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HW_1_Key

# HW_1_Key - Miller Kierste Homework 1 Due 3:00 am Inst...

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Miller, Kierste – Homework 1 – Due: Sep 11 2007, 3:00 am – Inst: JEGilbert 1 This print-out should have 17 questions. Multiple-choice 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 + cos 2 x in terms of cos 2 x . 1. f ( x ) = 3 - 3 cos 2 x 2. f ( x ) = 2 - 3 cos 2 x 3. f ( x ) = 3 - 4 cos 2 x 4. f ( x ) = 3 - 2 cos 2 x correct 5. f ( x ) = 2 + 3 cos 2 x 6. f ( x ) = 4 + 3 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 ) + 1 2 (1 + cos 2 x ) . Consequently, f ( x ) = 3 - 2 cos 2 x . keywords: 002 (part 1 of 1) 10 points Simplify the expression f ( θ ) = 1 + cos 2 θ , eliminating the radical. 1. f ( θ ) = 2 cos 2 4 θ 2. f ( θ ) = 2 | cos θ | correct 3. f ( θ ) = 2 | sin θ | 4. f ( θ ) = 2 sin θ 5. f ( θ ) = 2 sin 2 4 θ 6. f ( θ ) = 2 cos θ Explanation: By the double angle formula cos 2 x = 2 cos 2 x - 1 . With x = 2 θ , therefore, we see that 1 + cos 2 θ = 2 cos 2 θ . Consequently, f ( θ ) = 2 | cos θ | , the absolute value being included to ensure that f ( θ ) 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

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Miller, Kierste – 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 correct 2. b = 4 3. b = - 2 4. b = 7 4 5. b = - 4 Explanation: As 1 2 3 - 1 - 2 - 3 1 2 3 shows, the given graph is that of y = 3 2 + 2 cos 3 2 πx , in other words, the graph of y = 2 cos 3 2 πx 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 θ a b is a parallelogram and BD is a diagonal. (i) Express the area of ABCD as a function of a, b and θ . 1. Area ABCD = 1 2 ab cos θ 2. Area ABCD = ab cos θ 3. Area ABCD = 2 ab sin θ 4. Area ABCD = ab sin θ correct 5. Area ABCD = 1 2 ab sin θ Explanation: In the parallelogram ABCD the triangles Δ ABD and Δ 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 . Now let h be the height of the perpendicular from D onto side AB of as shown in grey in
Miller, Kierste – Homework 1 – Due: Sep 11 2007, 3:00 am – Inst: JEGilbert 3 A C B D θ a b h Then h = a sin θ is the height of Δ ABD , while b is its base length. Thus Area ABD = 1 2 base × height = 1 2 ab sin θ.

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