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Splash Screen. Five-Minute Check (over Lesson 13–4) CCSS 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

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**Five-Minute Check (over Lesson 13–4)**CCSS 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 Lesson Menu**Find the exact value of sin 2 when cos = – and**180° < < 270°. A. B. C. D. 7 __ 8 5-Minute Check 1**Find the exact value of sin 2 when cos = – and**180° < < 270°. A. B. C. D. 7 __ 8 5-Minute Check 1**A.**B. C. D. 5-Minute Check 2**A.**B. C. D. 5-Minute Check 2**A.**B. C. D. Find the exact value of sin 67.5° by using half-angle formulas. 5-Minute Check 3**A.**B. C. D. Find the exact value of sin 67.5° by using half-angle formulas. 5-Minute Check 3**A.**B. C. D. Find the exact value of cos 22.5° by using double-angle formulas. 5-Minute Check 4**A.**B. C. D. Find the exact value of cos 22.5° by using double-angle formulas. 5-Minute Check 4**Find the exact value of tan by using double-angle**formulas. A. B. C. D. 5-Minute Check 5**Find the exact value of tan by using double-angle**formulas. A. B. C. D. 5-Minute Check 5**Simplify the expression tan x (cot x + tan x).**A. sin2x B. cos2x C. sec2x D. csc2x 5-Minute Check 6**Simplify the expression tan x (cot x + tan x).**A. sin2x B. cos2x C. sec2x D. csc2x 5-Minute Check 6**Content Standards**F.TF.8 Prove the Pythagorean identity sin2 (θ) + cos2 (θ) = 1 and use it to find sin (θ), cos (θ), or tan (θ) given sin (θ), cos (θ), or tan (θ) and the quadrant of the angle. Mathematical Practices 4 Model with mathematics. 6 Attend to precision. CCSS**You verified trigonometric identities.**• Solve trigonometric equations. • Find extraneous solutions from trigonometric equations. Then/Now**trigonometric equations**Vocabulary**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. Example 1**Solve Equations for a Given Interval**Now use the Zero Product Property. 2 sin – 1 = 0 sin + 1 = 0 2 sin = 1 sin = –1 = 270° = 30° or 150° Answer: Example 1**Solve Equations for a Given Interval**Now use the Zero Product Property. 2 sin – 1 = 0 sin + 1 = 0 2 sin = 1 sin = –1 = 270° = 30° or 150° Answer: Since 0°≤ ≤ 180°, the solutions are 30°, and 150°. Example 1**Find all solutions of sin2 + cos 2 – cos = 0 for**the interval 0≤ ≤ 360. A. 0°, 90°, 180° B. 0°, 180°, 270° C. 90°, 180°, 270° D. 0°, 90°, 270° Example 1**Find all solutions of sin2 + cos 2 – cos = 0 for**the interval 0≤ ≤ 360. A. 0°, 90°, 180° B. 0°, 180°, 270° C. 90°, 180°, 270° D. 0°, 90°, 270° Example 1**A. Solve cos + = sin2for all values of if** is measured in degrees. Look at the graph of y = cos – sin2 to find solutions of cos – sin2 = – Infinitely Many Solutions Example 2**Infinitely Many Solutions**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. Answer: Example 2**Infinitely Many Solutions**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. Answer: 60° + k ● 360° and 300° + k ● 360°, where k is measured in degrees. Example 2**Infinitely Many Solutions**B. Solve 2cos = –1 for all values of if is measured in radians. 2cos = –1 Example 2**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. Infinitely Many Solutions Answer: Example 2**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. Answer: , where k is any integer. Infinitely Many Solutions 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 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 Example 2**A.**B. C. D. B. Solve 2 sin = –2 for all values of if is measured in radians. Example 2**A.**B. C. D. B. Solve 2 sin = –2 for all values of if is measured in radians. Example 2**AMUSEMENT PARKSWhen 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 Example 3**Replace h with**Solve Trigonometric Equations Original equation Subtract 21 from each side. Divide each side by –20. Take the Arccosine. Example 3**The Arccosine of**Solve Trigonometric Equations Divide each side by 3. Answer: Example 3**The Arccosine of**Answer: Solve Trigonometric Equations Divide each side by 3. Example 3**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? A. about 7 seconds B. about 10 seconds C. about 13 seconds D. about 16 seconds Example 3**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? A. about 7 seconds B. about 10 seconds C. about 13 seconds D. about 16 seconds Example 3**sin = cos Divide each side by cos **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 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**Answer:**? ? ? ? Determine Whether a Solution Exists Example 4**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 – sin2 0 = 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 Example 4**?**? cos (0°) = 1 – sin (0°) cos (180°) = 1 – sin (180°) ? ? 1 = 1 – 0 –1 = 1 – 0 ? cos (90°) = 1 – sin (90°) ? 0 = 1 – 1 Determine Whether a Solution Exists Check cos = 1 – sin cos = 1 – sin 1 = 1 –1 = 1 cos = 1 – sin 0 = 0 Answer: Example 4**?**? cos (0°) = 1 – sin (0°) cos (180°) = 1 – sin (180°) ? ? 1 = 1 – 0 –1 = 1 – 0 ? cos (90°) = 1 – sin (90°) ? 0 = 1 – 1 Determine Whether a Solution Exists Check cos = 1 – sin cos = 1 – sin 1 = 1 –1 = 1 cos = 1 – sin 0 = 0 Answer: 0° and 90° Example 4**A.**B. C. D. A. Solve the equation cos = (1 – sin2 ) if 0 ≤ < 2. Example 4**A.**B. C. D. A. Solve the equation cos = (1 – sin2 ) if 0 ≤ < 2. Example 4**B. Solve the equation sin cos = sin2 if 0 ≤** < 360. A. 0°, 45°, 180°, 225° B. 0°, 90°, 180°, 270° C. 30°, 45°, 225°, 330° D. 30°, 90°, 180°, 330° Example 4**B. Solve the equation sin cos = sin2 if 0 ≤** < 360. A. 0°, 45°, 180°, 225° B. 0°, 90°, 180°, 270° C. 30°, 45°, 225°, 330° D. 30°, 90°, 180°, 330° Example 4

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