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Unformatted text preview: Gauthier, Joseph Homework 1 Due: Sep 11 2007, 3:00 am Inst: JEGilbert 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. YES, homework 1 is due AFTER homework 2 001 (part 1 of 1) 10 points Express the function f ( x ) = 5 sin 2 x 4 cos 2 x in terms of cos 2 x . 1. f ( x ) = 1 2 + 9 2 cos 2 x 2. f ( x ) = 9 + 1 2 cos 2 x 3. f ( x ) = 9 2 1 2 cos 2 x 4. f ( x ) = 9 2 9 2 cos 2 x 5. f ( x ) = 1 2 9 2 cos 2 x correct 6. f ( x ) = 1 2 9 cos 2 x Explanation: Since sin 2 x = 1 2 (1 cos 2 x ) and cos 2 x = 1 2 (1 + cos 2 x ) , we can rewrite f as f ( x ) = 5 2 (1 cos 2 x ) 2(1 + cos 2 x ) . Consequently, f ( x ) = 1 2 9 2 cos 2 x . keywords: 002 (part 1 of 1) 10 points Simplify the expression f ( ) = 1 cos 4 , eliminating the radical. 1. f ( ) = 2 sin 2 2. f ( ) = 2  cos 2  3. f ( ) = 2 cos 2 8 4. f ( ) = 2 sin 2 8 5. f ( ) = 2 cos 2 6. f ( ) = 2  sin 2  correct Explanation: By the double angle formula cos 2 x = 1 2 sin 2 x. With x = 2 , therefore, we see that 1 cos 4 = p 2 sin 2 2 . Consequently, f ( ) = 2  sin 2  , the absolute value being included to ensure that f ( ) 0. keywords: trig identity, simplify trig expres sion, double angle formula 003 (part 1 of 1) 10 points When 1 2 3 1 2 3 1 2 3 Gauthier, Joseph Homework 1 Due: Sep 11 2007, 3:00 am Inst: JEGilbert 2 is the graph of y = a + b cos mx, ( m > 0) , on [ 4 , 4], what is b ? 1. b = 2 2. b = 4 3. b = 7 4 4. b = 4 5. b = 2 correct Explanation: As 1 2 3 1 2 3 1 2 3 shows, the given graph is that of y = 3 2 + 2 cos 1 2 x, in other words, the graph of y = 2 cos 1 2 x shifted vertically by a term y = 3 2 . Thus b is given by b = 2 . keywords: graph, trig function, phase, ampli tude, period, vertical shift 004 (part 1 of 3) 10 points The figure ABCD in A C B D a b is a parallelogram and BD is a diagonal. (i) Express the area of ABCD as a function of a, b and . 1. Area ABCD = 1 2 ab cos 2. Area ABCD = 1 2 ab sin 3. Area ABCD = ab cos 4. Area ABCD = 2 ab sin 5. Area ABCD = ab sin correct Explanation: In the parallelogram ABCD the triangles ABD and BDC are congruent by SSS since side BD is common to both triangles, while side AD is congruent to side BC and side AB is congruent to side DC . Thus Area ABCD = Area ABD + Area BDC = 2 Area ABD ....
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 Fall '07
 Gilbert

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