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Find the exact value of (a) sin 15° and (b) tan. sin 15°. a. 7 π. 12. 2. 1. 2. 3. 2. 2. 2. 2. = ( ) – ( ). 6 – 2. =. 4. EXAMPLE 1. Evaluate a trigonometric expression. = sin ( 60° – 45° ). Substitute 60° – 45° for 15°.
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Find the exact value of (a) sin 15°and (b) tan . sin15° a. 7π 12 2 1 2 3 2 2 2 2 = ( ) – ( ) 6 – 2 = 4 EXAMPLE 1 Evaluate a trigonometric expression = sin (60° – 45°) Substitute 60° – 45° for 15°. Difference formula for sine = sin 60° cos 45° – cos 60° sin 45° Evaluate. Simplify.
π π 7π = tan ( + ) tan b. 4 3 Substitute + For . 12 tan + tan = π π 1– tan tan π π π π 3 4 3 4 4 3 3 + 1 7π = 1 – 3 1 12 = –2 – 3 EXAMPLE 1 Evaluate a trigonometric expression Sum formula for tangent. Evaluate. Simplify.
Find cos (a – b) given that cos a = – with π < a < and sinb = with 0 < b < . 5 5 3π 13 13 2 Using a Pythagorean identity and quadrant signs gives sina = and cosb = . 4 π 3 4 3 – 5 2 5 5 5 12 63 12 cos (a – b) = cosa cosb + sina sinb 65 13 13 = – ( ) + (– )( ) = – EXAMPLE 2 Use a difference formula SOLUTION Difference formula for cosine Substitute. Simplify.
6 3 6 2 6 2 2 π 3. tan 12 ANSWER ANSWER 2 + + 4 5π 12 4. cos ANSWER ANSWER + – 4 4 for Examples 1 and 2 GUIDED PRACTICE Find the exact value of the expression. 1. sin 105° 2. cos 75°
5. Find sin (a –b) given that sina = with 0 < a < and cosb = – with π < b < . 8 3π 17 2 ANSWER π 2 – 24 87 25 425 for Examples 1 and 2 GUIDED PRACTICE
cos (x + π) = cosx cosπ– sinx sinπ EXAMPLE 3 Simplify an expression Simplify the expression cos (x + π). Sum formula for cosine = (cos x)(–1) – (sinx)(0) Evaluate. = – cosx Simplify.
Solve sin ( x + ) + sin ( x – ) = 1 for 0 ≤ x < 2π. ANSWER π π 3 3 In the interval 0 ≤ x <2π, the only solution isx = . sin ( x + ) + sin ( x – ) = 1 1 1 π π sinx cos+ cosx sin+ sinx cos– cosx sin 2 2 π π π π π 3 3 = 1 2 3 3 3 3 sinx + cos x + sinx – cosx = 1 3 3 2 2 = 1 sinx EXAMPLE 4 Solve a trigonometric equation Write equation. Use formulas. Evaluate. Simplify.
π t π t 182 182 = –6cos ( ) + 12.1 = 2 sin ( – 1.35) + 12.1 EXAMPLE 5 Solve a multi-step problem Daylight Hours The number h of hours of daylight for Dallas, Texas, and Anchorage, Alaska, can be approximated by the equations below, where tis the time in days and t = 0 represents January 1. On which days of the year will the two cities have the same amount of daylight? Dallas: h1 Anchorage: h2
2 sin ( – 1.35) + 12.1 = – 6 cos ( ) + 12.1 sin ( – 1.35) = – 3 cos ( ) = – 3 cos ( ) sin ( ) cos 1.35– cos ( ) sin1.35 = – 3 cos ( ) sin ( ) (0.219)– cos ( ) (0.976) 0.219 sin ( ) = – 2.024 cos ( ) π t π t π t π t π t π t π t π t π t π t π t π t 182 182 182 182 182 182 182 182 182 182 182 182 EXAMPLE 5 Solve a multi-step problem SOLUTION Solve the equation h1 = h2 for t. STEP 1
tan ( ) – 1.463+ nπ – 84.76 + 182n 97, or on April 8 – 84.76 + 182(1) π t π t π t – 84.76 + 182(2) 279, or on October7 182 182 182 EXAMPLE 5 Solve a multi-step problem = – 9.242 = tan–1 (– 9.242)+ nπ t STEP 2 Find the days within one year (365 days) for which Dallas and Anchorage will have the same amount of daylight. t t
ANSWER ANSWER ANSWER sinx cosx tanx for Examples 3, 4, and 5 GUIDED PRACTICE Simplify the expression. 6. sin (x + 2π) 8. tan (x – π) 7.cos (x – 2π)
ANSWER about5.65 π t π t 75 75 for Examples 3, 4, and 5 GUIDED PRACTICE 9. Solve 6 cos ( ) + 5 = – 24 sin ( + 22) + 5 for 0 ≤ t < 2π.