Homework for Week 6 Math 232 Spring 2002This homework will not be collected. It is your responsibility to do as many problems as necessaryto understand the material (this includes doing extra problems if you need more practice). Werecommend that you read each section before attempting any exercises. Next week’s quiz will be asubset 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 AnswersCaution: The hints and answers below are not necessarily full solutions. Many of them would notbe considered complete on a quiz or test. Answers are not provided for problems whoseanswers can be found in the reading or problems whose answers are easy to checkusing a calculator.Section 10.12. 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◦=1sin 44◦≈10.694658≈ 1.439557.22. csc 30◦=hypopp=1√3/2=2√3.23. cot 45◦=adjopp=√2/2√2/2= 1. (Note:√22=1√2; use whichever you like best.)27. TYPO: Should say Figure 12.If you think of the 30◦angle as θ, then 5 is the “adj” side. Now tan 30◦=opp5, soopp = 5 tan 30◦= 5(1√3) =5√3. Also,5hyp= cos 30◦, and thus hyp =5cos 30◦=5√3/2=10√3.31. tan 63◦=x10, so x = 10 tan 63◦≈ 10(1.96261) = 19.6261.33. Isosceles means two sides have the same length (in this case 3). Draw the triangle and splitit into two 30–60–90 triangles. Each of these triangles has side lengths 3,32, and3√32(bymultiplying the sides of our favorite 30-60-90 triangle by three). Therefore the area of thetriangle is A =12bh =12(3√32+3√32)(32) =9√34.35. sin 32◦=h400, so h ≈ 211.97 feet.36. Suppose x is half of the distance between the stars (so x is the “opposite” side of bothtriangles). Then tan 1◦=x60, so x ≈ 1.0473; thus the stars are approximately 2x ≈ 2.0946light years apart.37. Proof (a): Suppose θ is an acute angle. Then: csc θ =hypopp(by definition of csc θ)=1opp/adj(algebra)=1sin θ. (by definition of sin θ)40. 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= (a1)2+(b1)2= a2+b2, which by the PythagoreanTheorem is equal to a2+ b2= 12= 1.Section 10.27. If (x, y) is on the unit circle and y = −13then by the Pythagorean Theorem we havex2+ (−13)2= 12, thus x = ±√83. Thus cos θ = x is√83if θ is in the fourth quadrant, andcos θ = x is −√83if θ is in the third quadrant. (θ cannot be in the first or second quadrant,since the y-coordinate corresponding to θ is negative.)10.π6radians (or 30◦).14. True.15. False.20. False.22. True.24. True.25. True.40. Your angle should terminate in the fourth quadrant, a little bit after3π2radians (since 5radians is just a bit more than3π2≈ 4.71239 radians).46. The reference triangle is a 30–60–90 triangle in the third quadrant, with a reference angleof 60◦.49.201π2= 100π +π2= 50(2π) +π2terminates at the same place asπ2radians (so no referencetriangle exists).55. The reference triangle for −π4is a 45–45–90 triangle in the fourth quadrant, with a referenceangle of 45◦. The signed side lengths of the reference triangle are 1 (hyp),√22(horizontal),and√22(vertical); in other words, −π4corresponds to the point (x, y) = (√22, −√22) on theunit circle. Therefore tan θ =sin θcos θ=yx=−√2/2√2/2= −1.67. csc11π7=1sin11π7≈ −1.0257. This makes sense because11π7is in the fourth quadrant, justa tiny further than3π2. Thus the y-coordinate corresponding to11π7should be just a littlebit more than −1 (closer to zero is “more”, since the y-coordinate is negative); thereforecsc11π7=1yshould be a negative number that is just a little bit less than −1 (like −1.0257,for example).72. sin θ =15means 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 circlewhose height is15; draw a picture!) Sinceπ2< θ < π, we know that θ terminates in the secondquadrant, and thus that the x-coordinate is negative. By the Pythagorean Theorem (or, ifyou like, by the fact that (x, y) is on the unit circle), we have x2+(15)2= 12; thus x = ±√245.Since we know x is negative, we must have x = −√245. Therefore cos θ = x =