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 7 TYPO Should say Figure 8 a True b False c True All using the fact that opp adj 1 1 439557 16 csc 44 sin144 0 694658 hyp 1 2 22 csc 30 opp 3 2 3 adj opp 2 2 2 2 23 cot 45 27 TYPO Should say Figure 12 If you think of the 30 angle as then 5 is the adj side Now tan 30 opp 5 so 5 5 opp 5 tan 30 5 13 53 Also hyp cos 30 and thus hyp cos530 3 2 31 33 35 36 37 40 1 Note 2 2 1 2 use whichever you like best 10 3 x tan 63 10 so x 10 tan 63 10 1 96261 19 6261 Isosceles means two sides have the same length in this case 3 Draw the triangle and split 3 3 3 it into two 30 60 90 triangles Each of these triangles has side lengths 3 2 and 2 by multiplying the sides of our favorite 30 60 90 triangle by three Therefore the area of the 3 3 3 9 3 1 3 3 1 triangle is A 2 bh 2 2 2 2 4 h sin 32 400 so h 211 97 feet Suppose x is half of the distance between the stars so x is the opposite side of both x triangles Then tan 1 60 so x 1 0473 thus the stars are approximately 2x 2 0946 light years apart Proof a Suppose is an acute angle Then csc hyp by definition of csc opp 1 opp adj algebra 1 sin by definition of sin Proof Suppose a and b are the lengths of the legs of the triangle with a as the side opposite to Then sin 2 cos 2 a1 2 1b 2 a2 b2 which by the Pythagorean Theorem is equal to a2 b2 12 1 Section 10 2 7 If x y is on the unit circle and y 13 then by the Pythagorean Theorem we have x2 13 2 12 thus x 8 3 Thus cos x is 8 3 if is in the fourth quadrant and 8 3 10 14 15 20 22 24 25 40 46 49 55 cos x is if is in the third quadrant cannot be in the first or second quadrant since the y coordinate corresponding to is negative 6 radians or 30 True False False True True True Your angle should terminate in the fourth quadrant a little bit after 3 2 radians since 5 radians is just a bit more than 3 4 71239 radians 2 The reference triangle is a 30 60 90 triangle in the third quadrant with a reference angle of 60 201 2 100 2 50 2 2 terminates at the same place as 2 radians so no reference triangle exists The reference triangle for 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 and 2 2 vertical in other words 4 corresponds to the point x y unit circle Therefore tan 11 7 1 sin cos y x 2 2 2 2 2 horizontal 2 2 2 22 on the 1 1 0257 This makes sense because 11 7 67 csc 72 11 a tiny further than 3 2 Thus the y coordinate corresponding to 7 should be just a little bit more than 1 closer to zero is more since the y coordinate is negative therefore 1 csc 11 7 y should be a negative number that is just a little bit less than 1 like 1 0257 for example sin 51 means that corresponds to a point x 15 on the unit circle for some x Visually the point on the unit circle corresponding to is one of the two points on the unit circle whose height is 15 draw a picture Since 2 we know that terminates in the second quadrant and thus that the x coordinate is negative By the Pythagorean Theorem or if 24 1 2 2 2 you like by the fact that x y is on the unit circle we have x 5 1 thus x 5 sin 11 7 is in the fourth quadrant just Since we know x is negative we must have x 24 5 Therefore cos x 24 5