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Lesson Menu 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.D 5-Minute Check 1 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.D 5-Minute Check 2 A. B. C. D.

Over Lesson 14–4 A.A B.B C.C D.D 5-Minute Check 3 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.D 5-Minute Check 4 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.D 5-Minute Check 5 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.D 5-Minute Check 6 A.sin 2 x B.cos 2 x C.sec 2 x D.csc 2 x Simplify the expression tan x (cot x + tan x).

Then/Now You verified trigonometric identities. (Lessons 14–2 through 14–4) Solve trigonometric equations. Find extraneous solutions from trigonometric equations.

Vocabulary trigonometric equations

Example 1 Solve Equations for a Given Interval Solve 2cos 2  – 1 = sin  if 0  ≤   180 . 2cos 2  – 1=sin  Original equation 2(1 – sin 2  ) – 1 – sin  =0Subtract sin  from each side. 2 – 2 sin 2  – 1 – sin  =0Distributive Property –2 sin 2  – sin  + 1=0Simplify. 2 sin 2  + sin  – 1=0Divide each side by –1. (2 sin  – 1)(sin  + 1)=0Factor.

Example 1 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.D Example 1 A.0°, 90°, 180° B.0°, 180°, 270° C.90°, 180°, 270° D.0°, 90°, 270° Find all solutions of sin 2  + cos 2  – cos  = 0 for the interval 0  ≤  ≤ 360 .

Example 2 Infinitely Many Solutions Look at the graph of y = cos  – sin 2  to find solutions of cos  – sin 2  = – A. Solve cos  + = sin 2  for all values of  if  is measured in degrees.

Example 2 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.

Example 2 Infinitely Many Solutions 2cos  = –1 B. Solve 2cos  = –1 for all values of  if  is measured in radians.

Example 2 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 Example 2 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.D Example 2 B. Solve 2 sin  = –2 for all values of  if  is measured in radians. A. B. C. D.

Example 3 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 3  t. How long after the Ferris wheel starts will your seat first be meters above the ground?

Example 3 Solve Trigonometric Equations Original equation Replace h with Subtract 21 from each side. Divide each side by –20. Take the Arccosine.

Example 3 Solve Trigonometric Equations Divide each side by 3 . The Arccosine of Answer:

A.A B.B C.C D.D Example 3 A.about 7 seconds B.about 10 seconds C.about 13 seconds D.about 16 seconds 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 3  t. How long after the Ferris wheel starts will your seat first be 11 meters above the ground?

Example 4 Determine Whether a Solution Exists A. Solve the equation sin  cos  = cos 2  if 0 ≤  ≤ 2 . sin  cos  = cos 2  Original equation sin  cos  – cos 2  = 0Subtract cos 2  from each side. cos  (sin  – cos  )= 0Factor. cos  = 0orsin  – cos  = 0Zero Product Property sin  = cos  Divide each side by cos 

Example 4 Determine Whether a Solution Exists Divide each side by cos . tan  = 1

Example 4 Determine Whether a Solution Exists Check ? ? ? ?

Example 4 Determine Whether a Solution Exists ? ? ? ? Answer:

Example 4 Determine Whether a Solution Exists B. Solve the equation cos  = 1 – sin  if 0° ≤  < 360°. cos  = 1 – sin  Original equation cos 2  = (1 – sin  ) 2 Square each side. 1 – sin 2  = (1 – 2 sin  + sin 2  )cos 2  = 1 – sin 2  0= 2 sin 2  – 2 sin  Simplify. 0= sin  (2 sin  – 2)Factor. sin  = 0or2 sin  – 2 = 0Zero ProductProperty  = 0  or 180  sin  = 1Solve for sin   = 90 

Example 4 Determine Whether a Solution Exists Check 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=00=0 0=1 – 1 ? Answer:0° and 90°

A.A B.B C.C D.D Example 4 A. Solve the equation cos  = (1 – sin 2  ) if 0 ≤  < 2 . A. B. C. D.

A.A B.B C.C D.D Example 4 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  = sin 2  if 0  ≤  < 360 .

Example 5 Solve Trigonometric Equations by Using Identities Solve tan 4  – 4 sec 2  = –7 for all values of  if  is measured in degrees. tan 4  – 4 sec 2  = –7Original equation (tan 2  ) 2 – 4(1 + tan 2  )= – 7sec 2  = 1 + tan 2  (tan 2  ) 2 – 4 – 4 tan 2  = –7Distribute. (tan 2  ) 2 – 4 tan 2  + 3= 0Add 7 to each side. (tan 2  – 3)(tan 2  – 1)= 0Factor. (tan 2  – 3) = 0or(tan 2  – 1) = 0Zero Product Property

Example 5 Solve Trigonometric Equations by Using Identities Answer:  = 60° + 180°k,  = °k, and  = 45° + 90°k, where k is any integer. tan 2  = 3ortan 2  = 1  = 60°, 120°, 180°,   = 45°, 135°, 225°,  tan  = ortan  =  1

A.A B.B C.C D.D Example 5 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 sin 4  – 2sin 2  + 6 = 5 for all values of  if  is measured in degrees.

End of the Lesson