Gilbert_Hmwk02sol

# Gilbert_Hmwk02sol - moseley (cmm3869) HW02 Gilbert (56380)...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: moseley (cmm3869) 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 ( ) = (3 cos - sin ) 2 . 1. f ( ) = 3 + 4 sin2 + 5 cos2 2. f ( ) =- 3 + 5 sin2 + 4 cos2 3. f ( ) = 4- 3 sin2 + 5 cos2 4. f ( ) = 5- 3 sin2 + 4 cos2 correct 5. f ( ) = 5 + 4 sin2 - 3 cos2 6. f ( ) = 4 + 4 sin2 - 3 cos2 Explanation: After expansion we see that f ( ) = 9 cos 2 - 6 sin cos + sin 2 . But cos 2 = 1 2 (1 + cos 2 ) , sin 2 = 1 2 (1- cos 2 ) , while 2 sin cos = sin 2 . Consequently, f ( ) = 5- 3 sin 2 + 4 cos2 . 002 10.0 points When 1 2 3- 1- 2- 3- 1- 2 is the graph of y = a + b cos mx, ( m &gt; 0) , on [- 4 , 4], what is b ? 1. b = 3 2 correct 2. b = 5 4 3. b =- 3 4. b =- 3 2 5. b = 3 Explanation: As 1 2 3- 1- 2- 3- 1- 2 shows, the given graph is that of y =- 1 + 3 2 cos 3 2 x , in other words, the graph of y = 3 2 cos 3 2 x shifted vertically by a term y =- 1. Thus b is given by b = 3 2 . 003 (part 1 of 3) 10.0 points The figure ABCD in moseley (cmm3869) HW02 Gilbert (56380) 2 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 = ab cos 2. Area ABCD = 1 2 ab sin 3. Area ABCD = 1 2 ab cos 4. Area ABCD = ab sin correct 5. Area ABCD = 2 ab sin 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 . Now let h be the height of the perpendicular from D onto side AB of as shown in grey in A C B D a b h Then h = a sin is the height of ABD , while b is its base length. Thus Area ABD = 1 2 base height = 1 2 ab sin . Consequently, Area ABCD = ab sin . 004 (part 2 of 3) 10.0 points (ii) Find when a = 6, b = 8 and the length of the diagonal BD is 7. 1. = 60 . 91 2. = 59 . 91 3. = 58 . 91 4. = 57 . 91 correct 5. = 61 . 91 Explanation: To determine we use the Law of Cosines: with c the length of diagonal BD , c 2 = a 2 + b 2- 2 ab cos . When a = 6 , b = 8 , c = 7 , therefore, cos = 36 + 64- 49 96 , moseley (cmm3869) HW02 Gilbert (56380) 3 so = arccos 17 32 = 57 . 91 . 005 (part 3 of 3) 10.0 points (iii) Then find the area of parallelogram ABCD . 1. Area ABCD = 40 . 92 sq. units 2. Area ABCD = 41 . 17 sq. units 3. Area ABCD = 41 . 42 sq. units 4. Area ABCD = 40 . 67 sq. units correct 5. Area ABCD = 41 . 67 sq. units Explanation: By parts (i) and (ii) Area ABCD = ab sin = 48 sin 57 . 91 ....
View Full Document

## This note was uploaded on 04/12/2011 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.

### Page1 / 10

Gilbert_Hmwk02sol - moseley (cmm3869) HW02 Gilbert (56380)...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online