splash screen. lesson menu five-minute check (over lesson 14–4) then/now new vocabulary example...
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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
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Find the exact value of sin 2 when
cos = – and 180° < < 270°.__7
8
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B.
C.
D.
Over Lesson 14–4
A. A
B. B
C. C
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A.
B.
C.
D.
Over Lesson 14–4
A. A
B. B
C. C
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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
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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
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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
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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
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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.