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Material Type: Assignment; Professor: Taalman; Class: CALCULUS WITH FUNCTIONS II; Subject: Mathematics; University: James Madison University; Term: Unknown 1989;
Typology: Assignments
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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.
√ 2 / 2 √ 2 / 2 = 1. (Note:
√ 2 2 =^ √^1 2 ; use whichever you like best.)
√ 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
sin 32◦^ = 400 h , so h ≈ 211 .97 feet.
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.
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)
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.
√ 8
√ 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.)
√ 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 .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).
√ 24
Since we know x is negative, we must have x = −
√ 24
√ 24