hw 2 - rhodes (ajr2283) – HW02 – Gilbert – (56380) 1...

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Unformatted text preview: rhodes (ajr2283) – HW02 – Gilbert – (56380) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Use double angle formulas to simplify the expression f ( θ ) = (5 cos θ + sin θ ) 2 . 1. f ( θ ) = 12 + 5 sin 2 θ + 13 cos2 θ 2. f ( θ ) =- 5 + 13 sin2 θ + 12 cos 2 θ 3. f ( θ ) = 5 + 12 sin2 θ + 13 cos2 θ 4. f ( θ ) = 13 + 12 sin 2 θ + 5 cos2 θ 5. f ( θ ) = 12 + 12 sin 2 θ + 5 cos2 θ 6. f ( θ ) = 13+5 sin 2 θ +12 cos 2 θ correct Explanation: After expansion we see that f ( θ ) = 25 cos 2 θ + 10 sin θ cos θ + sin 2 θ . But cos 2 θ = 1 2 (1 + cos 2 θ ) , sin 2 θ = 1 2 (1- cos 2 θ ) , while 2 sin θ cos θ = sin 2 θ . Consequently, f ( θ ) = 13 + 5 sin 2 θ + 12 cos2 θ . 002 10.0 points When 1 2 3- 1- 2- 3 1 2 3- 1 is the graph of y = a + b cos mx, ( m > 0) , on [- 4 , 4], what is b ? 1. b = 3 2 2. b =- 2 3. b = 2 correct 4. b =- 4 5. b = 4 Explanation: As 1 2 3- 1- 2- 3 1 2 3- 1 shows, the given graph is that of y = 1 + 2 cos 3 2 πx , in other words, the graph of y = 2 cos 3 2 πx shifted vertically by a term y = 1. Thus b is given by b = 2 . 003 (part 1 of 3) 10.0 points The figure ABCD in rhodes (ajr2283) – HW02 – Gilbert – (56380) 2 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 = 2 ab sin θ 2. Area ABCD = ab cos θ 3. Area ABCD = 1 2 ab sin θ 4. Area ABCD = 1 2 ab cos θ 5. Area ABCD = ab sin θ correct 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 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. Consequently, Area ABCD = ab sin θ . 004 (part 2 of 3) 10.0 points (ii) Find θ when a = 2, b = 6 and the length of the diagonal BD is 5. 1. θ = 54 . 32 ◦ 2. θ = 55 . 32 ◦ 3. θ = 53 . 32 ◦ 4. θ = 51 . 32 ◦ correct 5. θ = 52 . 32 ◦ Explanation: To determine θ we use the Law of Cosines: with c the length of diagonal BD , c 2 = a 2 + b 2- 2 ab cos θ. When a = 2 , b = 6 , c = 5 , therefore, cos θ = 4 + 36- 25 24 , rhodes (ajr2283) – HW02 – Gilbert – (56380) 3 so θ = arccos 5 8 = 51 . 32 ◦ . 005 (part 3 of 3) 10.0 points (iii) Then find the area of parallelogram ABCD . 1. Area ABCD = 10 . 37 sq. units 2. Area ABCD = 9 . 87 sq. units 3. Area ABCD = 9 . 37 sq. units correct 4. Area ABCD = 9 . 62 sq. units 5. Area ABCD = 10 . 12 sq. units Explanation: By parts (i) and (ii) Area ABCD = ab sin θ = 12 sin 51 . 32 ◦ ....
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This note was uploaded on 09/27/2010 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas at Austin.

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hw 2 - rhodes (ajr2283) – HW02 – Gilbert – (56380) 1...

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