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Theorem 6.34. Let \( f : S \rightarrow T \) be a function and let \( A \) and \( B \) be subsets of \( S \) . Then\n\n1. \( f\left( {A \cap B}\right) \subseteq f\left( A\right) \cap f\left( B\right) \)\n\n2. \( f\left( {A \cup B}\right) = f\left( A\right) \cup f\left( B\right) \) | Proof. We will prove Part (1). The proof of Part (2) is Exercise (5).\n\nAssume that \( f : S \rightarrow T \) is a function and let \( A \) and \( B \) be subsets of \( S \) . We will prove that \( f\left( {A \cap B}\right) \subseteq f\left( A\right) \cap f\left( B\right) \) by proving that for all \( y \in T \), if \... | No |
Theorem 6.35. Let \( f : S \rightarrow T \) be a function and let \( C \) and \( D \) be subsets of \( T \). Then\n\n\[ \text{1.}{f}^{-1}\left( {C \cap D}\right) = {f}^{-1}\left( C\right) \cap {f}^{-1}\left( D\right) \]\n\n\[ \text{2.}{f}^{-1}\left( {C \cup D}\right) = {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\righ... | Proof. We will prove Part (2). The proof of Part (1) is Exercise (6).\n\nAssume that \( f : S \rightarrow T \) is a function and that \( C \) and \( D \) are subsets of \( T \). We will prove that \( {f}^{-1}\left( {C \cup D}\right) = {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \) by proving that each set is a... | No |
Theorem 6.36. Let \( f : S \rightarrow T \) be a function and let \( A \) be a subset of \( S \) and let \( C \) be a subset of \( T \). Then\n\n1. \( A \subseteq {f}^{-1}\left( {f\left( A\right) }\right) \) 2. \( f\left( {{f}^{-1}\left( C\right) }\right) \subseteq C \n | Proof. We will prove Part (1). The proof of Part (2) is Exercise (7).\n\nTo prove Part (1), we will prove that for all \( a \in S \), if \( a \in A \), then \( a \in {f}^{-1}\left( {f\left( A\right) }\right) \). So let \( a \in A \). Then, by definition, \( f\left( a\right) \in f\left( A\right) \). We know that \( f\le... | No |
Theorem 7.6. Let \( R \) be a relation from the set \( A \) to the set \( B \) . Then\n\n- The domain of \( {R}^{-1} \) is the range of \( R \) . That is, \( \operatorname{dom}\left( {R}^{-1}\right) = \operatorname{range}\left( R\right) \) .\n\n- The range of \( {R}^{-1} \) is the domain of \( R \) . That is, range \( ... | To prove the first part of Theorem 7.6, observe that the goal is to prove that two sets are equal,\n\n\[ \operatorname{dom}\left( {R}^{-1}\right) = \operatorname{range}\left( R\right) \]\n\nOne way to do this is to prove that each is a subset of the other. To prove that \( \operatorname{dom}\left( {R}^{-1}\right) \subs... | No |
Let \( M \) be the relation on \( \mathbb{Z} \) defined as follows:\n\nFor \( a, b \in \mathbb{Z},{aMb} \) if and only if \( a \) is a multiple of \( b \) .\n\nSo \( {aMb} \) if and only if there exists a \( k \in \mathbb{Z} \) such that \( a = {bk} \) . | - The relation \( M \) is reflexive on \( \mathbb{Z} \) since for each \( x \in \mathbb{Z}, x = x \cdot 1 \) and, hence, \( {xMx} \).\n\n- Notice that \( {4M2} \), but \( {2M4} \) . So there exist integers \( x \) and \( y \) such that \( {xMy} \) but \( {yMx} \) . Hence, the relation \( M \) is not symmetric.\n\n- Now... | Yes |
Theorem 7.10. Let \( n \in \mathbb{N} \) and let \( a, b \in \mathbb{Z} \) . Then \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) if and only if \( a \) and \( b \) have the same remainder when divided by \( n \) . | Proof. Let \( n \in \mathbb{N} \) and let \( a, b \in \mathbb{Z} \) . We will first prove that if \( a \) and \( b \) have the same remainder when divided by \( n \), then \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) . So assume that \( a \) and \( b \) have the same remainder when divided by \( n \), and le... | Yes |
Theorem 7.14. Let \( A \) be a nonempty set and let \( \sim \) be an equivalence relation on the set \( A \) . Then,\n\n1. For each \( a \in A, a \in \left\lbrack a\right\rbrack \) .\n\n2. For each \( a, b \in A, a \sim b \) if and only if \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) .\n\n3. For each... | Proof. Let \( A \) be a nonempty set and assume that \( \sim \) is an equivalence relation on \( A \) . To prove the first part of the theorem, let \( a \in A \) . Since \( \sim \) is an equivalence relation on \( A \), it is reflexive on \( A \) . Thus, \( a \sim a \), and we can conclude that \( a \in \left\lbrack a\... | Yes |
Theorem 7.18. Let \( \sim \) be an equivalence relation on the nonempty set \( A \) . Then the collection \( \mathcal{C} \) of all equivalence classes determined by \( \sim \) is a partition of the set \( A \) . | Proof. Let \( \sim \) be an equivalence relation on the nonempty set \( A \), and let \( \mathcal{C} \) be the collection of all equivalence classes determined by \( \sim \) . That is,\n\n\[ \mathcal{C} = \{ \left\lbrack a\right\rbrack \mid a \in A\} . \]\n\nWe will use Theorem 7.14 to prove that \( \mathcal{C} \) is a... | Yes |
Lemma 8.1. Let \( c \) and \( d \) be integers, not both equal to zero. If \( q \) and \( r \) are integers such that \( c = d \cdot q + r \), then \( \gcd \left( {c, d}\right) = \gcd \left( {d, r}\right) \) . | Proof. Let \( c \) and \( d \) be integers, not both equal to zero. Assume that \( q \) and \( r \) are integers such that \( c = d \cdot q + r \) . For ease of notation, we will let\n\n\[ m = \gcd \left( {c, d}\right) \text{ and }n = \gcd \left( {d, r}\right) . \]\n\nNow, \( m \) divides \( c \) and \( m \) divides \(... | Yes |
Theorem 8.8. Let \( a \) and \( b \) be integers, not both 0 . Then \( \gcd \left( {a, b}\right) \) can be written as a linear combination of \( a \) and \( b \) . That is, there exist integers \( u \) and \( v \) such that \( \gcd \left( {a, b}\right) = {au} + {bv}. \) | We will not give a formal proof of this theorem. Hopefully, the examples and activities provide evidence for its validity. The idea is to use the steps of the Euclidean Algorithm in reverse order to write \( \gcd \left( {a, b}\right) \) as a linear combination of \( a \) and \( b \) . For example, assume the completed ... | No |
Proposition 5.16 Let \( a, b \), and \( t \) be integers with \( t \neq 0 \) . If \( t \) divides \( a \) and \( t \) divides \( b \), then for all integers \( x \) and \( y, t \) divides \( \left( {{ax} + {by}}\right) \) . | Proof. Let \( a, b \), and \( t \) be integers with \( t \neq 0 \), and assume that \( t \) divides \( a \) and \( t \) divides \( b \) . We will prove that for all integers \( x \) and \( y, t \) divides \( \left( {{ax} + {by}}\right) \) .\n\nSo let \( x \in \mathbb{Z} \) and let \( y \in \mathbb{Z} \) . Since \( t \)... | Yes |
Theorem 8.11. Let \( a \) and \( b \) be nonzero integers, and let \( p \) be a prime number.\n\n1. If \( a \) and \( b \) are relatively prime, then there exist integers \( m \) and \( n \) such that \( {am} + {bn} = 1 \) . That is,1 can be written as linear combination of \( a \) and \( b \) . | Part (1) of Theorem 8.11 is actually a corollary of Theorem 8.9. | Yes |
Theorem 8.12. Let \( a, b \) be nonzero integers and let \( c \) be an integer. If \( a \) and \( b \) are relatively prime and \( a \mid \left( {bc}\right) \), then \( a \mid c \) . | The explorations in Preview Activity 1 were related to this theorem. We will first explore the forward-backward process for the proof. The goal is to prove that \( a \mid c \) . A standard way to do this is to prove that there exists an integer \( q \) such that\n\n\[ c = {aq}\text{.} \]\n\n(1)\n\nSince we are given th... | Yes |
Theorem 8.16. There are infinitely many prime numbers. | Proof. We will use a proof by contradiction. We assume that there are only finitely many primes, and let\n\n\[ \n{p}_{1},{p}_{2},\ldots ,{p}_{m} \n\]\n\nbe the list of all the primes. Let\n\n\[ \nM = {p}_{1}{p}_{2}\cdots {p}_{m} + 1. \n\]\n\n(1)\n\nNotice that \( M \neq 1 \) . So \( M \) is either a prime number or, by... | Yes |
Theorem 8.22. Let \( a, b \), and \( c \) be integers with \( a \neq 0 \) and \( b \neq 0 \), and let \( d = \gcd \left( {a, b}\right) \). 1. If \( d \) does not divide \( c \), then the linear Diophantine equation \( {ax} + {by} = c \) has no solution. 2. If \( d \) divides \( c \), then the linear Diophantine equatio... | Proof. The proof of Part (1) is Exercise (1). For Part (2), we let \( a, b \), and \( c \) be integers with \( a \neq 0 \) and \( b \neq 0 \), and let \( d = \gcd \left( {a, b}\right) \). We also assume that \( d \mid c \). Since \( d = \gcd \left( {a, b}\right) \), Theorem 8.8 tells us that \( d \) is a linear combina... | No |
Theorem 9.3. Any set equivalent to a finite nonempty set \( A \) is a finite set and has the same cardinality as \( A \) . | Proof. Suppose that \( A \) is a finite nonempty set, \( B \) is a set, and \( A \approx B \) . Since \( A \) is a finite set, there exists a \( k \in \mathbb{N} \) such that \( A \approx {\mathbb{N}}_{k} \) . We also have assumed that \( A \approx B \) and so by part (b) of Theorem 9.1 (in Preview Activity 2), we can ... | Yes |
Lemma 9.4. If \( A \) is a finite set and \( x \notin A \), then \( A \cup \{ x\} \) is a finite set and \( \operatorname{card}\left( {A\cup \{ x\} }\right) = \operatorname{card}\left( A\right) + 1. \) | Proof. Let \( A \) be a finite set and assume \( \operatorname{card}\left( A\right) = k \), where \( k = 0 \) or \( k \in \mathbb{N} \) . Assume \( x \notin A \) .
If \( A = \varnothing \), then \( \operatorname{card}\left( A\right) = 0 \) and \( A \cup \{ x\} = \{ x\} \), which is equivalent to \( {\mathbb{N}}_{1} \)... | No |
Lemma 9.5. For each natural number \( m \), if \( A \subseteq {\mathbb{N}}_{m} \), then \( A \) is a finite set and \( \operatorname{card}\left( A\right) \leq m \) . | Proof. We will use a proof using induction on \( m \) . For each \( m \in \mathbb{N} \), let \( P\left( m\right) \) be, \ | No |
Theorem 9.6. If \( S \) is a finite set and \( A \) is a subset of \( S \), then \( A \) is a finite set and \( \operatorname{card}\left( A\right) \leq \operatorname{card}\left( S\right) . \) | Proof. Let \( S \) be a finite set and assume that \( A \) is a subset of \( S \) . If \( A = \varnothing \), then \( A \) is a finite set and \( \operatorname{card}\left( A\right) \leq \operatorname{card}\left( S\right) \) . So we assume that \( A \neq \varnothing \) .\n\nSince \( S \) is finite, there exists a biject... | Yes |
Corollary 9.8. A finite set is not equivalent to any of its proper subsets. | Proof. Let \( B \) be a finite set and assume that \( A \) is a proper subset of \( B \) . Since \( A \) is a proper subset of \( B \), there exists an element \( x \) in \( B - A \) . This means that \( A \) is a subset of \( B - \{ x\} \) . Hence, by Theorem 9.6,\n\n\[ \operatorname{card}\left( A\right) \leq \operato... | Yes |
Theorem 9.9 (The Pigeonhole Principle). Let \( A \) and \( B \) be finite sets. If card \( \left( A\right) > \) \( \operatorname{card}\left( B\right) \), then any function \( f : A \rightarrow B \) is not an injection. | Proof. Let \( A \) and \( B \) be finite sets. We will prove the contrapositive of the theorem, which is, if there exists a function \( f : A \rightarrow B \) that is an injection, then \( \operatorname{card}\left( A\right) \leq \) \( \operatorname{card}\left( B\right) \) .\n\nSo assume that \( f : A \rightarrow B \) i... | Yes |
Theorem 9.10. Let \( A \) and \( B \) be sets.\n\n1. If \( A \) is infinite and \( A \approx B \), then \( B \) is infinite. | Proof. We will prove part (1). The proof of part (2) is exercise (3) on page 473.\n\nTo prove part (1), we use a proof by contradiction and assume that \( A \) is an infinite set, \( A \approx B \), and \( B \) is not infinite. That is, \( B \) is a finite set. Since \( A \approx B \) and \( B \) is finite, Theorem 9.3... | No |
Theorem 9.13. The set \( \mathbb{Z} \) of integers is countably infinite, and so card \( \left( \mathbb{Z}\right) = {\aleph }_{0} \) . | Proof. To prove that \( \mathbb{N} \approx \mathbb{Z} \), we will use the following function: \( f : \mathbb{N} \rightarrow \mathbb{Z} \) ,\n\nwhere\n\[ f\left( n\right) = \left\{ \begin{array}{ll} \frac{n}{2} & \text{ if }n\text{ is even } \\ \frac{1 - n}{2} & \text{ if }n\text{ is odd. } \end{array}\right. \]\n\nFrom... | Yes |
Theorem 9.14. The set of positive rational numbers is countably infinite. | Proof. We can write all the positive rational numbers in a two-dimensional array as shown in Figure 9.2. The top row in Figure 9.2 represents the numerator of the rational number, and the left column represents the denominator. We follow the arrows in Figure 9.2 to define \( f : \mathbb{N} \rightarrow {\mathbb{Q}}^{ + ... | Yes |
Theorem 9.15. If \( A \) is a countably infinite set, then \( A \cup \{ x\} \) is a countably infinite set. | Proof. Let \( A \) be a countably infinite set. Then there exists a bijection \( f : \mathbb{N} \rightarrow A \) . Since \( x \) is either in \( A \) or not in \( A \), we can consider two cases.\n\nIf \( x \in A \), then \( A \cup \{ x\} = A \) and \( A \cup \{ x\} \) is countably infinite.\n\nIf \( x \notin A \), def... | No |
Theorem 9.16. If \( A \) is a countably infinite set and \( B \) is a finite set, then \( A \cup B \) is a countably infinite set. | ## Proof. Exercise (5) on page 473. | No |
Theorem 9.17. If \( A \) and \( B \) are disjoint countably infinite sets, then \( A \cup B \) is a countably infinite set. | Proof. Let \( A \) and \( B \) be countably infinite sets and let \( f : \mathbb{N} \rightarrow A \) and \( g : \mathbb{N} \rightarrow B \) be bijections. Define \( h : \mathbb{N} \rightarrow A \cup B \) by\n\n\[ h\left( n\right) = \left\{ \begin{array}{ll} f\left( \frac{n + 1}{2}\right) & \text{ if }n\text{ is odd } \... | No |
Theorem 9.18. The set \( \mathbb{Q} \) of all rational numbers is countably infinite. | Proof. Exercise (7) on page 474. | No |
Theorem 9.19. Every subset of the natural numbers is countable. | Proof. Let \( B \) be a subset of \( \mathbb{N} \) . If \( B \) is finite, then \( B \) is countable. So we next assume that \( B \) is infinite. We will next give a recursive definition of a function \( g : \mathbb{N} \rightarrow B \) and then prove that \( g \) is a bijection.\n\n- Let \( g\left( 1\right) \) be the s... | No |
Corollary 9.20. Every subset of a countable set is countable. | Proof. Exercise (12) on page 475. | No |
Theorem 9.24. Let \( a \) and \( b \) be real numbers with \( a < b \) . The open interval \( \left( {a, b}\right) \) is uncountable and has cardinality \( \mathbf{c} \) . | Progress Check 9.25 (Proof of Theorem 9.24)\n\n1. In Part (3) of Progress Check 9, we proved that if \( b \in \mathbb{R} \) and \( b > 0 \), then the open interval \( \left( {0,1}\right) \) is equivalent to the open interval \( \left( {0, b}\right) \) . Now let \( a \) and \( b \) be real numbers with \( a < b \) . Fin... | No |
Theorem 9.26. The set of real numbers \( \mathbb{R} \) is uncountable and has cardinality \( c \) . | Proof. Let \( f : \left( {-\frac{\pi }{2},\frac{\pi }{2}}\right) \rightarrow \mathbb{R} \) be defined by \( f\left( x\right) = \tan x \), for each \( x \in \mathbb{R} \) . The function \( f \) is a bijection and, hence, \( \left( {-\frac{\pi }{2},\frac{\pi }{2}}\right) \approx \mathbb{R} \) . So by Theorem 9.24, \( \ma... | Yes |
Theorem 9.27 (Cantor’s Theorem). For every set \( A, A \) and \( \mathcal{P}\left( A\right) \) do not have the same cardinality. | Proof. Let \( A \) be a set. If \( A = \varnothing \), then \( \mathcal{P}\left( A\right) = \{ \varnothing \} \), which has cardinality 1 . Therefore, \( \varnothing \) and \( \mathcal{P}\left( \varnothing \right) \) do not have the same cardinality.\n\nNow suppose that \( A \neq \varnothing \), and let \( f : A \right... | Yes |
Corollary 9.28. \( \mathcal{P}\left( \mathbb{N}\right) \) is an infinite set that is not countably infinite. | Proof. Since \( \mathcal{P}\left( \mathbb{N}\right) \) contains the infinite subset \( \{ \{ 1\} ,\{ 2\} ,\{ 3\} \ldots \} \), we can use Theorem 9.10, to conclude that \( \mathcal{P}\left( \mathbb{N}\right) \) is an infinite set. By Cantor’s Theorem (Theorem 9.27), \( \mathbb{N} \) and \( \mathcal{P}\left( \mathbb{N}\... | Yes |
Proposition 4.11. For all natural numbers \( n \) with \( n \geq 8 \), there exist non-negative integers \( x \) and \( y \) such that \( n = {3x} + {5y} \) . | Proof. (by mathematical induction) Let \( {\mathbb{Z}}^{ * } = \{ x \in \mathbb{Z} \mid x \geq 0\} \), and for each natural number \( n \), let \( P\left( n\right) \) be,\ | No |
Theorem 8.12. Let \( a, b \), and \( c \) be integers. If \( a \) and \( b \) are relatively prime and \( a \mid \left( {bc}\right) \), then \( a \mid c \) . | Proof. Let \( a, b \), and \( c \) be integers. Assume that \( a \) and \( b \) are relatively prime and \( a \mid \left( {bc}\right) \) . We will prove that \( a \) divides \( c \) .\n\nSince \( a \) divides \( {bc} \), there exists an integer \( k \) such that\n\n\[ {bc} = {ak}.\text{.}\]\n\n(1)\n\nIn addition, we ar... | Yes |
1. Convert \( \alpha \) to the DMS system. Round your answer to the nearest second. | To convert \( \alpha \) to the DMS system, we start with \( {111.371}^{ \circ } = {111}^{ \circ } + {0.371}^{ \circ } \) . Next we convert \( {0.371}^{ \circ }\left( \frac{{60}^{\prime }}{{1}^{ \circ }}\right) = {22.26}^{\prime } \) . Writing \( {22.26}^{\prime } = {22}^{\prime } + {0.26}^{\prime } \), we convert \( {0... | Yes |
Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. | 1. To graph \( \alpha = {60}^{ \circ } \), we draw an angle with its initial side on the positive \( x \) -axis and rotate counter-clockwise \( \frac{{60}^{ \circ }}{{360}^{ \circ }} = \frac{1}{6} \) of a revolution. We see that \( \alpha \) is a Quadrant I angle. To find angles which are coterminal, we look for angles... | Yes |
Graph each of the (oriented) angles below in standard position and classify them according to where their terminal side lies. Find three coterminal angles, at least one of which is positive and one of which is negative. | 1. The angle \( \alpha = \frac{\pi }{6} \) is positive, so we draw an angle with its initial side on the positive \( x \) -axis and rotate counter-clockwise \( \frac{\left( \pi /6\right) }{2\pi } = \frac{1}{12} \) of a revolution. Thus \( \alpha \) is a Quadrant I angle. Coterminal angles \( \theta \) are of the form \... | Yes |
Sketch the oriented arc on the Unit Circle corresponding to each of the following real numbers. | 1. The arc associated with \( t = \frac{3\pi }{4} \) is the arc on the Unit Circle which subtends the angle \( \frac{3\pi }{4} \) in radian measure. Since \( \frac{3\pi }{4} \) is \( \frac{3}{8} \) of a revolution, we have an arc which begins at the point \( \left( {1,0}\right) \) proceeds counter-clockwise up to midwa... | Yes |
Assuming that the surface of the Earth is a sphere, any point on the Earth can be thought of as an object traveling on a circle which completes one revolution in (approximately) 24 hours. The path traced out by the point during this 24 hour period is the Latitude of that point. Lakeland Community College is at \( {41.6... | Solution. To use the formula \( v = {r\omega } \), we first need to compute the angular velocity \( \omega \) . The earth makes one revolution in 24 hours, and one revolution is \( {2\pi } \) radians, so \( \omega = \frac{{2\pi }\text{ radians }}{{24}\text{ hours }} = \frac{\pi }{{12}\text{ hours }} \), where, once aga... | Yes |
Find the cosine and sine of the following angles.\n\n1. \( \theta = {270}^{ \circ } \) 2. \( \theta = - \pi \) 3. \( \theta = {45}^{ \circ } \) 4. \( \theta = \frac{\pi }{6} \) 5. \( \theta = {60}^{ \circ } \) | 1. To find \( \cos \left( {270}^{ \circ }\right) \) and \( \sin \left( {270}^{ \circ }\right) \), we plot the angle \( \theta = {270}^{ \circ } \) in standard position and find the point on the terminal side of \( \theta \) which lies on the Unit Circle. Since \( {270}^{ \circ } \) represents \( \frac{3}{4} \) of a cou... | No |
If \( \theta \) is a Quadrant II angle with \( \sin \left( \theta \right) = \frac{3}{5} \), find \( \cos \left( \theta \right) \) . | When we substitute \( \sin \left( \theta \right) = \frac{3}{5} \) into The Pythagorean Identity, \( {\cos }^{2}\left( \theta \right) + {\sin }^{2}\left( \theta \right) = 1 \), we obtain \( {\cos }^{2}\left( \theta \right) + \frac{9}{25} = 1 \) . Solving, we find \( \cos \left( \theta \right) = \pm \frac{4}{5} \) . Sinc... | Yes |
Find the cosine and sine of the following angles.\n\n1. \( \theta = {225}^{ \circ } \) 2. \( \theta = \frac{11\pi }{6} \) 3. \( \theta = - \frac{5\pi }{4} \) 4. \( \theta = \frac{7\pi }{3} \) | Solution.\n\n1. We begin by plotting \( \theta = {225}^{ \circ } \) in standard position and find its terminal side overshoots the negative \( x \) -axis to land in Quadrant III. Hence, we obtain \( \theta \) ’s reference angle \( \alpha \) by subtracting: \( \alpha = \theta - {180}^{ \circ } = {225}^{ \circ } - {180}^... | No |
Suppose \( \alpha \) is an acute angle with \( \cos \left( \alpha \right) = \frac{5}{13} \). Find \( \sin \left( \alpha \right) \) and use this to plot \( \alpha \) in standard position. | Proceeding as in Example 10.2.2, we substitute \( \cos \left( \alpha \right) = \frac{5}{13} \) into \( {\cos }^{2}\left( \alpha \right) + {\sin }^{2}\left( \alpha \right) = 1 \) and find \( \sin \left( \alpha \right) = \pm \frac{12}{13} \). Since \( \alpha \) is an acute (and therefore Quadrant I) angle, \( \sin \left(... | Yes |
Find all of the angles which satisfy the given equation.\n\n1. \( \cos \left( \theta \right) = \frac{1}{2} \) 2. \( \sin \left( \theta \right) = - \frac{1}{2} \) 3. \( \cos \left( \theta \right) = 0 \) . | Solution. Since there is no context in the problem to indicate whether to use degrees or radians, we will default to using radian measure in our answers to each of these problems. This choice will be justified later in the text when we study what is known as Analytic Trigonometry. In those sections to come, radian meas... | No |
Theorem 10.3. If \( Q\left( {x, y}\right) \) is the point on the terminal side of an angle \( \theta \), plotted in standard position, which lies on the circle \( {x}^{2} + {y}^{2} = {r}^{2} \) then \( x = r\cos \left( \theta \right) \) and \( y = r\sin \left( \theta \right) \) . | Moreover, \[ \cos \left( \theta \right) = \frac{x}{r} = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}\text{ and }\sin \left( \theta \right) = \frac{y}{r} = \frac{y}{\sqrt{{x}^{2} + {y}^{2}}} \] | Yes |
1. Suppose that the terminal side of an angle \( \theta \), when plotted in standard position, contains the point \( Q\left( {4, - 2}\right) \) . Find \( \sin \left( \theta \right) \) and \( \cos \left( \theta \right) \) . | 1. Using Theorem 10.3 with \( x = 4 \) and \( y = - 2 \), we find \( r = \sqrt{{\left( 4\right) }^{2} + {\left( -2\right) }^{2}} = \sqrt{20} = 2\sqrt{5} \) so that \( \cos \left( \theta \right) = \frac{x}{r} = \frac{4}{2\sqrt{5}} = \frac{2\sqrt{5}}{5} \) and \( \sin \left( \theta \right) = \frac{y}{r} = \frac{-2}{2\sqr... | Yes |
Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates. | Solution. From Example 10.1.5, we take \( r = {2960} \) miles and and \( \omega = \frac{\pi }{{12}\text{ hours }} \) . Hence, the equations of motion are \( x = r\cos \left( {\omega t}\right) = {2960}\cos \left( {\frac{\pi }{12}t}\right) \) and \( y = r\sin \left( {\omega t}\right) = {2960}\sin \left( {\frac{\pi }{12}t... | Yes |
Find the measure of the missing angle and the lengths of the missing sides of: | Solution. The first and easiest task is to find the measure of the missing angle. Since the sum of angles of a triangle is \( {180}^{ \circ } \), we know that the missing angle has measure \( {180}^{ \circ } - {30}^{ \circ } - {90}^{ \circ } = {60}^{ \circ } \) . We now proceed to find the lengths of the remaining two ... | Yes |
1. \( \sec \left( {60}^{ \circ }\right) \) | According to Theorem 10.6, \( \sec \left( {60}^{ \circ }\right) = \frac{1}{\cos \left( {60}^{ \circ }\right) } \) . Hence, \( \sec \left( {60}^{ \circ }\right) = \frac{1}{\left( 1/2\right) } = 2 \) . | Yes |
Find all angles which satisfy the given equation.\n\n1. \( \sec \left( \theta \right) = 2 \) | 1. To solve \( \sec \left( \theta \right) = 2 \), we convert to cosines and get \( \frac{1}{\cos \left( \theta \right) } = 2 \) or \( \cos \left( \theta \right) = \frac{1}{2} \) . This is the exact same equation we solved in Example 10.2.5, number 1, so we know the answer is: \( \theta = \frac{\pi }{3} + {2\pi k} \) or... | Yes |
1. \( \frac{1}{\csc \left( \theta \right) } = \sin \left( \theta \right) \) | To verify \( \frac{1}{\csc \left( \theta \right) } = \sin \left( \theta \right) \), we start with the left side. Using \( \csc \left( \theta \right) = \frac{1}{\sin \left( \theta \right) } \), we get:\n\n\[ \frac{1}{\csc \left( \theta \right) } = \frac{1}{\frac{1}{\sin \left( \theta \right) }} = \sin \left( \theta \rig... | Yes |
Theorem 10.9. Suppose \( Q\left( {x, y}\right) \) is the point on the terminal side of an angle \( \theta \) (plotted in standard position) which lies on the circle of radius \( r,{x}^{2} + {y}^{2} = {r}^{2} \) . Then:\n\n\[ \text{-}\sec \left( \theta \right) = \frac{r}{x} = \frac{\sqrt{{x}^{2} + {y}^{2}}}{x}\text{, pr... | ## Solution.\n\n1. Since \( x = 3 \) and \( y = - 4, r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{\left( 3\right) }^{2} + {\left( -4\right) }^{2}} = \sqrt{25} = 5 \) . Theorem 10.9 tells us \( \cos \left( \theta \right) = \frac{3}{5},\sin \left( \theta \right) = - \frac{4}{5},\sec \left( \theta \right) = \frac{5}{3},\csc \lef... | Yes |
Theorem 10.11. Domains and Ranges of the Circular Functions | - The function \( f\left( t\right) = \cos \left( t\right) \; \bullet \) The function \( g\left( t\right) = \sin \left( t\right) \)\n\n- has domain \( \left( {-\infty ,\infty }\right) \; - \) has domain \( \left( {-\infty ,\infty }\right) \)\n\n- has range \( \left\lbrack {-1,1}\right\rbrack \; - \) has range \( \left\l... | Yes |
Even / Odd Identities: For all applicable angles \( \theta \) , \n\n\[ \text{-}\cos \left( {-\theta }\right) = \cos \left( \theta \right) \; \bullet \sin \left( {-\theta }\right) = - \sin \left( \theta \right) \; \bullet \tan \left( {-\theta }\right) = - \tan \left( \theta \right) \]\n\n\[ \text{-}\sec \left( {-\theta ... | In light of the Quotient and Reciprocal Identities, Theorem 10.6, it suffices to show \( \cos \left( {-\theta }\right) = \cos \left( \theta \right) \) and \( \sin \left( {-\theta }\right) = - \sin \left( \theta \right) \) . The remaining four circular functions can be expressed in terms of \( \cos \left( \theta \right)... | No |
Find the exact value of \( \cos \left( {15}^{ \circ }\right) \) . | In order to use Theorem 10.13 to find \( \cos \left( {15}^{ \circ }\right) \), we need to write \( {15}^{ \circ } \) as a sum or difference of angles whose cosines and sines we know. One way to do so is to write \( {15}^{ \circ } = {45}^{ \circ } - {30}^{ \circ } \). \n\n\[ \cos \left( {15}^{ \circ }\right) = \cos \lef... | Yes |
Rewrite \( {\sin }^{2}\left( \theta \right) {\cos }^{2}\left( \theta \right) \) as a sum and difference of cosines to the first power. | Solution. We begin with a straightforward application of Theorem 10.18\n\n\[ \n{\sin }^{2}\left( \theta \right) {\cos }^{2}\left( \theta \right) = \left( \frac{1 - \cos \left( {2\theta }\right) }{2}\right) \left( \frac{1 + \cos \left( {2\theta }\right) }{2}\right) \n\]\n\n\[ \n= \frac{1}{4}\left( {1 - {\cos }^{2}\left(... | Yes |
1. Write \( \cos \left( {2\theta }\right) \cos \left( {6\theta }\right) \) as a sum. | 1. Identifying \( \alpha = {2\theta } \) and \( \beta = {6\theta } \), we find\n\n\[ \cos \left( {2\theta }\right) \cos \left( {6\theta }\right) = \frac{1}{2}\left\lbrack {\cos \left( {{2\theta } - {6\theta }}\right) + \cos \left( {{2\theta } + {6\theta }}\right) }\right\rbrack \]\n\n\[ = \;\frac{1}{2}\cos \left( {-{4\... | Yes |
Graph one cycle of the following functions. State the period of each.\n\n1. \( f\left( x\right) = 3\cos \left( \frac{{\pi x} - \pi }{2}\right) + 1 \) 2. \( g\left( x\right) = \frac{1}{2}\sin \left( {\pi - {2x}}\right) + \frac{3}{2} \) | Solution.\n\n1. We set the argument of the cosine, \( \frac{{\pi x} - \pi }{2} \), equal to each of the values: \( 0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},{2\pi } \) and solve for \( x \) . We summarize the results below.\n\n<table><tr><td>\( a \)</td><td>\( \frac{{\pi x} - \pi }{2} = a \)</td><td>\( x \)</td></tr><tr><t... | No |
1. Find a cosine function whose graph matches the graph of \( y = f\left( x\right) \) . | We fit the data to a function of the form \( C\left( x\right) = A\cos \left( {{\omega x} + \phi }\right) + B \) . Since one cycle is graphed over the interval \( \left\lbrack {-1,5}\right\rbrack \), its period is \( 5 - \left( {-1}\right) = 6 \) . According to Theorem 10.23, \( 6 = \frac{2\pi }{\omega } \), so that \( ... | Yes |
Consider the function \( f\left( x\right) = \cos \left( {2x}\right) - \sqrt{3}\sin \left( {2x}\right) \). Find a formula for \( f\left( x\right) \): 1. in the form \( C\left( x\right) = A\cos \left( {{\omega x} + \phi }\right) + B \) for \( \omega > 0 \) | The key to this problem is to use the expanded forms of the sinusoid formulas and match up corresponding coefficients. Equating \( f\left( x\right) = \cos \left( {2x}\right) - \sqrt{3}\sin \left( {2x}\right) \) with the expanded form of \( C\left( x\right) = A\cos \left( {{\omega x} + \phi }\right) + B \), we get\n\n\[... | Yes |
Graph one cycle of the following functions. State the period of each.\n\n1. \( f\left( x\right) = 1 - 2\sec \left( {2x}\right) \) 2. \( g\left( x\right) = \frac{\csc \left( {\pi - {\pi x}}\right) - 5}{3} \) | Solution.\n\n1. To graph \( y = 1 - 2\sec \left( {2x}\right) \), we follow the same procedure as in Example 10.5.1. First, we set the argument of secant, \( {2x} \), equal to the ’quarter marks’ \( 0,\frac{\pi }{2},\pi ,\frac{3\pi }{2} \) and \( {2\pi } \) and solve for \( x \) .\n\n<table><tr><td>\\( a \\)</td><td>\\(... | Yes |
Graph one cycle of the following functions. Find the period.\n\n1. \( f\left( x\right) = 1 - \tan \left( \frac{x}{2}\right) \) | 1. We proceed as we have in all of the previous graphing examples by setting the argument of tangent in \( f\left( x\right) = 1 - \tan \left( \frac{x}{2}\right) \), namely \( \frac{x}{2} \), equal to each of the ’quarter marks’ \( - \frac{\pi }{2}, - \frac{\pi }{4},0,\frac{\pi }{4} \) and \( \frac{\pi }{2} \), and solv... | Yes |
Example 10.6.6. \( {}^{7} \) The roof on the house below has a ’ \( 6/{12} \) pitch’. This means that when viewed from the side, the roof line has a rise of 6 feet over a run of 12 feet. Find the angle of inclination from the bottom of the roof to the top of the roof. Express your answer in decimal degrees, rounded to ... | Solution. If we divide the side view of the house down the middle, we find that the roof line forms the hypotenuse of a right triangle with legs of length 6 feet and 12 feet. Using Theorem 10.10, we find the angle of inclination, labeled \( \theta \) below, satisfies \( \tan \left( \theta \right) = \frac{6}{12} = \frac... | Yes |
Find all angles \( \theta \) for which \( \sin \left( \theta \right) = \frac{1}{3} \) . | If \( \sin \left( \theta \right) = \frac{1}{3} \), then the terminal side of \( \theta \), when plotted in standard position, intersects the Unit Circle at \( y = \frac{1}{3} \) . Geometrically, we see that this happens at two places: in Quadrant I and Quadrant II. If we let \( \alpha \) denote the acute solution to th... | Yes |
Solve the following equations and check your answers analytically. List the solutions which lie in the interval \( \lbrack 0,{2\pi }) \) and verify them using a graphing utility.\n\n1. \( \cos \left( {2x}\right) = - \frac{\sqrt{3}}{2} \) | The solutions to \( \cos \left( u\right) = - \frac{\sqrt{3}}{2} \) are \( u = \frac{5\pi }{6} + {2\pi k} \) or \( u = \frac{7\pi }{6} + {2\pi k} \) for integers \( k \) . Since the argument of cosine here is \( {2x} \), this means \( {2x} = \frac{5\pi }{6} + {2\pi k} \) or \( {2x} = \frac{7\pi }{6} + {2\pi k} \) for in... | Yes |
1. \( 3{\sin }^{3}\left( x\right) = {\sin }^{2}\left( x\right) \) | We resist the temptation to divide both sides of \( 3{\sin }^{3}\left( x\right) = {\sin }^{2}\left( x\right) \) by \( {\sin }^{2}\left( x\right) \) (What goes wrong if you do?) and instead gather all of the terms to one side of the equation and factor.\n\n\[ 3{\sin }^{3}\left( x\right) = {\sin }^{2}\left( x\right) \]\n... | Yes |
1. \( 2\sin \left( x\right) \leq 1 \) | We begin solving \( 2\sin \left( x\right) \leq 1 \) by collecting all of the terms on one side of the equation and zero on the other to get \( 2\sin \left( x\right) - 1 \leq 0 \) . Next, we let \( f\left( x\right) = 2\sin \left( x\right) - 1 \) and note that our original inequality is equivalent to solving \( f\left( x... | Yes |
Express the domain of the following functions using extended interval notation. 1. \( f\left( x\right) = \csc \left( {{2x} + \frac{\pi }{3}}\right) \) | To find the domain of \( f\left( x\right) = \csc \left( {{2x} + \frac{\pi }{3}}\right) \), we rewrite \( f \) in terms of sine as \( f\left( x\right) = \frac{1}{\sin \left( {{2x} + \frac{\pi }{3}}\right) } \) . Since the sine function is defined everywhere, our only concern comes from zeros in the denominator. Solving ... | Yes |
1. \( \arcsin \left( {2x}\right) = \frac{\pi }{3} \) | To solve \( \arcsin \left( {2x}\right) = \frac{\pi }{3} \), we first note that \( \frac{\pi }{3} \) is in the range of the arcsine function (so a solution exists!) Next, we exploit the inverse property of sine and arcsine from Theorem 10.26\n\n\[ \arcsin \left( {2x}\right) = \frac{\pi }{3} \]\n\n\[ \sin \left( {\arcsin... | Yes |
Recall from Exercise 55 in Section 10.1 that The Giant Wheel at Cedar Point is a circle with diameter 128 feet which sits on an 8 foot tall platform making its overall height 136 feet. It completes two revolutions in 2 minutes and 7 seconds. Assuming that the riders are at the edge of the circle, find a sinusoid which ... | Solution. We sketch the problem situation below and assume a counter-clockwise rotation. \( {}^{3} \)\n\n\n\n\( {}^{3} \) Otherwise, we could just observe the motion of the wheel from the other side.\n\n\n\nWe know f... | Yes |
1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data. | 1. To get a feel for the data, we plot it below.\n\n\n\nThe data certainly appear sinusoidal, \( {}^{5} \) but when it comes down to it, fitting a sinusoid to data manually is not an exact science. We do our best to ... | Yes |
Theorem 11.1. Equation for Free Undamped Harmonic Motion: Suppose an object of mass \( m \) is suspended from a spring with spring constant \( k \) . If the initial displacement from the equilibrium position is \( {x}_{0} \) and the initial velocity of the object is \( {v}_{0} \), then the displacement \( x \) from the... | \[ \text{-}\omega = \sqrt{\frac{k}{m}}\text{and}A = \sqrt{{x}_{0}^{2} + {\left( \frac{{v}_{0}}{\omega }\right) }^{2}} \] \[ \text{-}A\sin \left( \phi \right) = {x}_{0}\text{and}{A\omega }\cos \left( \phi \right) = {v}_{0}\text{.} \] | Yes |
1. If the object is attached to the spring and released 3 feet below the equilibrium position from rest, find the equation of motion of the object, \( x\left( t\right) \) . When does the object first pass through the equilibrium position? Is the object heading upwards or downwards at this instant? | Solution. In order to use the formulas in Theorem 11.1, we first need to determine the spring constant \( k \) and the mass of the object \( m \) . To find \( k \), we use Hooke’s Law \( F = {kd} \) . We know the object weighs 64 lbs. and stretches the spring \( 8 \) ft.. Using \( F = {64} \) and \( d = 8 \), we get \(... | Yes |
Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, find the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree. | For definitiveness, we label the triangle below.  \( b = 4 \) To find the length of the missing side \( a \), we use the Pythagorean Theorem to get \( {a}^{2} + {4}^{2} = {7}^{2} \) which then yields \( a = \sqrt{33}... | Yes |
Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs \( \left( {\alpha, a}\right) \) , \( \left( {\beta, b}\right) \) and \( \left( {\gamma, c}\right) \), the following ratios hold\n\n\[ \frac{\sin \left( \alpha \right) }{a} = \frac{\sin \left( \beta \right) }{b} = \frac{\sin \left( \gamma \r... | The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle \( \bigtriangleup {ABC} \) below, all of whose angles are acute, with angle-side opposite pairs \( \left( {\alpha, a}\right) ,\left( {\beta, b}\right) \) and \( \left( {\gamma, c}\right) \) . If we drop an altitude f... | Yes |
Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle.\n\n1. \( \alpha = {120}^{ \circ }, a = 7 \) units, \( \beta = {45}^{ \circ } \) | Knowing an angle-side opposite pair, namely \( \alpha \) and \( a \), we may proceed in using the Law of Sines. Since \( \beta = {45}^{ \circ } \), we use \( \frac{b}{\sin \left( {45}^{ \circ }\right) } = \frac{7}{\sin \left( {120}^{ \circ }\right) } \) so \( b = \frac{7\sin \left( {45}^{ \circ }\right) }{\sin \left( {... | Yes |
Theorem 11.3. Suppose \( \left( {\alpha, a}\right) \) and \( \left( {\gamma, c}\right) \) are intended to be angle-side pairs in a triangle where \( \alpha, a \) and \( c \) are given. Let \( h = c\sin \left( \alpha \right) \n- If \( a < h \), then no triangle exists which satisfies the given criteria.\n- If \( a = h \... | Theorem 11.3 is proved on a case-by-case basis. If \( a < h \), then \( a < c\sin \left( \alpha \right) \) . If a triangle were to exist, the Law of Sines would have \( \frac{\sin \left( \gamma \right) }{c} = \frac{\sin \left( \alpha \right) }{a} \) so that \( \sin \left( \gamma \right) = \frac{c\sin \left( \alpha \rig... | No |
Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is \( {30}^{ \circ } \) and at the second point the angle is \( {45}^{ \circ } \). Assuming a straight coastline, find the distance ... | We sketch the problem below with the first observation point labeled as \( P \) and the second as \( Q \). In order to use the Law of Sines to find the distance \( d \) from \( Q \) to the island, we first need to find the measure of \( \beta \) which is the angle opposite the side of length \( 5\mathrm{{miles}} \). To... | Yes |
Find the area of the triangle in Example 11.2.2 number 1. | Solution. From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose \( A = \frac{1}{2}{ac}\sin \left( \beta \right) \) from Theorem 11.4 because it uses the most pieces of given information. We are given \( a = 7 \) and \( \beta... | Yes |
Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle.\n\n1. \( \beta = {50}^{ \circ }, a = 7 \) units, \( c = 2 \) units | 1. We are given the lengths of two sides, \( a = 7 \) and \( c = 2 \), and the measure of the included angle, \( \beta = {50}^{ \circ } \) . With no angle-side opposite pair to use, we apply the Law of Cosines. We get \( {b}^{2} = {7}^{2} + {2}^{2} - 2\left( 7\right) \left( 2\right) \cos \left( {50}^{ \circ }\right) \)... | Yes |
A researcher wishes to determine the width of a vernal pond as drawn below. From a point \( P \), he finds the distance to the eastern-most point of the pond to be 950 feet, while the distance to the western-most point of the pond from \( P \) is 1000 feet. If the angle between the two lines of sight is \( {60}^{ \circ... | Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to find the length of the missing side opposite the given angle. Calling this length \( w \) (for width), we get \( {w}^{2} = {950}^{2} + {1000}^{2} - 2\left( {950}\right) \left( {1000}\right) \cos \... | Yes |
Find the area enclosed of the triangle in Example 11.3.1 number 2. | Solution. We are given \( a = 4, b = 7 \) and \( c = 5 \) . Using these values, we find \( s = \frac{1}{2}\left( {4 + 7 + 5}\right) = 8 \) , \( \left( {s - a}\right) = 8 - 4 = 4,\left( {s - b}\right) = 8 - 7 = 1 \) and \( \left( {s - c}\right) = 8 - 5 = 3 \) . Using Heron’s Formula, we get \( A = \sqrt{s\left( {s - a}\... | Yes |
For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has \( r > 0 \) and the other with \( r < 0 \) . | 1. Whether we move 2 units along the polar axis and then rotate \( {240}^{ \circ } \) or rotate \( {240}^{ \circ } \) then move out 2 units from the pole, we plot \( P\left( {2,{240}^{ \circ }}\right) \) below.\n\n !... | Yes |
Theorem 11.7. Conversion Between Rectangular and Polar Coordinates: Suppose \( P \) is represented in rectangular coordinates as \( \left( {x, y}\right) \) and in polar coordinates as \( \left( {r,\theta }\right) \). Then\n\n\[ \text{-}x = r\cos \left( \theta \right) \text{and}y = r\sin \left( \theta \right) \]\n\n\[ \... | In the case \( r > 0 \), Theorem 11.7 is an immediate consequence of Theorem 10.3 along with the quotient identity \( \tan \left( \theta \right) = \frac{\sin \left( \theta \right) }{\cos \left( \theta \right) } \). If \( r < 0 \), then we know an alternate representation for \( \left( {r,\theta }\right) \) is \( \left(... | Yes |
Convert each point in rectangular coordinates given below into polar coordinates with \( r \geq 0 \) and \( 0 \leq \theta < {2\pi } \) . Use exact values if possible and round any approximate values to two decimal places. Check your answer by converting them back to rectangular coordinates. | 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking the time to plot the points before we do any calculations. Plotting \( P\left( {2, - 2\sqrt{3}}\right) \) shows that it lies in Quadrant IV. With \( x = 2 \) and \( y = - 2\sqrt{3} \), we get \( {r}^{2} = {x}^{2} + {y}^{2} =... | Yes |
Graph the following polar equations.\n\n1. \( r = 4 \) | In the equation \( r = 4,\theta \) is free. The graph of this equation is, therefore, all points which have a polar coordinate representation \( \left( {4,\theta }\right) \), for any choice of \( \theta \) . Graphically this translates into tracing out all of the points 4 units away from the origin. This is exactly the... | Yes |
Graph the following polar equations.\n\n1. \( r = 4 - 2\sin \left( \theta \right) \) 2. \( r = 2 + 4\cos \left( \theta \right) \) 3. \( r = 5\sin \left( {2\theta }\right) \) 4. \( {r}^{2} = {16}\cos \left( {2\theta }\right) \n | Solution.\n\n1. We first plot the fundamental cycle of \( r = 4 - 2\sin \left( \theta \right) \) on the \( {\theta r} \) -axes. To help us visualize what is going on graphically, we divide up \( \left\lbrack {0,{2\pi }}\right\rbrack \) into the usual four subintervals \( \left\lbrack {0,\frac{\pi }{2}}\right\rbrack ,\l... | No |
Sketch the region in the \( {xy} \) -plane described by the following sets.\n\n1. \( \left\{ {\left( {r,\theta }\right) \mid 0 \leq r \leq 5\sin \left( {2\theta }\right) ,0 \leq \theta \leq \frac{\pi }{2}}\right\} \) | Solution. Our first step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this example are found in either Example 11.5.2 or Example 11.5.3, most of the work is done for us.\n\n1. We know from Example 11.5.2 number 3 that ... | Yes |
Theorem 11.9. Rotation of Axes: Suppose the positive \( x \) and \( y \) axes are rotated counterclockwise through an angle \( \theta \) to produce the axes \( {x}^{\prime } \) and \( {y}^{\prime } \), respectively. Then the coordinates \( P\left( {x, y}\right) \) and \( P\left( {{x}^{\prime },{y}^{\prime }}\right) \) ... | \[ \left\{ {\begin{array}{l} x = {x}^{\prime }\cos \left( \theta \right) - {y}^{\prime }\sin \left( \theta \right) \\ y = {x}^{\prime }\sin \left( \theta \right) + {y}^{\prime }\cos \left( \theta \right) \end{array}\;\text{ and }\left\{ \begin{array}{l} {x}^{\prime } = x\cos \left( \theta \right) + y\sin \left( \theta ... | Yes |
1. Let \( P\left( {x, y}\right) = \left( {2, - 4}\right) \) and find \( P\left( {{x}^{\prime },{y}^{\prime }}\right) \) . Check your answer algebraically and graphically. | 1. If \( P\left( {x, y}\right) = \left( {2, - 4}\right) \) then \( x = 2 \) and \( y = - 4 \) . Using these values for \( x \) and \( y \) along with \( \theta = \frac{\pi }{3} \), Theorem 11.9 gives \( {x}^{\prime } = x\cos \left( \theta \right) + y\sin \left( \theta \right) = 2\cos \left( \frac{\pi }{3}\right) + \lef... | Yes |
Graph the following equations.\n\n1. \( 5{x}^{2} + {26xy} + 5{y}^{2} - {16x}\sqrt{2} + {16y}\sqrt{2} - {104} = 0 \) | Since the equation \( 5{x}^{2} + {26xy} + 5{y}^{2} - {16x}\sqrt{2} + {16y}\sqrt{2} - {104} = 0 \) is already given to us in the form required by Theorem 11.10, we identify \( A = 5, B = {26} \) and \( C = 5 \) so that \( \cot \left( {2\theta }\right) = \frac{A - C}{B} = \frac{5 - 5}{26} = 0 \) . This means \( \cot \lef... | Yes |
Example 11.6.3. Use Theorem 11.11 to classify the graphs of the following non-degenerate conics. | Solution. This is a straightforward application of Theorem 11.11.\n\n1. We have \( A = {21}, B = {10}\sqrt{3} \) and \( C = {31} \) so \( {B}^{2} - {4AC} = {\left( {10}\sqrt{3}\right) }^{2} - 4\left( {21}\right) \left( {31}\right) = - {2304} < 0 \) . Theorem 11.11 predicts the graph is an ellipse, which checks with our... | Yes |
Sketch the graphs of the following equations.\n\n1. \( r = \frac{4}{1 - \sin \left( \theta \right) } \) | From \( r = \frac{4}{1 - \sin \left( \theta \right) } \), we first note \( e = 1 \) which means we have a parabola on our hands. Since \( {ed} = 4 \), we have \( d = 4 \) and considering the form of the equation, this puts the directrix at \( y = - 4 \) . Since the focus is at \( \left( {0,0}\right) \), we know that th... | Yes |
Theorem 11.13. Given constants \( \ell > 0, e \geq 0 \) and \( \phi \), the graph of the equation\n\n\[ r = \frac{\ell }{1 - e\cos \left( {\theta - \phi }\right) } \]\n\n is a conic section with eccentricity \( e \) and one focus at \( \left( {0,0}\right) \) . | - If \( e = 0 \), the graph is a circle centered at \( \left( {0,0}\right) \) with radius \( \ell \) .\n\n- If \( e \neq 0 \), then the conic has a focus at \( \left( {0,0}\right) \) and the directrix contains the point with polar coordinates \( \left( {-d,\phi }\right) \) where \( d = \frac{\ell }{e} \) .\n\n- If \( 0... | Yes |
For each of the following complex numbers find \( \operatorname{Re}\left( z\right) ,\operatorname{Im}\left( z\right) ,\left| z\right| ,\arg \left( z\right) \) and \( \operatorname{Arg}\left( z\right) \) . Plot \( z \) in the complex plane. | 1. For \( z = \sqrt{3} - i = \sqrt{3} + \left( {-1}\right) i \), we have \( \operatorname{Re}\left( z\right) = \sqrt{3} \) and \( \operatorname{Im}\left( z\right) = - 1 \) . To find \( \left| z\right| ,\arg \left( z\right) \) and \( \operatorname{Arg}\left( z\right) \), we need to find a polar representation \( \left( ... | Yes |
Theorem 11.14. Properties of the Modulus: Let \( z \) and \( w \) be complex numbers.\n\n- \( \left| z\right| \) is the distance from \( z \) to 0 in the complex plane\n\n- \( \left| z\right| \geq 0 \) and \( \left| z\right| = 0 \) if and only if \( z = 0 \)\n\n- \( \left| z\right| = \sqrt{\operatorname{Re}{\left( z\ri... | To prove the first three properties in Theorem 11.14, suppose \( z = a + {bi} \) where \( a \) and \( b \) are real numbers. To determine \( \left| z\right| \), we find a polar representation \( \left( {r,\theta }\right) \) with \( r \geq 0 \) for the point \( \left( {a, b}\right) \) . From Section 11.4, we know \( {r}... | Yes |
Theorem 11.15. Properties of the Argument: Let \( z \) be a complex number.\n\n- If \( \operatorname{Re}\left( z\right) \neq 0 \) and \( \theta \in \arg \left( z\right) \), then \( \tan \left( \theta \right) = \frac{\operatorname{Im}\left( z\right) }{\operatorname{Re}\left( z\right) } \) .\n\n- If \( \operatorname{Re}\... | To prove Theorem 11.15, suppose \( z = a + {bi} \) for real numbers \( a \) and \( b \) . By definition, \( a = \operatorname{Re}\left( z\right) \) and \( b = \operatorname{Im}\left( z\right) \), so the point associated with \( z \) is \( \left( {a, b}\right) = \left( {\operatorname{Re}\left( z\right) ,\operatorname{Im... | Yes |
Theorem 11.16. Products, Powers and Quotients Complex Numbers in Polar Form:\n\nSuppose \( z \) and \( w \) are complex numbers with polar forms \( z = \left| z\right| \operatorname{cis}\left( \alpha \right) \) and \( w = \left| w\right| \operatorname{cis}\left( \beta \right) \) . Then\n\n- Product Rule: \( {zw} = \lef... | The proof of Theorem 11.16 requires a healthy mix of definition, arithmetic and identities. We first start with the product rule.\n\n\[ {zw} = \left\lbrack {\left| z\right| \operatorname{cis}\left( \alpha \right) }\right\rbrack \left\lbrack {\left| w\right| \operatorname{cis}\left( \beta \right) }\right\rbrack \]\n\n\[... | Yes |
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