1 – sin (0°) cos (180°) = 1 – sin (180°)

1 = 1 –1 = 1

1 = 1 – 0 –1 = 1 – 0? ?

cos = 1 – sin

?cos (90°) = 1 – sin (90°)

0 = 0

0 = 1 – 1?

Answer: 0° and 90°

A. A

B. B

C. C

D. D0% 0%0%0%

A. Solve the equation cos = (1 – sin2 ) if 0 ≤ < 2.

A.

B.

C.

D.

A. A

B. B

C. C

D. D A B C D

0% 0%0%0%

A. 0°, 45°, 180°, 225°

B. 0°, 90°, 180°, 270°

C. 30°, 45°, 225°, 330°

D. 30°, 90°, 180°, 330°

B. Solve the equation sin cos = sin2 if 0 ≤ < 360.

Solve Trigonometric Equations by Using Identities

Solve tan4 – 4 sec2 = –7 for all values of if is measured in degrees.

tan4 – 4 sec2 = –7 Original equation

(tan2 )2 – 4(1 + tan2 ) = – 7 sec2 = 1 + tan2 (tan2 )2 – 4 – 4 tan2 = –7 Distribute.

(tan2 )2 – 4 tan2 + 3 = 0 Add 7 to each side.

(tan2 – 3)(tan2 – 1) = 0 Factor.(tan2 – 3) = 0 or (tan2 – 1) = 0 Zero Product

Property

Solve Trigonometric Equations by Using Identities

Answer: = 60° + 180°k, = 120 + 180°k, and = 45° + 90°k, where k is any integer.

tan2 = 3 or tan2 = 1

= 60°, 120°, 180°, = 45°, 135°, 225°,

tan = or tan = 1

A. A

B. B

C. C

D. D0% 0%0%0%

A. = 45° + 180°k, where k is any integer.

B. = 90° + 180°k, where k is any integer.

C. = 45° + 90°k, where k is any integer.

D. = 135° + 45°k, where k is any integer.

Solve sin4 – 2sin2 + 6 = 5 for all values of if is measured in degrees.