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Five-Minute Check (over Chapter 13) Then/Now New Vocabulary Key Concept: Basic Trigonometric Identities Example 1: Use Trigonometric Identities Example 2: Simplify an Expression Example 3: Real-World Example: Simplify and Use an Expression Lesson Menu

Rewrite 240° in radians. A. B. C. D. A B C D 5-Minute Check 1

A B C D Find the exact value of tan (–120°). A. B. C. D. 5-Minute Check 2

Solve ΔABC shown. Round measures of sides to the nearest tenth and measures of angles to the nearest degree. A. c = 5.3, A = 26°, B = 118° B. c = 5.3, A = 118°, B = 26° C. c = 6, A = 26°, B = 118° D. c = 6, A = 118°, B = 26° A B C D 5-Minute Check 3

Georgia is watching a space shuttle lift off from 0. 5 mile away Georgia is watching a space shuttle lift off from 0.5 mile away. When she looks up at the shuttle at a 60° angle, how high is the space shuttle? A. 0.16 mi B. 0.87 mi C. 31.42 mi D. 62.83 mi A B C D 5-Minute Check 4

A B C D Find the amplitude and period of y = 2 cos A. 2 B. 2; 4 5-Minute Check 5

Which measure is needed to solve ΔABC using the Law of Cosines? A. side a B. side b C. angle B D. angle C A B C D 5-Minute Check 6

You evaluated trigonometric functions. (Lesson 13–7) Use trigonometric identities to find trigonometric values. Use trigonometric identities to simplify expressions. Then/Now

trigonometric identity Vocabulary

Concept

A. Find tan  if sec  = –2 and 180 <  < 270. Use Trigonometric Identities A. Find tan  if sec  = –2 and 180 <  < 270. tan2 + 1 = sec2 Trigonometric identity tan2 = sec2 – 1 Subtract 1 from each side. tan2 = (–2)2 – 1 Substitute –2 for sec . tan2 = 4 – 1 Square –2. tan2 = 3 Subtract. Take the square root of each side. Answer: Since  is in the third quadrant, tan  is positive. Thus, tan θ = Example 1A

sin2 + cos2 = 1 Trigonometric identity Use Trigonometric Identities B. sin2 + cos2 = 1 Trigonometric identity Subtract. Take the square root of each side. Example 1B

Answer: Since  is in the second quadrant, sin  is positive. Thus, . Use Trigonometric Identities Answer: Since  is in the second quadrant, sin  is positive. Thus, . Example 1B

A. B. C. D. A B C D Example 1A

A. 5 B. C. D. A B C D Example 1B

Simplify sin  (csc  – sin  ). Simplify an Expression Simplify sin  (csc  – sin  ). Distributive Property = 1 – sin2  Simplify. = cos2  1 – sin2  = cos2  Answer: cos2  Example 2

Simplify tan  cot . A. –1 B. 0 C. 1 D. –2 A B C D Example 2

Simplify and Use an Expression A. LIGHTING The amount of light that a source provides to a surface is called the illuminance. The illuminance E in foot candles on a surface is related to the distance R in feet from the light source. The formula where I is the intensity of the light source measured in candles and θ is the angle between the light beam and a line perpendicular to the surface, can be used in situations in which lighting is important, as in photography. Solve the illuminance formula in terms of R. Example 3A

Multiply each side by ER2. Simplify and Use an Expression Original equation Multiply each side by ER2. Divide each side by E. Example 3A

Multiply each side by cos . Simplify and Use an Expression Multiply each side by cos . Take the square root of each side. Answer: Example 3A

B. Is the equation in part A equivalent to Simplify and Use an Expression B. Is the equation in part A equivalent to Original equation Cross products Divide each side by E. Example 3B

Take the square root of each side. Simplify and Use an Expression Take the square root of each side. Answer: Example 3B

A. Solve the formula A. B. C. D. A B C D Example 3A

B. Is the equation equivalent to ? A. yes B. no A B Example 3

End of the Lesson