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Solution for Homework Questions - Calculus with Functions II | MATH 232, Assignments of Mathematics

Material Type: Assignment; Professor: Taalman; Class: CALCULUS WITH FUNCTIONS II; Subject: Mathematics; University: James Madison University; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 02/13/2009

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Homework for Week 6 Math 232 Spring 2002
This homework will not be collected. It is your responsibility to do as many problems as necessary
to understand the material (this includes doing extra problems if you need more practice). We
recommend that you read each section before attempting any exercises. Next week’s quiz will be a
subset of the problems below.
Section 10.1 1, 2, 4, 5, 7, 9, 16, 22, 23, 24, 27, 31, 33, 35, 36, 37, 38, 39, 40.
Section 10.2 2, 5, 7, 8, 9, 10, 12, 14, 17, 20, 22, 24, 25, 29, 32, 40, 46, 49, 55, 56, 57, 62,
67, 72, 73, 75.
Selected Hints and Answers
Caution: The hints and answers below are not necessarily full solutions. Many of them would not
be considered complete on a quiz or test. Answers are not provided for problems whose
answers can be found in the reading or problems whose answers are easy to check
using a calculator.
Section 10.1
2. They’re all equal to cot θ!
7. TYPO: Should say Figure 8.
(a) True; (b) False; (c) True. (All using the fact that opp >adj.)
16. csc 44=1
sin 441
0.694658 1.439557.
22. csc 30=hyp
opp =1
3/2=2
3.
23. cot 45=adj
opp =2/2
2/2= 1. (Note: 2
2=1
2; use whichever you like best.)
27. TYPO: Should say Figure 12.
If you think of the 30angle as θ, then 5 is the “adj” side. Now tan 30=opp
5, so
opp = 5 tan30= 5( 1
3) = 5
3. Also, 5
hyp = cos 30, and thus hyp = 5
cos 30=5
3/2=10
3.
31. tan 63=x
10 , so x= 10 tan 6310(1.96261) = 19.6261.
33. Isosceles means two sides have the same length (in this case 3). Draw the triangle and split
it into two 30–60–90 triangles. Each of these triangles has side lengths 3, 3
2, and 33
2(by
multiplying the sides of our favorite 30-60-90 triangle by three). Therefore the area of the
triangle is A=1
2bh =1
2(33
2+33
2)(3
2) = 93
4.
35. sin 32=h
400 , so h211.97 feet.
36. Suppose xis half of the distance between the stars (so xis the “opposite” side of both
triangles). Then tan 1=x
60 , so x1.0473; thus the stars are approximately 2x2.0946
light years apart.
37. Proof (a): Suppose θis an acute angle. Then: csc θ=hyp
opp (by definition of csc θ)
=1
opp/adj (algebra)
=1
sin θ. (by definition of sin θ)
40. Proof: Suppose aand bare the lengths of the legs of the triangle (with aas the side
“opposite” to θ). Then (sin θ)2+(cos θ)2= ( a
1)2+ ( b
1)2=a2+b2, which by the Pythagorean
Theorem is equal to a2+b2= 12= 1.
pf2

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Download Solution for Homework Questions - Calculus with Functions II | MATH 232 and more Assignments Mathematics in PDF only on Docsity!

Homework for Week 6 Math 232 Spring 2002

This homework will not be collected. It is your responsibility to do as many problems as necessary to understand the material (this includes doing extra problems if you need more practice). We recommend that you read each section before attempting any exercises. Next week’s quiz will be a subset of the problems below.

Section 10.1 1, 2, 4, 5, 7, 9, 16, 22, 23, 24, 27, 31, 33, 35, 36, 37, 38, 39, 40. Section 10.2 2, 5, 7, 8, 9, 10, 12, 14, 17, 20, 22, 24, 25, 29, 32, 40, 46, 49, 55, 56, 57, 62, 67, 72, 73, 75.

Selected Hints and Answers

Caution: The hints and answers below are not necessarily full solutions. Many of them would not be considered complete on a quiz or test. Answers are not provided for problems whose answers can be found in the reading or problems whose answers are easy to check using a calculator.

Section 10.

  1. They’re all equal to cot θ!
  2. TYPO: Should say Figure 8. (a) True; (b) False; (c) True. (All using the fact that opp > adj.)
  3. csc 44◦^ = (^) sin 44^1 ◦ ≈ (^0). 6946581 ≈ 1 .439557.
  4. csc 30◦^ = hypopp = √^13 / 2 = √^23.
  5. cot 45◦^ = (^) oppadj =

√ 2 / 2 √ 2 / 2 = 1. (Note:

√ 2 2 =^ √^1 2 ; use whichever you like best.)

  1. TYPO: Should say Figure 12. If you think of the 30◦^ angle as θ, then 5 is the “adj” side. Now tan 30◦^ = opp 5 , so opp = 5 tan 30◦^ = 5( √^13 ) = √^53. Also, (^) hyp^5 = cos 30◦, and thus hyp = (^) cos 30^5 ◦ = √^53 / 2 = √^103.
  2. tan 63◦^ = 10 x , so x = 10 tan 63◦^ ≈ 10(1.96261) = 19.6261.
  3. Isosceles means two sides have the same length (in this case 3). Draw the triangle and split it into two 30–60–90 triangles. Each of these triangles has side lengths 3, 32 , and 3

√ 3 2 (by multiplying the sides of our favorite 30-60-90 triangle by three). Therefore the area of the triangle is A = 12 bh = 12 ( 3

√ 3 2 +^

3 √ 3 2 )(^

3 2 ) =^

9 √ 3

  1. sin 32◦^ = 400 h , so h ≈ 211 .97 feet.

  2. Suppose x is half of the distance between the stars (so x is the “opposite” side of both triangles). Then tan 1◦^ = 60 x , so x ≈ 1 .0473; thus the stars are approximately 2x ≈ 2. 0946 light years apart.

  3. Proof (a): Suppose θ is an acute angle. Then: csc θ = hypopp (by definition of csc θ) = (^) opp^1 /adj (algebra) = (^) sin^1 θ. (by definition of sin θ)

  4. Proof: Suppose a and b are the lengths of the legs of the triangle (with a as the side “opposite” to θ). Then (sin θ)^2 + (cos θ)^2 = ( a 1 )^2 + ( 1 b )^2 = a^2 + b^2 , which by the Pythagorean Theorem is equal to a^2 + b^2 = 1^2 = 1.

Section 10.

  1. If (x, y) is on the unit circle and y = − 13 then by the Pythagorean Theorem we have x^2 + (− 13 )^2 = 1^2 , thus x = ±

√ 8

  1. Thus cos^ θ^ =^ x^ is

√ 8 3 if^ θ^ is in the fourth quadrant, and cos θ = x is −

√ 8 3 if^ θ^ is in the third quadrant. (θ^ cannot be in the first or second quadrant, since the y-coordinate corresponding to θ is negative.)

  1. π 6 radians (or 30◦).
  2. True.
  3. False.
  4. False.
  5. True.
  6. True.
  7. True.
  8. Your angle should terminate in the fourth quadrant, a little bit after 32 π radians (since 5 radians is just a bit more than 32 π ≈ 4 .71239 radians).
  9. The reference triangle is a 30–60–90 triangle in the third quadrant, with a reference angle of 60◦.
  10. 2012 π= 100π + π 2 = 50(2π) + π 2 terminates at the same place as π 2 radians (so no reference triangle exists).
  11. The reference triangle for − π 4 is a 45–45–90 triangle in the fourth quadrant, with a reference angle of 45◦. The signed side lengths of the reference triangle are 1 (hyp),

√ 2 2 (horizontal), and

√ 2 2 (vertical); in other words,^ −^ π 4 corresponds to the point (x, y) = (

√ 2 2 ,^ −

√ 2 2 ) on the unit circle. Therefore tan θ = sin cos^ θθ = yx = −

√ 2 / 2 √ 2 / 2 = −1.

  1. csc 117 π = (^) sin 111 π 7

≈ − 1 .0257. This makes sense because 117 π is in the fourth quadrant, just a tiny further than 32 π. Thus the y-coordinate corresponding to 117 π should be just a little bit more than −1 (closer to zero is “more”, since the y-coordinate is negative); therefore csc 117 π = (^) y^1 should be a negative number that is just a little bit less than −1 (like − 1 .0257, for example).

  1. sin θ = 15 means that θ corresponds to a point (x, 15 ) on the unit circle, for some x. (Visually, the point on the unit circle corresponding to θ is one of the two points on the unit circle whose height is 15 ; draw a picture!) Since π 2 < θ < π, we know that θ terminates in the second quadrant, and thus that the x-coordinate is negative. By the Pythagorean Theorem (or, if you like, by the fact that (x, y) is on the unit circle), we have x^2 +( 15 )^2 = 1^2 ; thus x = ±

√ 24

Since we know x is negative, we must have x = −

√ 24

  1. Therefore cos^ θ^ =^ x^ =^ −

√ 24