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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. \( f\left( {A \cup B}\right) = f\left( A\right) \cup f\left( B\right) \)
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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 \( y \in f\left( {A \cap B}\right) \), then \( y \in f\left( A\right) \cap f\left( B\right) \) .\n\nWe assume that \( y \in f\left( {A \cap B}\right) \) . This means that there exists an \( x \in A \cap B \) such that \( f\left( x\right) = y \) . Since \( x \in A \cap B \), we conclude that \( x \in A \) and \( x \in B \) .\n\n- Since \( x \in A \) and \( f\left( x\right) = y \), we conclude that \( y \in f\left( A\right) \).\n\n- Since \( x \in B \) and \( f\left( x\right) = y \), we conclude that \( y \in f\left( B\right) \).\n\nSince \( y \in f\left( A\right) \) and \( y \in f\left( B\right), y \in f\left( A\right) \cap f\left( B\right) \) . This proves that if \( y \in \) \( f\left( {A \cap B}\right) \), then \( y \in f\left( A\right) \cap f\left( B\right) \) . Hence \( f\left( {A \cap B}\right) \subseteq f\left( A\right) \cap f\left( B\right) \) .
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No
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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\[ \text{2.}{f}^{-1}\left( {C \cup D}\right) = {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \]
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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 subset of the other.\n\nWe start by letting \( x \) be an element of \( {f}^{-1}\left( {C \cup D}\right) \). This means that \( f\left( x\right) \) is an element of \( C \cup D \). Hence,\n\n\[ f\left( x\right) \in C\text{ or }f\left( x\right) \in D. \]\n\nIn the case where \( f\left( x\right) \in C \), we conclude that \( x \in {f}^{-1}\left( C\right) \), and hence that \( x \in {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \). In the case where \( f\left( x\right) \in D \), we see that \( x \in {f}^{-1}\left( D\right) \), and hence that \( x \in {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \). So in both cases, \( x \in {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \), and we have proved that \( {f}^{-1}\left( {C \cup D}\right) \subseteq {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \).\n\nWe now let \( t \in {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \). This means that\n\n\[ t \in {f}^{-1}\left( C\right) \text{ or }t \in {f}^{-1}\left( D\right) . \]\n\n- In the case where \( t \in {f}^{-1}\left( C\right) \), we conclude that \( f\left( t\right) \in C \) and hence that \( f\left( t\right) \in C \cup D \). This means that \( t \in {f}^{-1}\left( {C \cup D}\right) \).\n\n- Similarly, when \( t \in {f}^{-1}\left( D\right) \), it follows that \( f\left( t\right) \in D \) and hence that \( f\left( t\right) \in C \cup D \). This means that \( t \in {f}^{-1}\left( {C \cup D}\right) \).\n\nThese two cases prove that if \( t \in {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \), then \( t \in {f}^{-1}\left( {C \cup D}\right) \). Therefore, \( {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \subseteq {f}^{-1}\left( {C \cup D}\right) \).\n\nSince we have now proved that each of the two sets is a subset of the other set, we can conclude that \( {f}^{-1}\left( {C \cup D}\right) = {f}^{-1}\left( C\right) \cup {f}^{-1}\left( D\right) \).
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No
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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
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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 \( f\left( A\right) \subseteq T \), and so \( {f}^{-1}\left( {f\left( A\right) }\right) \subseteq S \). Notice that\n\n\[ \n{f}^{-1}\left( {f\left( A\right) }\right) = \{ x \in S \mid f\left( x\right) \in f\left( A\right) \} .\n\]\n\nSince \( f\left( a\right) \in f\left( A\right) \), we use this to conclude that \( a \in {f}^{-1}\left( {f\left( A\right) }\right) \). This proves that if \( a \in A \), then \( a \in {f}^{-1}\left( {f\left( A\right) }\right) \), and hence that \( A \subseteq {f}^{-1}\left( {f\left( A\right) }\right) \).
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No
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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 \( \left( {R}^{-1}\right) = \operatorname{dom}\left( R\right) \) .\n\n- The inverse of \( {R}^{-1} \) is \( R \) . That is, \( {\left( {R}^{-1}\right) }^{-1} = R \) .
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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 \( \operatorname{dom}\left( {R}^{-1}\right) \subseteq \operatorname{range}\left( R\right) \), we can start by choosing an arbitrary element of \( \operatorname{dom}\left( {R}^{-1}\right) \) . So let \( y \in \operatorname{dom}\left( {R}^{-1}\right) \) . The goal now is to prove that \( y \in \operatorname{range}\left( R\right) \) . What does it mean to say that \( y \in \operatorname{dom}\left( {R}^{-1}\right) \) ? It means that there exists an \( x \in A \) such that\n\n\[ \left( {y, x}\right) \in {R}^{-1}\text{.} \]\n\nNow what does it mean to say that \( \left( {y, x}\right) \in {R}^{-1} \) ? It means that \( \left( {x, y}\right) \in R \) . What does this tell us about \( y \) ?
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No
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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 \( a = {bk} \) .
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- 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 assume that \( {xMy} \) and \( {yMz} \) . Then there exist integers \( p \) and \( q \) such that\n\n\[ x = {yp}\text{and}y = {zq}\text{.} \]\n\nUsing the second equation to make a substitution in the first equation, we see that \( x = z\left( {pq}\right) \) . Since \( {pq} \in \mathbb{Z} \), we have shown that \( x \) is a multiple of \( z \) and hence \( {xMz} \) . Therefore, \( M \) is a transitive relation.\n\nThe relation \( M \) is reflexive on \( \mathbb{Z} \) and is transitive, but since \( M \) is not symmetric, it is not an equivalence relation on \( \mathbb{Z} \).
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Yes
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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 \( n \) .
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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 let \( r \) be this common remainder. Then, by Theorem 3.31,\n\n\[ a \equiv r\left( {\;\operatorname{mod}\;n}\right) \text{ and }b \equiv r\left( {\;\operatorname{mod}\;n}\right) . \]\n\nSince congruence modulo \( n \) is an equivalence relation, it is a symmetric relation. Hence, since \( b \equiv r\left( {\;\operatorname{mod}\;n}\right) \), we can conclude that \( r \equiv b\left( {\;\operatorname{mod}\;n}\right) \) . Combining this with the fact that \( a \equiv r\left( {\;\operatorname{mod}\;n}\right) \), we now have\n\n\[ a \equiv r\left( {\;\operatorname{mod}\;n}\right) \text{ and }r \equiv b\left( {\;\operatorname{mod}\;n}\right) . \]\n\nWe can now use the transitive property to conclude that \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) . This proves that if \( a \) and \( b \) have the same remainder when divided by \( n \), then \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) .\n\nWe will now prove that if \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \), then \( a \) and \( b \) have the same remainder when divided by \( n \) . Assume that \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \), and let \( r \) be the least nonnegative remainder when \( b \) is divided by \( n \) . Then \( 0 \leq r < n \) and, by Theorem 3.31,\n\n\[ b \equiv r\left( {\;\operatorname{mod}\;n}\right) . \]\n\nNow, using the facts that \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \) and \( b \equiv r\left( {\;\operatorname{mod}\;n}\right) \), we can use the transitive property to conclude that\n\n\[ a \equiv r\left( {\;\operatorname{mod}\;n}\right) . \]\n\nThis means that there exists an integer \( q \) such that \( a - r = {nq} \) or that\n\n\[ a = {nq} + r. \]\n\nSince we already know that \( 0 \leq r < n \), the last equation tells us that \( r \) is the least nonnegative remainder when \( a \) is divided by \( n \) . Hence we have proven that if \( a \equiv b\left( {\;\operatorname{mod}\;n}\right) \), then \( a \) and \( b \) have the same remainder when divided by \( n \) .
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Yes
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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 \( a, b \in A,\left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) or \( \left\lbrack a\right\rbrack \cap \left\lbrack b\right\rbrack = \varnothing \) .
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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 \( a \in \left\lbrack a\right\rbrack \) .\n\nThe second part of this theorem is a biconditional statement. We will prove it by proving two conditional statements. We will first prove that if \( a \sim b \), then \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) . So let \( a, b \in A \) and assume that \( a \sim b \) . We will now prove that the two sets \( \left\lbrack a\right\rbrack \) and \( \left\lbrack b\right\rbrack \) are equal. We will do this by proving that each is a subset of the other.\n\nFirst, assume that \( x \in \left\lbrack a\right\rbrack \) . Then, by definition, \( x \sim a \) . Since we have assumed that \( a \sim b \), we can use the transitive property of \( \sim \) to conclude that \( x \sim b \), and this means that \( x \in \left\lbrack b\right\rbrack \) . This proves that \( \left\lbrack a\right\rbrack \subseteq \left\lbrack b\right\rbrack \) .\n\nWe now assume that \( y \in \left\lbrack b\right\rbrack \) . This means that \( y \sim b \), and hence by the symmetric property, that \( b \sim y \) . Again, we are assuming that \( a \sim b \) . So we have\n\n\[a \sim b\text{and}b \sim y\text{.}\]\n\nWe use the transitive property to conclude that \( a \sim y \) and then, using the symmetric property, we conclude that \( y \sim a \) . This proves that \( y \in \left\lbrack a\right\rbrack \) and, hence, that \( \left\lbrack b\right\rbrack \subseteq \left\lbrack a\right\rbrack \) . This means that we can conclude that if \( a \sim b \), then \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) .\n\nWe must now prove that if \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \), then \( a \sim b \) . Let \( a, b \in A \) and assume that \( \left\lbrack a\right\rbrack = \left\lbrack b\right\rbrack \) . Using the first part of the theorem, we know that \( a \in \left\lbrack a\right\rbrack \) and since the two sets are equal, this tells us that \( a \in \left\lbrack b\right\rbrack \) . Hence by the definition of \( \left\lbrack b\right\rbrack \), we conclude that \( a \sim b \) . This completes the proof of the second part of the theorem.
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Yes
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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 \( A \) .
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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 partition of \( A \) .\n\nPart (1) of Theorem 7.14 states that for each \( a \in A, a \in \left\lbrack a\right\rbrack \) . In terms of the equivalence classes, this means that each equivalence class is nonempty since each element of \( A \) is in its own equivalence class. Consequently, \( \mathcal{C} \), the collection of all equivalence classes determined by \( \sim \), satisfies the first two conditions of the definition of a partition.\n\nWe must now show that the collection \( \mathcal{C} \) of all equivalence classes determined by \( \sim \) satisfies the third condition for being a partition. That is, we need to show that any two equivalence classes are either equal or are disjoint. However, this is exactly the result in Part (3) of Theorem 7.14.\n\nHence, we have proven that the collection \( \mathcal{C} \) of all equivalence classes determined by \( \sim \) is a partition of the set \( A \) .
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Yes
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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 \( \gcd \left( {c, d}\right) = \gcd \left( {d, r}\right) \) .
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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 \( d \) . Consequently, there exist integers \( x \) and \( y \) such that \( c = {mx} \) and \( d = {my} \) . Hence,\n\n\[ r = c - d \cdot q \]\n\n\[ r = {mx} - \left( {my}\right) q \]\n\n\[ r = m\left( {x - {yq}}\right) . \]\n\nBut this means that \( m \) divides \( r \) . Since \( m \) divides \( d \) and \( m \) divides \( r, m \) is less than or equal to \( \gcd \left( {d, r}\right) \) . Thus, \( m \leq n \) .\n\nUsing a similar argument, we see that \( n \) divides \( d \) and \( n \) divides \( r \) . Since \( c = d \cdot q + r \), we can prove that \( n \) divides \( c \) . Hence, \( n \) divides \( c \) and \( n \) divides \( d \) . Thus, \( n \leq \gcd \left( {c, d}\right) \) or \( n \leq m \) . We now have \( m \leq n \) and \( n \leq m \) . Hence, \( m = n \) and \( \gcd \left( {c, d}\right) = \gcd \left( {d, r}\right) \) .
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Yes
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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 \( \gcd \left( {a, b}\right) = {au} + {bv}. \)
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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 table for the Euclidean Algorithm is\n\n<table><thead><tr><th>Step</th><th>Original Pair</th><th>Equation from Division Algorithm</th><th>New Pair</th></tr></thead><tr><td>1</td><td>\( \left( {a, b}\right) \)</td><td>\( a = b \cdot {q}_{1} + {r}_{1} \)</td><td>\( \left( {b,{r}_{1}}\right) \)</td></tr><tr><td>2</td><td>\( \left( {b,{r}_{1}}\right) \)</td><td>\( b = {r}_{1} \cdot {q}_{2} + {r}_{2} \)</td><td>\( \left( {{r}_{1},{r}_{2}}\right) \)</td></tr><tr><td>3</td><td>\( \left( {{r}_{1},{r}_{2}}\right) \)</td><td>\( {r}_{1} = {r}_{2} \cdot {q}_{3} + {r}_{3} \)</td><td>\( \left( {{r}_{2},{r}_{3}}\right) \)</td></tr><tr><td>4</td><td>\( \left( {{r}_{2},{r}_{3}}\right) \)</td><td>\( {r}_{2} = {r}_{3} \cdot {q}_{4} + 0 \)</td><td></td></tr></table>\n\nWe can use Step 3 to write \( {r}_{3} = \gcd \left( {a, b}\right) \) as a linear combination of \( {r}_{1} \) and \( {r}_{2} \) . We can then solve the equation in Step 2 for \( {r}_{2} \) and use this to write \( {r}_{3} = \gcd \left( {a, b}\right) \) as a linear combination of \( {r}_{1} \) and \( b \) . We can then use the equation in Step 1 to solve for \( {r}_{1} \) and use this to write \( {r}_{3} = \gcd \left( {a, b}\right) \) as a linear combination of \( a \) and \( b \) .\n\nIn general, if we can write \( {r}_{p} = \gcd \left( {a, b}\right) \) as a linear combination of a pair in a given row, then we can use the equation in the preceding step to write \( {r}_{p} = \) \( \gcd \left( {a, b}\right) \) as a linear combination of the pair in this preceding row.\n\nThe notational details of this induction argument get quite involved. Many mathematicians prefer to prove Theorem 8.8 using a property of the natural numbers called the Well-Ordering Principle. The Well-Ordering Principle for the natural numbers states that any nonempty set of natural numbers must contain a least element. It can be proven that the Well-Ordering Principle is equivalent to the Principle of Mathematical Induction.
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No
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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 \( \left( {{ax} + {by}}\right) \) .
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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 \( t \) divides \( a \), there exists an integer \( m \) such that \( a = {mt} \) and since \( t \) divides \( b \), there exists an integer \( n \) such that \( b = {nt} \) . Using substitution and algebra, we then see that\n\n\[ {ax} + {by} = \left( {mt}\right) x + \left( {nt}\right) y \]\n\n\[ = t\left( {{mx} + {ny}}\right) \]\n\nSince \( \left( {{mx} + {ny}}\right) \) is an integer, the last equation proves that \( t \) divides \( {ax} + {by} \) and this proves that for all integers \( x \) and \( y, t \) divides \( \left( {{ax} + {by}}\right) \) .
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Yes
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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 \( b \) .
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Part (1) of Theorem 8.11 is actually a corollary of Theorem 8.9.
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Yes
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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 \( a \mid c \) .
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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 that \( a \mid \left( {bc}\right) \), there exists an integer \( k \) such that\n\n\[ {bc} = {ak}\text{.} \]\n\n(2)\n\nIt may seem tempting to divide both sides of equation (2) by \( b \), but if we do so, we run into problems with the fact that the integers are not closed under division. Instead, we look at the other part of the hypothesis, which is that \( a \) and \( b \) are relatively prime. This means that \( \gcd \left( {a, b}\right) = 1 \) . How can we use this? This means that \( a \) and \( b \) have no common factors except for 1 . In light of equation (2), it seems reasonable that any factor of \( a \) must also be a factor of \( c \) . But how do we formalize this?\n\nOne conclusion that we can use is that since \( \gcd \left( {a, b}\right) = 1 \), by Theorem 8.11, there exist integers \( m \) and \( n \) such that\n\n\[ {am} + {bn} = 1\text{.} \]\n\n(3)\n\nWe may consider solving equation (3) for \( b \) and substituting this into equation (2). The problem, again, is that in order to solve equation (3) for \( b \), we need to divide by \( n \) .\n\nBefore doing anything else, we should look at the goal in equation (1). We need to introduce \( c \) into equation (3). One way to do this is to multiply both sides of equation (3) by \( c \) . (This keeps us in the system of integers since the integers are closed under multiplication.) This gives\n\n\[ \left( {{am} + {bn}}\right) c = 1 \cdot c \]\n\n\[ {acm} + {bcn} = c\text{.} \]\n\n(4)\n\nNotice that the left side of equation (4) contains a term, \( {bcn} \), that contains \( {bc} \) . This means that we can use equation (2) and substitute \( {bc} = {ak} \) in equation (4). After doing this, we can factor the left side of the equation to prove that \( a \mid c \) .
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Yes
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Theorem 8.16. There are infinitely many prime numbers.
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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 the Fundamental Theorem of Arithmetic, \( M \) is a product of prime numbers. In either case, \( M \) has a factor that is a prime number. Since we have listed all the prime numbers, this means that there exists a natural number \( j \) with \( 1 \leq j \leq m \) such that \( {p}_{j} \mid M \) . Now, we can rewrite equation (1) as follows:\n\n\[ \n1 = M - {p}_{1}{p}_{2}\cdots {p}_{m} \n\]\n\n(2)\n\nWe have proved \( {p}_{j} \mid M \), and since \( {p}_{j} \) is one of the prime factors of \( {p}_{1}{p}_{2}\cdots {p}_{m} \) , we can also conclude that \( {p}_{j} \mid \left( {{p}_{1}{p}_{2}\cdots {p}_{m}}\right) \) . Since \( {p}_{j} \) divides both of the terms on the right side of equation (2), we can use this equation to conclude that \( {p}_{j} \) divides 1 . This is a contradiction since a prime number is greater than 1 and cannot divide 1. Hence, our assumption that there are only finitely many primes is false, and so there must be infinitely many primes.
|
Yes
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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 equation \( {ax} + {by} = c \) has infinitely many solutions. In addition, if \( \left( {{x}_{0},{y}_{0}}\right) \) is a particular solution of this equation, then all the solutions of this equation can be written in the form \( x = {x}_{0} + \frac{b}{d}k\;\text{ and }\;y = {y}_{0} - \frac{a}{d}k, \) for some integer \( k \).
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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 combination of \( a \) and \( b \). So there exist integers \( s \) and \( t \) such that \( d = {as} + {bt}. \) Since \( d \mid c \), there exists an integer \( m \) such that \( c = {dm} \). We can now multiply both sides of equation (1) by \( m \) and obtain \( {dm} = \left( {{as} + {bt}}\right) m \) \( c = a\left( {sm}\right) + b\left( {tm}\right) . \) This means that \( x = {sm}, y = {tm} \) is a solution of \( {ax} + {by} = c \), and we have proved that the Diophantine equation \( {ax} + {by} = c \) has at least one solution. Now let \( x = {x}_{0}, y = {y}_{0} \) be any particular solution of \( {ax} + {by} = c \), let \( k \in \mathbb{Z} \), and let \( x = {x}_{0} + \frac{b}{d}k\;y = {y}_{0} - \frac{a}{d}k. \) We now verify that for each \( k \in \mathbb{Z} \), the equations in (2) produce a solution of \( {ax} + {by} = c \). \( {ax} + {by} = a\left( {{x}_{0} + \frac{b}{d}k}\right) + b\left( {{y}_{0} - \frac{a}{d}k}\right) \) \( = a{x}_{0} + \frac{ab}{d}k + b{y}_{0} - \frac{ab}{d}k \) \( = a{x}_{0} + b{y}_{0} \) \( = c\text{.} \) This proves that the Diophantine equation \( {ax} + {by} = c \) has infinitely many solutions. We now show that every solution of this equation can be written in the form described in (2). So suppose that \( x \) and \( y \) are integers such that \( {ax} + {by} = c \). Then \( \left( {{ax} + {by}}\right) - \left( {a{x}_{0} + b{y}_{0}}\right) = c - c = 0, \) and this equation can be rewritten in the following form: \( a\left( {x - {x}_{0}}\right) = b\left( {{y}_{0} - y}\right) . \) Dividing both sides of this equation by \( d \), we obtain \( \left( \frac{a}{d}\right) \left( {x - {x}_{0}}\right) = \left( \frac{b}{d}\right) \left( {{y}_{0} - y}\right) . \) This implies that \( \frac{a}{d}\text{ divides }\left( \frac{b}{d}\right) \left( {{y}_{0} - y}\right) . \) However, by Exercise (7) in Section 8.2, \( \gcd \left( {\frac{a}{d},\frac{b}{d}}\right) = 1 \), and so by Theorem 8.12, we can conclude that \( \frac{a}{d} \) divides \( \left( {{y}_{0} - y}\right) \). This means that there exists an integer \( k \) such that \( {y}_{0} - y = \frac{a}{d}k \), and solving for \( y \) gives \( y = {y}_{0} - \frac{a}{d}k \) Substituting this value for \( y \) in equation (3) and solving for \( x \) yields \( x = {x}_{0} + \frac{b}{d}k \) This proves that every solution of the Diophantine equation \( {ax} + {by} = c \) can be written in the form prescribed in (2).
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No
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Theorem 9.3. Any set equivalent to a finite nonempty set \( A \) is a finite set and has the same cardinality as \( A \) .
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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 conclude that \( B \approx A \) . Since \( A \approx {\mathbb{N}}_{k} \), we can use part (c) of Theorem 9.1 to conclude that \( B \approx {\mathbb{N}}_{k} \) . Thus, \( B \) is finite and has the same cardinality as \( A \) .
|
Yes
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Lemma 9.4. If \( A \) is a finite set and \( x \notin A \), then \( A \cup \{ x\} \) is a finite set and \( \operatorname{card}\left( {A\cup \{ x\} }\right) = \operatorname{card}\left( A\right) + 1. \)
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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 \( {\mathbb{N}}_{1} \) . Thus, \( A \cup \{ x\} \) is finite with cardinality 1, which equals \( \operatorname{card}\left( A\right) + 1 \) .
If \( A \neq \varnothing \), then \( A \approx {\mathbb{N}}_{k} \), for some \( k \in \mathbb{N} \) . This means that \( \operatorname{card}\left( A\right) = k \) , and there exists a bijection \( f : A \rightarrow {\mathbb{N}}_{k} \) . We will now use this bijection to define a function \( g : A \cup \{ x\} \rightarrow {\mathbb{N}}_{k + 1} \) and then prove that the function \( g \) is a bijection. We define \( g : A \cup \{ x\} \rightarrow {\mathbb{N}}_{k + 1} \) as follows: For each \( t \in A \cup \{ x\} \) ,
\[
g\left( t\right) = \left\{ \begin{array}{l} f\left( t\right) \text{ if }t \in A \\ k + 1\text{ if }t = x. \end{array}\right.
\]
To prove that \( g \) is an injection, we let \( {x}_{1},{x}_{2} \in A \cup \{ x\} \) and assume \( {x}_{1} \neq {x}_{2} \) .
- If \( {x}_{1},{x}_{2} \in A \), then since \( f \) is a bijection, \( f\left( {x}_{1}\right) \neq f\left( {x}_{2}\right) \), and this implies that \( g\left( {x}_{1}\right) \neq g\left( {x}_{2}\right) \) .
- If \( {x}_{1} = x \), then since \( {x}_{2} \neq {x}_{1} \), we conclude that \( {x}_{2} \neq x \) and hence \( {x}_{2} \in A \) . So \( g\left( {x}_{1}\right) = k + 1 \), and since \( f\left( {x}_{2}\right) \in {\mathbb{N}}_{k} \) and \( g\left( {x}_{2}\right) = f\left( {x}_{2}\right) \), we can conclude that \( g\left( {x}_{1}\right) \neq g\left( {x}_{2}\right) \) .
- The case where \( {x}_{2} = x \) is handled similarly to the previous case.
This proves that the function \( g \) is an injection. The proof that \( g \) is a surjection is Exercise (1). Since \( g \) is a bijection, we conclude that \( A \cup \{ x\} \approx {\mathbb{N}}_{k + 1} \), and
\[
\operatorname{card}\left( {A\cup \{ x\} }\right) = k + 1.
\]
Since \( \operatorname{card}\left( A\right) = k \), we have proved that \( \operatorname{card}\left( {A\cup \{ x\} }\right) = \operatorname{card}\left( A\right) + 1 \) .
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No
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Lemma 9.5. For each natural number \( m \), if \( A \subseteq {\mathbb{N}}_{m} \), then \( A \) is a finite set and \( \operatorname{card}\left( A\right) \leq m \) .
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Proof. We will use a proof using induction on \( m \) . For each \( m \in \mathbb{N} \), let \( P\left( m\right) \) be, \
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No
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Theorem 9.6. If \( S \) is a finite set and \( A \) is a subset of \( S \), then \( A \) is a finite set and \( \operatorname{card}\left( A\right) \leq \operatorname{card}\left( S\right) . \)
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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 bijection \( f : S \rightarrow {\mathbb{N}}_{k} \) for some \( k \in \mathbb{N} \) . In this case, \( \operatorname{card}\left( S\right) = k \) . We need to show that \( A \) is equivalent to a finite set. To do this, we define \( g : A \rightarrow f\left( A\right) \) by\n\n\[ g\left( x\right) = f\left( x\right) \text{ for each }x \in A. \]\n\nSince \( f \) is an injection, we conclude that \( g \) is an injection. Now let \( y \in f\left( A\right) \) . Then there exists an \( a \in A \) such that \( f\left( a\right) = y \) . But by the definition of \( g \), this means that \( g\left( a\right) = y \), and hence \( g \) is a surjection. This proves that \( g \) is a bijection.\n\nHence, we have proved that \( A \approx f\left( A\right) \) . Since \( f\left( A\right) \) is a subset of \( {\mathbb{N}}_{k} \), we use Lemma 9.5 to conclude that \( f\left( A\right) \) is finite and \( \operatorname{card}\left( {f\left( A\right) }\right) \leq k \) . In addition, by Theorem 9.3, \( A \) is a finite set and \( \operatorname{card}\left( A\right) = \operatorname{card}\left( {f\left( A\right) }\right) \) . This proves that \( A \) is a finite set and \( \operatorname{card}\left( A\right) \leq \operatorname{card}\left( S\right) \) .
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Yes
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Corollary 9.8. A finite set is not equivalent to any of its proper subsets.
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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\[ \operatorname{card}\left( A\right) \leq \operatorname{card}\left( {B-\{ x\} }\right) .\n\]\n\nAlso, by Corollary 9.7\n\n\[ \operatorname{card}\left( {B-\{ x\} }\right) = \operatorname{card}\left( B\right) - 1.\n\]\n\nHence, we may conclude that \( \operatorname{card}\left( A\right) \leq \operatorname{card}\left( B\right) - 1 \) and that\n\n\[ \operatorname{card}\left( A\right) < \operatorname{card}\left( B\right) \text{.} \]\n\nTheorem 9.3 implies that \( B ≉ A \) . This proves that a finite set is not equivalent to any of its proper subsets.
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Yes
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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.
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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 \( f : A \rightarrow B \) is an injection. As in Theorem 9.6, we define a function \( g : A \rightarrow f\left( A\right) \) by\n\n\[ g\left( x\right) = f\left( x\right) \text{ for each }x \in A. \]\n\nAs we saw in Theorem 9.6, the function \( g \) is a bijection. But then \( A \approx f\left( A\right) \) and \( f\left( A\right) \subseteq B \) . Hence,\n\n\[ \operatorname{card}\left( A\right) = \operatorname{card}\left( {f\left( A\right) }\right) \text{and}\operatorname{card}\left( {f\left( A\right) }\right) \leq \operatorname{card}\left( B\right) \text{.} \]\n\nHence, \( \operatorname{card}\left( A\right) \leq \operatorname{card}\left( B\right) \), and this proves the contrapositive. Hence, if \( \operatorname{card}\left( A\right) > \) \( \operatorname{card}\left( B\right) \), then any function \( f : A \rightarrow B \) is not an injection.
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Yes
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Theorem 9.10. Let \( A \) and \( B \) be sets.\n\n1. If \( A \) is infinite and \( A \approx B \), then \( B \) is infinite.
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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 on page 455 implies that \( A \) is a finite set. This is a contradiction to the assumption that \( A \) is infinite. We have therefore proved that if \( A \) is infinite and \( A \approx B \), then \( B \) is infinite.
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No
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Theorem 9.13. The set \( \mathbb{Z} \) of integers is countably infinite, and so card \( \left( \mathbb{Z}\right) = {\aleph }_{0} \) .
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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 our work in Preview Activity 2, it appears that if \( n \) is an even natural number, then \( f\left( n\right) > 0 \), and if \( n \) is an odd natural number, then \( f\left( n\right) \leq 0 \) . So it seems reasonable to use cases to prove that \( f \) is a surjection and that \( f \) is an injection. To prove that \( f \) is a surjection, we let \( y \in \mathbb{Z} \) .\n\n- If \( y > 0 \), then \( {2y} \in \mathbb{N} \), and\n\[ f\left( {2y}\right) = \frac{2y}{2} = y \]\n\n- If \( y \leq 0 \), then \( - {2y} \geq 0 \) and \( 1 - {2y} \) is an odd natural number. Hence,\n\[ f\left( {1 - {2y}}\right) = \frac{1 - \left( {1 - {2y}}\right) }{2} = \frac{2y}{2} = y. \]\n\nThese two cases prove that if \( y \in \mathbb{Z} \), then there exists an \( n \in \mathbb{N} \) such that \( f\left( n\right) = y \) . Hence, \( f \) is a surjection.\n\nTo prove that \( f \) is an injection, we let \( m, n \in \mathbb{N} \) and assume that \( f\left( m\right) = f\left( n\right) \) . First note that if one of \( m \) and \( n \) is odd and the other is even, then one of \( f\left( m\right) \) and \( f\left( n\right) \) is positive and the other is less than or equal to 0 . So if \( f\left( m\right) = f\left( n\right) \), then both \( m \) and \( n \) must be even or both \( m \) and \( n \) must be odd.\n\n- If both \( m \) and \( n \) are even, then\n\[ f\left( m\right) = f\left( n\right) \text{ implies that }\frac{m}{2} = \frac{n}{2} \]\nand hence that \( m = n \) .\n\n- If both \( m \) and \( n \) are odd, then\n\[ f\left( m\right) = f\left( n\right) \text{ implies that }\frac{1 - m}{2} = \frac{1 - n}{2}. \]\nFrom this, we conclude that \( 1 - m = 1 - n \) and hence that \( m = n \) . This proves that if \( f\left( m\right) = f\left( n\right) \), then \( m = n \) and hence that \( f \) is an injection.\n\nSince \( f \) is both a surjection and an injection, we see that \( f \) is a bijection and, therefore, \( \mathbb{N} \approx \mathbb{Z} \) . Hence, \( \mathbb{Z} \) is countably infinite and \( \operatorname{card}\left( \mathbb{Z}\right) = {\aleph }_{0} \) .
|
Yes
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Theorem 9.14. The set of positive rational numbers is countably infinite.
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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 \( f : \mathbb{N} \rightarrow {\mathbb{Q}}^{ + } \) . The idea is to start in the upper left corner of the table and move to successive diagonals as follows:\n\n- We start with all fractions in which the sum of the numerator and denominator is \( 2\left( \right. \) only \( \left. \frac{1}{1}\right) \) . So \( f\left( 1\right) = \frac{1}{1} \) .\n\n- We next use those fractions in which the sum of the numerator and denominator is 3 . So \( f\left( 2\right) = \frac{2}{1} \) and \( f\left( 3\right) = \frac{1}{2} \) .\n\n- We next use those fractions in which the sum of the numerator and denominator is 4. So \( f\left( 4\right) = \frac{1}{3}, f\left( 5\right) = \frac{3}{1} \) . We skipped \( \frac{2}{2} \) since \( \frac{2}{2} = \frac{1}{1} \) . In this way, we will ensure that the function \( f \) is a one-to-one function.\n\nWe now continue with successive diagonals omitting fractions that are not in lowest terms. This process guarantees that the function \( f \) will be an injection and a surjection. Therefore, \( \mathbb{N} \approx {\mathbb{Q}}^{ + } \) and \( \operatorname{card}\left( {\mathbb{Q}}^{ + }\right) = {\aleph }_{0} \) .
|
Yes
|
Theorem 9.15. If \( A \) is a countably infinite set, then \( A \cup \{ x\} \) is a countably infinite set.
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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 \( x \notin A \), define \( g : \mathbb{N} \rightarrow A \cup \{ x\} \) by\n\n\[ g\left( n\right) = \left\{ \begin{array}{ll} x & \text{ if }n = 1 \\ f\left( {n - 1}\right) & \text{ if }n > 1 \end{array}\right. \]\n\nThe proof that the function \( g \) is a bijection is Exercise (4). Since \( g \) is a bijection, we have proved that \( A \cup \{ x\} \approx \mathbb{N} \) and hence, \( A \cup \{ x\} \) is countably infinite.
|
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.
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## Proof. Exercise (5) on page 473.
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No
|
Theorem 9.17. If \( A \) and \( B \) are disjoint countably infinite sets, then \( A \cup B \) is a countably infinite set.
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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\[ h\left( n\right) = \left\{ \begin{array}{ll} f\left( \frac{n + 1}{2}\right) & \text{ if }n\text{ is odd } \\ g\left( \frac{n}{2}\right) & \text{ if }n\text{ is even. } \end{array}\right. \]\n\nIt is left as Exercise (6) on page 474 to prove that the function \( h \) is a bijection.
|
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.
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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 smallest natural number in \( B \) .\n\n- For each \( n \in \mathbb{N} \), the set \( B - \{ g\left( 1\right), g\left( 2\right) ,\ldots, g\left( n\right) \} \) is not empty since \( B \) is infinite. Define \( g\left( {n + 1}\right) \) to be the smallest natural number in \( B - \{ g\left( 1\right), g\left( 2\right) ,\ldots, g\left( n\right) \} \) .\n\nThe proof that the function \( g \) is a bijection is Exercise (11) on page 474.
|
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 \( \mathbf{c} \) .
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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 \( a < b \) . Find a function\n\n\[ f : \left( {0,1}\right) \rightarrow \left( {a, b}\right) \]\n\nthat is a bijection and conclude that the open interval \( \left( {0,1}\right) \approx \left( {a, b}\right) \) .\n\nHint: Find a linear function that passes through the points \( \left( {0, a}\right) \) and \( \left( {1, b}\right) \) . Use this to define the function \( f \) . Make sure you prove that this function \( f \) is a bijection.
|
No
|
Theorem 9.26. The set of real numbers \( \mathbb{R} \) is uncountable and has cardinality \( c \) .
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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, \( \mathbb{R} \) is uncountable and has cardinality \( \mathbf{c} \) .
|
Yes
|
Theorem 9.27 (Cantor’s Theorem). For every set \( A, A \) and \( \mathcal{P}\left( A\right) \) do not have the same cardinality.
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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 \( f : A \rightarrow \mathcal{P}\left( A\right) \) . We will show that \( f \) cannot be a surjection, and hence there is no bijection from \( A \) to \( \mathcal{P}\left( A\right) \) . This will prove that \( A \) is not equivalent to \( \mathcal{P}\left( A\right) \) . Define\n\n\[ S = \{ x \in A \mid x \notin f\left( x\right) \} . \]\n\nAssume that there exists a \( t \) in \( A \) such that \( f\left( t\right) = S \) . Now, either \( t \in S \) or \( t \notin S \) .\n\n- If \( t \in S \), then \( t \in \{ x \in A \mid x \notin f\left( x\right) \} \) . By the definition of \( S \), this means that \( t \notin f\left( t\right) \) . However, \( f\left( t\right) = S \) and so we conclude that \( t \notin S \) . But now we have \( t \in S \) and \( t \notin S \) . This is a contradiction.\n\n- If \( t \notin S \), then \( t \notin \{ x \in A \mid x \notin f\left( x\right) \} \) . By the definition of \( S \), this means that \( t \in f\left( t\right) \) . However, \( f\left( t\right) = S \) and so we conclude that \( t \in S \) . But now we have \( t \notin S \) and \( t \in S \) . This is a contradiction.\n\nSo in both cases we have arrived at a contradiction. This means that there does not exist a \( t \) in \( A \) such that \( f\left( t\right) = S \) . Therefore, any function from \( A \) to \( \mathcal{P}\left( A\right) \) is not a surjection and hence not a bijection. Hence, \( A \) and \( \mathcal{P}\left( A\right) \) do not have the same cardinality.
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Yes
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Corollary 9.28. \( \mathcal{P}\left( \mathbb{N}\right) \) is an infinite set that is not countably infinite.
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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 \( \mathcal{P}\left( \mathbb{N}\right) \) do not have the same cardinality. Therefore, \( \mathcal{P}\left( \mathbb{N}\right) \) is not countable and hence is an uncountable set.
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Yes
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Proposition 4.11. For all natural numbers \( n \) with \( n \geq 8 \), there exist non-negative integers \( x \) and \( y \) such that \( n = {3x} + {5y} \) .
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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,\
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No
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Theorem 8.12. Let \( a, b \), and \( c \) be integers. If \( a \) and \( b \) are relatively prime and \( a \mid \left( {bc}\right) \), then \( a \mid c \) .
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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 are assuming that \( a \) and \( b \) are relatively prime and hence \( \gcd \left( {a, b}\right) = \) 1. So by Theorem 8.9, there exist integers \( m \) and \( n \) such that\n\n\[ {am} + {bn} = 1.\text{.}\]\n\n(2)\n\nWe now multiply both sides of equation (2) by \( c \) . This gives\n\n\[ \left( {{am} + {bn}}\right) c = 1 \cdot c \]\n\n\[ {acm} + {bcn} = c \]\n\n(3)\n\nWe can now use equation (1) to substitute \( {bc} = {ak} \) in equation (3) and obtain\n\n\[ {acm} + {akn} = c.\]\n\nIf we now factor the left side of this last equation, we see that \( a\left( {{cm} + {kn}}\right) = c \) . Since \( \left( {{cm} + {kn}}\right) \) is an integer, this proves that \( a \) divides \( c \) . Hence, we have proven that if \( a \) and \( b \) are relatively prime and \( a \mid \left( {bc}\right) \), then \( a \mid c \) .
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Yes
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1. Convert \( \alpha \) to the DMS system. Round your answer to the nearest second.
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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 \( {0.26}^{\prime }\left( \frac{{60}^{\prime \prime }}{{1}^{\prime }}\right) = {15.6}^{\prime \prime } \) . Hence,\n\n\[ \n{111.371}^{ \circ } = {111}^{ \circ } + {0.371}^{ \circ } \n\] \n\n\[ \n= {111}^{ \circ } + {22.26}^{\prime } \n\] \n\n\[ \n= {111}^{ \circ } + {22}^{\prime } + {0.26}^{\prime } \n\] \n\n\[ \n= {111}^{ \circ } + {22}^{\prime } + {15.6}^{\prime \prime } \n\] \n\n\[ \n= {111}^{ \circ }{22}^{\prime }{15.6}^{\prime \prime } \n\] \n\nRounding to seconds, we obtain \( \alpha \approx {111}^{ \circ }{22}^{\prime }{16}^{\prime \prime } \) .
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Yes
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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.
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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 \( \theta \) of the form \( \theta = \alpha + {360}^{ \circ } \cdot k \), for some integer \( k \) . When \( k = 1 \), we get \( \theta = {60}^{ \circ } + {360}^{ \circ } = {420}^{ \circ } \) . Substituting \( k = - 1 \) gives \( \theta = {60}^{ \circ } - {360}^{ \circ } = - {300}^{ \circ } \) . Finally, if we let \( k = 2 \), we get \( \theta = {60}^{ \circ } + {720}^{ \circ } = {780}^{ \circ } \) .
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Yes
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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.
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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 \( \theta = \alpha + {2\pi } \cdot k \), for some integer \( k \) . To make the arithmetic a bit easier, we note that \( {2\pi } = \frac{12\pi }{6} \), thus when \( k = 1 \), we get \( \theta = \frac{\pi }{6} + \frac{12\pi }{6} = \frac{13\pi }{6} \) . Substituting \( k = - 1 \) gives \( \theta = \frac{\pi }{6} - \frac{12\pi }{6} = - \frac{11\pi }{6} \) and when we let \( k = 2 \), we get \( \theta = \frac{\pi }{6} + \frac{24\pi }{6} = \frac{25\pi }{6} \) .
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Yes
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Sketch the oriented arc on the Unit Circle corresponding to each of the following real numbers.
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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 midway through Quadrant II.\n\n2. Since one revolution is \( {2\pi } \) radians, and \( t = - {2\pi } \) is negative, we graph the arc which begins at \( \left( {1,0}\right) \) and proceeds clockwise for one full revolution.\n\n3. Like \( t = - {2\pi }, t = - 2 \) is negative, so we begin our arc at \( \left( {1,0}\right) \) and proceed clockwise around the unit circle. Since \( \pi \approx {3.14} \) and \( \frac{\pi }{2} \approx {1.57} \), we find that rotating 2 radians clockwise from the point \( \left( {1,0}\right) \) lands us in Quadrant III. To more accurately place the endpoint, we proceed as we did in Example 10.1.1, successively halving the angle measure until we find \( \frac{5\pi }{8} \approx {1.96} \) which tells us our arc extends just a bit beyond the quarter mark into Quadrant III.\n\n4. Since 117 is positive, the arc corresponding to \( t = {117} \) begins at \( \left( {1,0}\right) \) and proceeds counterclockwise. As 117 is much greater than \( {2\pi } \), we wrap around the Unit Circle several times before finally reaching our endpoint. We approximate \( \frac{117}{2\pi } \) as 18.62 which tells us we complete 18 revolutions counter-clockwise with 0.62, or just shy of \( \frac{5}{8} \) of a revolution to spare. In other words, the terminal side of the angle which measures 117 radians in standard position is just short of being midway through Quadrant III.
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Yes
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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 \( {41.628}^{ \circ } \) north latitude, and it can be shown \( {}^{19} \) that the radius of the earth at this Latitude is approximately 2960 miles. Find the linear velocity, in miles per hour, of Lakeland Community College as the world turns.
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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 again, we are using the fact that radians are real numbers and are dimensionless. (For simplicity's sake, we are also assuming that we are viewing the rotation of the earth as counterclockwise so \( \omega > 0 \) .) Hence, the linear velocity is \[ v = {2960}\text{ miles } \cdot \frac{\pi }{{12}\text{ hours }} \approx {775}\frac{\text{ miles }}{\text{ hour }} \]
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Yes
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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. \( \theta = {60}^{ \circ } \)
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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 counter-clockwise revolution, the terminal side of \( \theta \) lies along the negative \( y \) -axis. Hence, the point we seek is \( \left( {0, - 1}\right) \) so that \( \cos \left( {270}^{ \circ }\right) = 0 \) and \( \sin \left( {270}^{ \circ }\right) = - 1 \) .
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No
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If \( \theta \) is a Quadrant II angle with \( \sin \left( \theta \right) = \frac{3}{5} \), find \( \cos \left( \theta \right) \) .
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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 \( \cos \left( \theta \right) = \pm \frac{4}{5} \) . Since \( \theta \) is a Quadrant II angle, its terminal side, when plotted in standard position, lies in Quadrant II. Since the \( x \) -coordinates are negative in Quadrant II, \( \cos \left( \theta \right) \) is too. Hence, \( \cos \left( \theta \right) = - \frac{4}{5} \) .
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Yes
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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. \( \theta = \frac{7\pi }{3} \)
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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: \( \alpha = \theta - {180}^{ \circ } = {225}^{ \circ } - {180}^{ \circ } = {45}^{ \circ } \) . Since \( \theta \) is a Quadrant III angle, both \( \cos \left( \theta \right) < 0 \) and \( \sin \left( \theta \right) < 0 \) . The Reference Angle Theorem yields: \( \cos \left( {225}^{ \circ }\right) = - \cos \left( {45}^{ \circ }\right) = - \frac{\sqrt{2}}{2} \) and \( \sin \left( {225}^{ \circ }\right) = - \sin \left( {45}^{ \circ }\right) = - \frac{\sqrt{2}}{2}. \)
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No
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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.
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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, \( \sin \left( \alpha \right) \) is positive. Hence, \( \sin \left( \alpha \right) = \frac{12}{13} \). To plot \( \alpha \) in standard position, we begin our rotation on the positive \( x \) -axis to the ray which contains the point \( \left( {\cos \left( \alpha \right) ,\sin \left( \alpha \right) }\right) = \left( {\frac{5}{13},\frac{12}{13}}\right) \).
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Yes
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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. \( \cos \left( \theta \right) = 0 \) .
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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 measure will be the only appropriate angle measure so it is worth the time to become \
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No
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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 \( y = r\sin \left( \theta \right) \) .
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Moreover, \[ \cos \left( \theta \right) = \frac{x}{r} = \frac{x}{\sqrt{{x}^{2} + {y}^{2}}}\text{ and }\sin \left( \theta \right) = \frac{y}{r} = \frac{y}{\sqrt{{x}^{2} + {y}^{2}}} \]
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Yes
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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 \( \cos \left( \theta \right) \) .
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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\sqrt{5}} = - \frac{\sqrt{5}}{5}. \
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Yes
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Suppose we are in the situation of Example 10.1.5. Find the equations of motion of Lakeland Community College as the earth rotates.
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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 \( y = r\sin \left( {\omega t}\right) = {2960}\sin \left( {\frac{\pi }{12}t}\right) \), where \( x \) and \( y \) are measured in miles and \( t \) is measured in hours.
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Yes
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Find the measure of the missing angle and the lengths of the missing sides of:
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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 sides of the triangle. Let \( c \) denote the length of the hypotenuse of the triangle. By Theorem 10.4, we have \( \cos \left( {30}^{ \circ }\right) = \frac{7}{c} \), or \( c = \frac{7}{\cos \left( {30}^{ \circ }\right) } \) . Since \( \cos \left( {30}^{ \circ }\right) = \frac{\sqrt{3}}{2} \), we have, after the usual fraction gymnastics, \( c = \frac{{14}\sqrt{3}}{3} \) . At this point, we have two ways to proceed to find the length of the side opposite the \( {30}^{ \circ } \) angle, which we’ll denote \( b \) . We know the length of the adjacent side is 7 and the length of the hypotenuse is \( \frac{{14}\sqrt{3}}{3} \), so we could use the Pythagorean Theorem to find the missing side and solve \( {\left( 7\right) }^{2} + {b}^{2} = {\left( \frac{{14}\sqrt{3}}{3}\right) }^{2} \) for \( b \) . Alternatively, we could use Theorem 10.4, namely that \( \sin \left( {30}^{ \circ }\right) = \frac{b}{c} \) . Choosing the latter, we find \( b = c\sin \left( {30}^{ \circ }\right) = \frac{{14}\sqrt{3}}{3} \cdot \frac{1}{2} = \frac{7\sqrt{3}}{3} \) . The triangle with all of its data is recorded below.
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Yes
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1. \( \sec \left( {60}^{ \circ }\right) \)
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According to Theorem 10.6, \( \sec \left( {60}^{ \circ }\right) = \frac{1}{\cos \left( {60}^{ \circ }\right) } \) . Hence, \( \sec \left( {60}^{ \circ }\right) = \frac{1}{\left( 1/2\right) } = 2 \) .
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Yes
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Find all angles which satisfy the given equation.\n\n1. \( \sec \left( \theta \right) = 2 \)
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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 \( \theta = \frac{5\pi }{3} + {2\pi k} \) for integers \( k \) .
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Yes
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1. \( \frac{1}{\csc \left( \theta \right) } = \sin \left( \theta \right) \)
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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\[ \frac{1}{\csc \left( \theta \right) } = \frac{1}{\frac{1}{\sin \left( \theta \right) }} = \sin \left( \theta \right) \]\n\nwhich is what we were trying to prove.
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Yes
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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\[ \text{-}\sec \left( \theta \right) = \frac{r}{x} = \frac{\sqrt{{x}^{2} + {y}^{2}}}{x}\text{, provided}x \neq 0\text{.} \]\n\n\[ \text{-}\csc \left( \theta \right) = \frac{r}{y} = \frac{\sqrt{{x}^{2} + {y}^{2}}}{y}\text{, provided}y \neq 0\text{.} \]\n\n\[ \text{-}\tan \left( \theta \right) = \frac{y}{x}\text{, provided}x \neq 0\text{.} \]\n\n\[ \text{-}\cot \left( \theta \right) = \frac{x}{y}\text{, provided}y \neq 0\text{.} \]
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## 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 \( \cos \left( \theta \right) = \frac{3}{5},\sin \left( \theta \right) = - \frac{4}{5},\sec \left( \theta \right) = \frac{5}{3},\csc \left( \theta \right) = - \frac{5}{4},\tan \left( \theta \right) = - \frac{4}{3} \) and \( \cot \left( \theta \right) = - \frac{3}{4} \) .\n\n2. In order to use Theorem 10.9, we need to find a point \( Q\left( {x, y}\right) \) which lies on the terminal side of \( \theta \), when \( \theta \) is plotted in standard position. We have that \( \cot \left( \theta \right) = - 4 = \frac{x}{y} \), and since \( \theta \) is a Quadrant IV angle, we also know \( x > 0 \) and \( y < 0 \) . Viewing \( - 4 = \frac{4}{-1} \), we may choose \( {}^{6}x = 4 \) and \( y = - 1 \) so that \( r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{\left( 4\right) }^{2} + {\left( -1\right) }^{2}} = \sqrt{17} \) . Applying Theorem 10.9 once more, we find \( \cos \left( \theta \right) = \frac{4}{\sqrt{17}} = \frac{4\sqrt{17}}{17},\sin \left( \theta \right) = - \frac{1}{\sqrt{17}} = - \frac{\sqrt{17}}{17},\sec \left( \theta \right) = \frac{\sqrt{17}}{4},\csc \left( \theta \right) = - \sqrt{17} \) and \( \tan \left( \theta \right) = - \frac{1}{4} \) .
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Yes
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Theorem 10.11. Domains and Ranges of the Circular Functions
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- 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 \( \left\lbrack {-1,1}\right\rbrack \)\n\n- The function \( F\left( t\right) = \sec \left( t\right) = \frac{1}{\cos \left( t\right) }\)\n\n- has domain \( \{ t : t \neq \frac{\pi }{2} + {\pi k} \), for integers \( k\} = \mathop{\bigcup }\limits_{{k = - \infty }}^{\infty }\left( {\frac{\left( {{2k} + 1}\right) \pi }{2},\frac{\left( {{2k} + 3}\right) \pi }{2}}\right) \)\n\n- has range \( \{ u : \left| u\right| \geq 1\} = \left( {-\infty , - 1\rbrack \cup \lbrack 1,\infty }\right) \)\n\n- The function \( G\left( t\right) = \csc \left( t\right) = \frac{1}{\sin \left( t\right) }\)\n\n- has domain \( \{ t : t \neq {\pi k} \), for integers \( k\} = \mathop{\bigcup }\limits_{{k = - \infty }}^{\infty }\left( {{k\pi },\left( {k + 1}\right) \pi }\right) \)\n\n- has range \( \{ u : \left| u\right| \geq 1\} = \left( {-\infty , - 1\rbrack \cup \lbrack 1,\infty }\right) \)\n\n- The function \( J\left( t\right) = \tan \left( t\right) = \frac{\sin \left( t\right) }{\cos \left( t\right) }\)\n\n- has domain \( \{ t : t \neq \frac{\pi }{2} + {\pi k} \), for integers \( k\} = \mathop{\bigcup }\limits_{{k = - \infty }}^{\infty }\left( {\frac{\left( {{2k} + 1}\right) \pi }{2},\frac{\left( {{2k} + 3}\right) \pi }{2}}\right) \)\n\n- has range \( \left( {-\infty ,\infty }\right) \)\n\n- The function \( K\left( t\right) = \cot \left( t\right) = \frac{\cos \left( t\right) }{\sin \left( t\right) }\)\n\n- has domain \( \{ t : t \neq {\pi k} \), for integers \( k\} = \mathop{\bigcup }\limits_{{k = - \infty }}^{\infty }\left( {{k\pi },\left( {k + 1}\right) \pi }\right) \)\n\n- has range \( \left( {-\infty ,\infty }\right) \)
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Yes
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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\[ \text{-}\sec \left( {-\theta }\right) = \sec \left( \theta \right) \]\n\n\[ \text{-}\csc \left( {-\theta }\right) = - \csc \left( \theta \right) \]\n\n\[ \text{-}\cot \left( {-\theta }\right) = - \cot \left( \theta \right) \]
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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 \( \cos \left( \theta \right) \) and \( \sin \left( \theta \right) \) so the proofs of their Even / Odd Identities are left as exercises. Consider an angle \( \theta \) plotted in standard position. Let \( {\theta }_{0} \) be the angle coterminal with \( \theta \) with \( 0 \leq {\theta }_{0} < {2\pi } \) . (We can construct the angle \( {\theta }_{0} \) by rotating counter-clockwise from the positive \( x \) -axis to the terminal side of \( \theta \) as pictured below.) Since \( \theta \) and \( {\theta }_{0} \) are coterminal, \( \cos \left( \theta \right) = \cos \left( {\theta }_{0}\right) \) and \( \sin \left( \theta \right) = \sin \left( {\theta }_{0}\right) \) .\n\n\n\nWe now consider the angles \( - \theta \) and \( - {\theta }_{0} \) . Since \( \theta \) is coterminal with \( {\theta }_{0} \), there is some integer \( k \) so that \( \theta = {\theta }_{0} + {2\pi } \cdot k \) . Therefore, \( - \theta = - {\theta }_{0} - {2\pi } \cdot k = - {\theta }_{0} + {2\pi } \cdot \left( {-k}\right) \) . Since \( k \) is an integer, so is \( \left( {-k}\right) \), which means \( - \theta \) is coterminal with \( - {\theta }_{0} \) . Hence, \( \cos \left( {-\theta }\right) = \cos \left( {-{\theta }_{0}}\right) \) and \( \sin \left( {-\theta }\right) = \sin \left( {-{\theta }_{0}}\right) \) . Let \( P \) and \( Q \) denote the points on the terminal sides of \( {\theta }_{0} \) and \( - {\theta }_{0} \), respectively, which lie on the Unit Circle. By definition, the coordinates of \( P \) are \( \left( {\cos \left( {\theta }_{0}\right) ,\sin \left( {\theta }_{0}\right) }\right) \) and the coordinates of \( Q \) are \( \left( {\cos \left( {-{\theta }_{0}}\right) ,\sin \left( {-{\theta }_{0}}\right) }\right) \) . Since \( {\theta }_{0} \) and \( - {\theta }_{0} \) sweep out congruent central sectors of the Unit Circle, it follows that the points \( P \) and \( Q \) are symmetric about the \( x \) -axis. Thus, \( \cos \left( {-{\theta }_{0}}\right) = \cos \left( {\theta }_{0}\right) \) and \( \sin \left( {-{\theta }_{0}}\right) = - \sin \left( {\theta }_{0}\right) \) . Since the cosines and sines of \( {\theta }_{0} \) and \( - {\theta }_{0} \) are the same as those for \( \theta \) and \( - \theta \), respectively, we get \( \cos \left( {-\theta }\right) = \cos \left( \theta \right) \) and \( \sin \left( {-\theta }\right) = - \sin \left( \theta \right) \), as required.
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No
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Find the exact value of \( \cos \left( {15}^{ \circ }\right) \) .
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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\[ \cos \left( {15}^{ \circ }\right) = \cos \left( {{45}^{ \circ } - {30}^{ \circ }}\right) \]\n\n\[ = \cos \left( {45}^{ \circ }\right) \cos \left( {30}^{ \circ }\right) + \sin \left( {45}^{ \circ }\right) \sin \left( {30}^{ \circ }\right) \]\n\n\[ = \left( \frac{\sqrt{2}}{2}\right) \left( \frac{\sqrt{3}}{2}\right) + \left( \frac{\sqrt{2}}{2}\right) \left( \frac{1}{2}\right) \]\n\n\[ = \frac{\sqrt{6} + \sqrt{2}}{4} \]
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Yes
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Rewrite \( {\sin }^{2}\left( \theta \right) {\cos }^{2}\left( \theta \right) \) as a sum and difference of cosines to the first power.
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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\[ \n= \frac{1}{4}\left( {1 - {\cos }^{2}\left( {2\theta }\right) }\right) \n\]\n\n\[ \n= \frac{1}{4} - \frac{1}{4}{\cos }^{2}\left( {2\theta }\right) \n\]\n\nNext, we apply the power reduction formula to \( {\cos }^{2}\left( {2\theta }\right) \) to finish the reduction\n\n\[ \n{\sin }^{2}\left( \theta \right) {\cos }^{2}\left( \theta \right) = \frac{1}{4} - \frac{1}{4}{\cos }^{2}\left( {2\theta }\right) \n\]\n\n\[ \n= \frac{1}{4} - \frac{1}{4}\left( \frac{1 + \cos \left( {2\left( {2\theta }\right) }\right) }{2}\right) \n\]\n\n\[ \n= \frac{1}{4} - \frac{1}{8} - \frac{1}{8}\cos \left( {4\theta }\right) \n\]\n\n\[ \n= \frac{1}{8} - \frac{1}{8}\cos \left( {4\theta }\right) \n\]
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Yes
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1. Write \( \cos \left( {2\theta }\right) \cos \left( {6\theta }\right) \) as a sum.
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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\[ = \;\frac{1}{2}\cos \left( {-{4\theta }}\right) + \frac{1}{2}\cos \left( {8\theta }\right) \]\n\n\[ = \frac{1}{2}\cos \left( {4\theta }\right) + \frac{1}{2}\cos \left( {8\theta }\right) \]\n\nwhere the last equality is courtesy of the even identity for cosine, \( \cos \left( {-{4\theta }}\right) = \cos \left( {4\theta }\right) \).
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Yes
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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. \( g\left( x\right) = \frac{1}{2}\sin \left( {\pi - {2x}}\right) + \frac{3}{2} \)
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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><td>0</td><td>\( \frac{{\pi x} - \pi }{2} = 0 \)</td><td>1</td></tr><tr><td>\( \frac{\pi }{2} \)</td><td>\( \frac{{\pi x} - \pi }{2} = \frac{\pi }{2} \)</td><td>2</td></tr><tr><td>π</td><td>\( \frac{{\pi x} - \pi }{2} = \pi \)</td><td>3</td></tr><tr><td>\( \frac{3\pi }{2} \)</td><td>\( \frac{{\pi x} - \pi }{2} = \frac{3\pi }{2} \)</td><td>4</td></tr><tr><td>\( {2\pi } \)</td><td>\( \frac{{\pi x} - \pi }{2} = {2\pi } \)</td><td>5</td></tr></table>\n\nNext, we substitute each of these \( x \) values into \( f\left( x\right) = 3\cos \left( \frac{{\pi x} - \pi }{2}\right) + 1 \) to determine the corresponding \( y \) -values and connect the dots in a pleasing wavelike fashion.\n\n<table><thead><tr><th>\( x \)</th><th>\( f\left( x\right) \)</th><th>\( \left( {x, f\left( x\right) }\right) \)</th></tr></thead><tr><td>1</td><td>4</td><td>\( \left( {1,4}\right) \)</td></tr><tr><td>2</td><td>1</td><td>\( \left( {2,1}\right) \)</td></tr><tr><td>3</td><td>\( - 2 \)</td><td>\( \left( {3, - 2}\right) \)</td></tr><tr><td>4</td><td>1</td><td>\( \left( {4,1}\right) \)</td></tr><tr><td>5</td><td>4</td><td>\( \left( {5,4}\right) \)</td></tr></table>\n\n\n\nOne cycle is graphed on \( \left\lbrack {1,5}\right\rbrack \) so the period is the length of that interval which is 4 .
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No
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1. Find a cosine function whose graph matches the graph of \( y = f\left( x\right) \) .
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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 \( \omega = \frac{\pi }{3} \) . Next, we see that the phase shift is -1, so we have \( - \frac{\phi }{\omega } = - 1 \), or \( \phi = \omega = \frac{\pi }{3} \) . To find the amplitude, note that the range of the sinusoid is \( \left\lbrack {-\frac{3}{2},\frac{5}{2}}\right\rbrack \) . As a result, the amplitude \( A = \frac{1}{2}\left\lbrack {\frac{5}{2} - \left( {-\frac{3}{2}}\right) }\right\rbrack = \frac{1}{2}\left( 4\right) = 2 \) . Finally, to determine the vertical shift, we average the endpoints of the range to find \( B = \frac{1}{2}\left\lbrack {\frac{5}{2} + \left( {-\frac{3}{2}}\right) }\right\rbrack = \frac{1}{2}\left( 1\right) = \frac{1}{2} \) . Our final answer is \( C\left( x\right) = 2\cos \left( {\frac{\pi }{3}x + \frac{\pi }{3}}\right) + \frac{1}{2}. \)
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Yes
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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 \( \omega > 0 \)
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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\[ \cos \left( {2x}\right) - \sqrt{3}\sin \left( {2x}\right) = A\cos \left( {\omega x}\right) \cos \left( \phi \right) - A\sin \left( {\omega x}\right) \sin \left( \phi \right) + B \]\n\nIt should be clear that we can take \( \omega = 2 \) and \( B = 0 \) to get\n\n\[ \cos \left( {2x}\right) - \sqrt{3}\sin \left( {2x}\right) = A\cos \left( {2x}\right) \cos \left( \phi \right) - A\sin \left( {2x}\right) \sin \left( \phi \right) \]\n\nTo determine \( A \) and \( \phi \), a bit more work is involved. We get started by equating the coefficients of the trigonometric functions on either side of the equation. On the left hand side, the coefficient of \( \cos \left( {2x}\right) \) is 1, while on the right hand side, it is \( A\cos \left( \phi \right) \). Since this equation is to hold for all real numbers, we must have \( {}^{8} \) that \( A\cos \left( \phi \right) = 1 \). Similarly, we find by equating the coefficients of \( \sin \left( {2x}\right) \) that \( A\sin \left( \phi \right) = \sqrt{3} \). What we have here is a system of nonlinear equations! We can temporarily eliminate the dependence on \( \phi \) by using the Pythagorean Identity. We know \( {\cos }^{2}\left( \phi \right) + {\sin }^{2}\left( \phi \right) = 1 \), so multiplying this by \( {A}^{2} \) gives \( {A}^{2}{\cos }^{2}\left( \phi \right) + {A}^{2}{\sin }^{2}\left( \phi \right) = {A}^{2} \). Since \( A\cos \left( \phi \right) = 1 \) and \( A\sin \left( \phi \right) = \sqrt{3} \), we get \( {A}^{2} = {1}^{2} + {\left( \sqrt{3}\right) }^{2} = 4 \) or \( A = \pm 2 \). Choosing \( A = 2 \), we have \( 2\cos \left( \phi \right) = 1 \) and \( 2\sin \left( \phi \right) = \sqrt{3} \) or, after some rearrangement, \( \cos \left( \phi \right) = \frac{1}{2} \) and \( \sin \left( \phi \right) = \frac{\sqrt{3}}{2} \). One such angle \( \phi \) which satisfies this criteria is \( \phi = \frac{\pi }{3} \). Hence, one way to write \( f\left( x\right) \) as a sinusoid is \( f\left( x\right) = 2\cos \left( {{2x} + \frac{\pi }{3}}\right) \). We can easily check our answer using the sum formula for cosine\n\n\[ f\left( x\right) = 2\cos \left( {{2x} + \frac{\pi }{3}}\right) \]\n\n\[ = \;2\;\left\lbrack {\cos \left( {2x}\right) \cos \left( \frac{\pi }{3}\right) - \sin \left( {2x}\right) \sin \left( \frac{\pi }{3}\right) }\right\rbrack \]\n\n\[ = \;2\left\lbrack {\cos \left( {2x}\right) \left( \frac{1}{2}\right) - \sin \left( {2x}\right) \left( \frac{\sqrt{3}}{2}\right) }\right\rbrack \]\n\n\[ = \cos \left( {2x}\right) - \sqrt{3}\sin \left( {2x}\right) \]\n\n\( {}^{8} \) This should remind you of equation coefficients of like powers of \( x \) in Section 8.6.
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Yes
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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. \( g\left( x\right) = \frac{\csc \left( {\pi - {\pi x}}\right) - 5}{3} \)
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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>\\( {2x} = a \\)</td><td>\\( x \\)</td></tr><tr><td>0</td><td>\\( {2x} = 0 \\)</td><td>0</td></tr><tr><td>\\( \\frac{\\pi }{2} \\)</td><td>\\( {2x} = \\frac{\\pi }{2} \\)</td><td>\\( \\frac{\\pi }{4} \\)</td></tr><tr><td>\\( \\pi \\)</td><td>\\( {2x} = \\pi \\)</td><td>\\( \\frac{\\pi }{2} \\)</td></tr><tr><td>\\( \\frac{3\\pi }{2} \\)</td><td>\\( {2x} = \\frac{3\\pi }{2} \\)</td><td>\\( \\frac{3\\pi }{4} \\)</td></tr><tr><td>\\( {2\\pi } \\)</td><td>\\( {2x} = {2\\pi } \\)</td><td>\\( \\pi \\)</td></tr></table>\n\nNext, we substitute these \( x \) values into \( f\left( x\right) \) . If \( f\left( x\right) \) exists, we have a point on the graph; otherwise, we have found a vertical asymptote. In addition to these points and asymptotes, we have graphed the associated cosine curve - in this case \( y = 1 - 2\cos \left( {2x}\right) \) - dotted in the picture below. Since one cycle is graphed over the interval \( \\left\\lbrack {0,\\pi }\\right\\rbrack \), the period is \( \\pi - 0 = \\pi \) .\n\n<table><thead><tr><th>\\( x \\)</th><th>\\( f\\left( x\\right) \\)</th><th>\\( \\left( {x, f\\left( x\\right) }\\right) \\)</th></tr></thead><tr><td>0</td><td>\\( - 1 \\)</td><td>\\( \\left( {0, - 1}\\right) \\)</td></tr><tr><td>\\( \\frac{\\pi }{4} \\)</td><td>undefined</td><td></td></tr><tr><td>\\( \\frac{\\pi }{2} \\)</td><td>3</td><td>\\( \\left( {\\frac{\\pi }{2},3}\\right) \\)</td></tr><tr><td>\\( \\frac{3\\pi }{4} \\)</td><td>undefined</td><td></td></tr><tr><td>\\( \\pi \\)</td><td>\\( - 1 \\)</td><td>\\( \\left( {\\pi , - 1}\\right) \\)</td></tr></table>\n\n\n\nOne cycle of \( y = 1 - 2\\sec \\left( {2x}\\right) \) .
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Yes
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Graph one cycle of the following functions. Find the period.\n\n1. \( f\left( x\right) = 1 - \tan \left( \frac{x}{2}\right) \)
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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 solving for \( x \) .\n\n<table><tr><td>\( a \)</td><td>\( \frac{x}{2} = a \)</td><td>\( x \)</td></tr><tr><td>\( - \frac{\pi }{2} \)</td><td>\( \frac{x}{2} = - \frac{\pi }{2} \)</td><td>\( - \pi \)</td></tr><tr><td>\( - \frac{\pi }{4} \)</td><td>\( \frac{x}{2} = - \frac{\pi }{4} \)</td><td>\( - \frac{\pi }{2} \)</td></tr><tr><td>0</td><td>\( \frac{x}{2} = 0 \)</td><td>0</td></tr><tr><td>\( \frac{\pi }{4} \)</td><td>\( \frac{x}{2} = \frac{\pi }{4} \)</td><td>\( \frac{\pi }{2} \)</td></tr><tr><td>\( \frac{\pi }{2} \)</td><td>\( \frac{x}{2} = \frac{\pi }{2} \)</td><td>\( \pi \)</td></tr></table>\n\nSubstituting these \( x \) -values into \( f\left( x\right) \), we find points on the graph and the vertical asymptotes.\n\n<table><thead><tr><th>\( x \)</th><th>\( f\left( x\right) \)</th><th>\( \left( {x, f\left( x\right) }\right) \)</th></tr></thead><tr><td>\( - \pi \)</td><td>undefined</td><td></td></tr><tr><td>\( - \frac{\pi }{2} \)</td><td>2</td><td>\( \left( {-\frac{\pi }{2},2}\right) \)</td></tr><tr><td>0</td><td>1</td><td>\( \left( {0,1}\right) \)</td></tr><tr><td>\( \frac{\pi }{2} \)</td><td>0</td><td>\( \left( {\frac{\pi }{2},0}\right) \)</td></tr><tr><td>π</td><td>undefined</td><td></td></tr></table>\n\n\n\nWe see that the period is \( \pi - \left( {-\pi }\right) = {2\pi } \).
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Yes
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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 the nearest hundredth of a degree.
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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 \( \tan \left( \theta \right) = \frac{6}{12} = \frac{1}{2} \) . Since \( \theta \) is an acute angle, we can use the arctangent function and we find \( \theta = \arctan \left( \frac{1}{2}\right) \) radians \( \approx {26.56}^{ \circ } \) .
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Yes
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Find all angles \( \theta \) for which \( \sin \left( \theta \right) = \frac{1}{3} \) .
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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 the equation, then all the solutions to this equation in Quadrant I are coterminal with \( \alpha \), and \( \alpha \) serves as the reference angle for all of the solutions to this equation in Quadrant II.\n\nSince \( \frac{1}{3} \) isn’t the sine of any of the ’common angles’ discussed earlier, we use the arcsine functions to express our answers. The real number \( t = \arcsin \left( \frac{1}{3}\right) \) is defined so it satisfies \( 0 < t < \frac{\pi }{2} \) with \( \sin \left( t\right) = \frac{1}{3} \) . Hence, \( \alpha = \arcsin \left( \frac{1}{3}\right) \) radians. Since the solutions in Quadrant I are all coterminal with \( \alpha \), we get part of our solution to be \( \theta = \alpha + {2\pi k} = \arcsin \left( \frac{1}{3}\right) + {2\pi k} \) for integers \( k \) . Turning our attention to Quadrant II, we get one solution to be \( \pi - \alpha \) . Hence, the Quadrant II solutions are \( \theta = \pi - \alpha + {2\pi k} = \pi - \arcsin \left( \frac{1}{3}\right) + {2\pi k} \), for integers \( k \) .
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Yes
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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. \( \cos \left( {2x}\right) = - \frac{\sqrt{3}}{2} \)
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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 integers \( k \) . Solving for \( x \) gives \( x = \frac{5\pi }{12} + {\pi k} \) or \( x = \frac{7\pi }{12} + {\pi k} \) for integers \( k \) . To check these answers analytically, we substitute them into the original equation. For any integer \( k \) we have\n\n\[ \cos \left( {2\left\lbrack {\frac{5\pi }{12} + {\pi k}}\right\rbrack }\right) = \cos \left( {\frac{5\pi }{6} + {2\pi k}}\right) \]\n\n\[ = \cos \left( \frac{5\pi }{6}\right) \]\n(the period of cosine is \( {2\pi } \) )\n\n\[ = - \frac{\sqrt{3}}{2} \]\n\nSimilarly, we find \( \cos \left( {2\left\lbrack {\frac{7\pi }{12} + {\pi k}}\right\rbrack }\right) = \cos \left( {\frac{7\pi }{6} + {2\pi k}}\right) = \cos \left( \frac{7\pi }{6}\right) = - \frac{\sqrt{3}}{2} \) . To determine which of our solutions lie in \( \lbrack 0,{2\pi }) \), we substitute integer values for \( k \) . The solutions we keep come from the values of \( k = 0 \) and \( k = 1 \) and are \( x = \frac{5\pi }{12},\frac{7\pi }{12},\frac{17\pi }{12} \) and \( \frac{19\pi }{12} \) . Using a calculator, we graph \( y = \cos \left( {2x}\right) \) and \( y = - \frac{\sqrt{3}}{2} \) over \( \lbrack 0,{2\pi }) \) and examine where these two graphs intersect. We see that the \( x \) -coordinates of the intersection points correspond to the decimal representations of our exact answers.
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Yes
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1. \( 3{\sin }^{3}\left( x\right) = {\sin }^{2}\left( x\right) \)
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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\n\[ 3{\sin }^{3}\left( x\right) - {\sin }^{2}\left( x\right) = 0 \]\n\n\[ {\sin }^{2}\left( x\right) \left( {3\sin \left( x\right) - 1}\right) = 0\;\text{Factor out}{\sin }^{2}\left( x\right) \text{from both terms.} \]\n\nWe get \( {\sin }^{2}\left( x\right) = 0 \) or \( 3\sin \left( x\right) - 1 = 0 \) . Solving for \( \sin \left( x\right) \), we find \( \sin \left( x\right) = 0 \) or \( \sin \left( x\right) = \frac{1}{3} \) . The solution to the first equation is \( x = {\pi k} \), with \( x = 0 \) and \( x = \pi \) being the two solutions which lie in \( \lbrack 0,{2\pi }) \) . To solve \( \sin \left( x\right) = \frac{1}{3} \), we use the arcsine function to get \( x = \arcsin \left( \frac{1}{3}\right) + {2\pi k} \) or \( x = \pi - \arcsin \left( \frac{1}{3}\right) + {2\pi k} \) for integers \( k \) . We find the two solutions here which lie in \( \lbrack 0,{2\pi }) \) to be \( x = \arcsin \left( \frac{1}{3}\right) \) and \( x = \pi - \arcsin \left( \frac{1}{3}\right) \) . To check graphically, we plot \( y = 3{\left( \sin \left( x\right) \right) }^{3} \) and \( y = {\left( \sin \left( x\right) \right) }^{2} \) and find the \( x \) -coordinates of the intersection points of these two curves. Some extra zooming is required near \( x = 0 \) and \( x = \pi \) to verify that these two curves do in fact intersect four times. \( {}^{6} \)
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Yes
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1. \( 2\sin \left( x\right) \leq 1 \)
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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 \( f\left( x\right) \leq 0 \) . We now look to see where, if ever, \( f \) is undefined and where \( f\left( x\right) = 0 \) . Since the domain of \( f \) is all real numbers, we can immediately set about finding the zeros of \( f \) . Solving \( f\left( x\right) = 0 \), we have \( 2\sin \left( x\right) - 1 = 0 \) or \( \sin \left( x\right) = \frac{1}{2} \) . The solutions here are \( x = \frac{\pi }{6} + {2\pi k} \) and \( x = \frac{5\pi }{6} + {2\pi k} \) for integers \( k \) . Since we are restricting our attention to \( \lbrack 0,{2\pi }) \), only \( x = \frac{\pi }{6} \) and \( x = \frac{5\pi }{6} \) are of concern to us. Next, we choose test values in \( \lbrack 0,{2\pi }) \) other than the zeros and determine if \( f \) is positive or negative there. For \( x = 0 \) we have \( f\left( 0\right) = - 1 \), for \( x = \frac{\pi }{2} \) we get \( f\left( \frac{\pi }{2}\right) = 1 \) and for \( x = \pi \) we get \( f\left( \pi \right) = - 1 \) . Since our original inequality is equivalent to \( f\left( x\right) \leq 0 \), we are looking for where the function is negative \( \left( -\right) \) or 0, and we get the intervals \( \left\lbrack {0,\frac{\pi }{6}}\right\rbrack \cup \left\lbrack {\frac{5\pi }{6},{2\pi }}\right) \) . We can confirm our answer graphically by seeing where the graph of \( y = 2\sin \left( x\right) \) crosses or is below the graph of \( y = 1 \) .
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Yes
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Express the domain of the following functions using extended interval notation. 1. \( f\left( x\right) = \csc \left( {{2x} + \frac{\pi }{3}}\right) \)
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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 \( \sin \left( {{2x} + \frac{\pi }{3}}\right) = 0 \), we get \( x = - \frac{\pi }{6} + \frac{\pi }{2}k \) for integers \( k \) . In set-builder notation, our domain is \( \left\{ {x : x \neq - \frac{\pi }{6} + \frac{\pi }{2}k\text{for integers}k}\right\} \) . To help visualize the domain, we follow the old mantra ’When in doubt, write it out!’ We get \( \left\{ {x : x \neq - \frac{\pi }{6},\frac{2\pi }{6}, - \frac{4\pi }{6},\frac{5\pi }{6}, - \frac{7\pi }{6},\frac{8\pi }{6},\ldots }\right\} \) , where we have kept the denominators 6 throughout to help see the pattern. Graphing the situation on a numberline, we have\n\n\n\nProceeding as we did on page 756 in Section 10.3.1, we let \( {x}_{k} \) denote the \( k \) th number excluded from the domain and we have \( {x}_{k} = - \frac{\pi }{6} + \frac{\pi }{2}k = \frac{\left( {{3k} - 1}\right) \pi }{6} \) for integers \( k \) . The intervals which comprise the domain are of the form \( \left( {{x}_{k},{x}_{k + 1}}\right) = \left( {\frac{\left( {{3k} - 1}\right) \pi }{6},\frac{\left( {{3k} + 2}\right) \pi }{6}}\right) \) as \( k \) runs through the integers. Using extended interval notation, we have that the domain is\n\n\[ \mathop{\bigcup }\limits_{{k = - \infty }}^{\infty }\left( {\frac{\left( {{3k} - 1}\right) \pi }{6},\frac{\left( {{3k} + 2}\right) \pi }{6}}\right) \]\n\nWe can check our answer by substituting in values of \( k \) to see that it matches our diagram.
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Yes
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1. \( \arcsin \left( {2x}\right) = \frac{\pi }{3} \)
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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\[ \sin \left( {\arcsin \left( {2x}\right) }\right) = \sin \left( \frac{\pi }{3}\right) \]\n\n\[ {2x} = \frac{\sqrt{3}}{2}\;\text{Since}\sin \left( {\arcsin \left( u\right) }\right) = u \]\n\n\[ x = \frac{\sqrt{3}}{4} \]\n\nGraphing \( y = \arcsin \left( {2x}\right) \) and the horizontal line \( y = \frac{\pi }{3} \), we see they intersect at \( \frac{\sqrt{3}}{4} \approx {0.4430} \).
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Yes
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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 describes the height of the passengers above the ground \( t \) seconds after they pass the point on the wheel closest to the ground.
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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 from the equations given on page 732 in Section 10.2.1 that the \( y \) -coordinate for counterclockwise motion on a circle of radius \( r \) centered at the origin with constant angular velocity (frequency) \( \omega \) is given by \( y = r\sin \left( {\omega t}\right) \) . Here, \( t = 0 \) corresponds to the point \( \left( {r,0}\right) \) so that \( \theta \), the angle measuring the amount of rotation, is in standard position. In our case, the diameter of the wheel is 128 feet, so the radius is \( r = {64} \) feet. Since the wheel completes two revolutions in 2 minutes and 7 seconds (which is 127 seconds) the period \( T = \frac{1}{2}\left( {127}\right) = \frac{127}{2} \) seconds. Hence, the angular frequency is \( \omega = \frac{2\pi }{T} = \frac{4\pi }{127} \) radians per second. Putting these two pieces of information together, we have that \( y = {64}\sin \left( {\frac{4\pi }{127}t}\right) \) describes the \( y \) -coordinate on the Giant Wheel after \( t \) seconds, assuming it is centered at \( \left( {0,0}\right) \) with \( t = 0 \) corresponding to the point \( Q \) . In order to find an expression for \( h \), we take the point \( O \) in the figure as the origin. Since the base of the Giant Wheel ride is 8 feet above the ground and the Giant Wheel itself has a radius of 64 feet, its center is 72 feet above the ground. To account for this vertical shift upward, \( {}^{4} \) we add 72 to our formula for \( y \) to obtain the new formula \( h = y + {72} = {64}\sin \left( {\frac{4\pi }{127}t}\right) + {72} \) . Next, we need to adjust things so that \( t = 0 \) corresponds to the point \( P \) instead of the point \( Q \) . This is where the phase comes into play. Geometrically, we need to shift the angle \( \theta \) in the figure back \( \frac{\pi }{2} \) radians. From Section 10.2.1, we know \( \theta = {\omega t} = \frac{4\pi }{127}t \), so we (temporarily) write the height in terms of \( \theta \) as \( h = {64}\sin \left( \theta \right) + {72} \) . Subtracting \( \frac{\pi }{2} \) from \( \theta \) gives the final answer \( h\left( t\right) = {64}\sin \left( {\theta - \frac{\pi }{2}}\right) + {72} = {64}\sin \left( {\frac{4\pi }{127}t - \frac{\pi }{2}}\right) + {72} \) . We can check the reasonableness of our answer by graphing \( y = h\left( t\right) \) over the interval \( \left\lbrack {0,\frac{127}{2}}\right\rbrack \) .\n\n
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Yes
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1. Find a sinusoid which models these data and use a graphing utility to graph your answer along with the data.
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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 find the constants \( A,\omega ,\phi \) and \( B \) so that the function \( H\left( t\right) = A\sin \left( {{\omega t} + \phi }\right) + B \) closely matches the data. We first go after the vertical shift \( B \) whose value determines the baseline. In a typical sinusoid, the value of \( B \) is the average of the maximum and minimum values. So here we take \( B = \frac{{3.3} + {21.8}}{2} = {12.55} \) . Next is the amplitude \( A \) which is the displacement from the baseline to the maximum (and minimum) values. We find \( A = {21.8} - {12.55} = {12.55} - {3.3} = {9.25} \) . At this point, we have \( H\left( t\right) = {9.25}\sin \left( {{\omega t} + \phi }\right) + {12.55} \) . Next, we go after the angular frequency \( \omega \) . Since the data collected is over the span of a year (12 months), we take the period \( T = {12} \) months. \( {}^{6} \) This\n\n---\n\n\( {}^{5} \) Okay, it appears to be the ’ \( \land \) ’ shape we saw in some of the graphs in Section 2.2. Just humor us.\n\n\( {}^{6} \) Even though the data collected lies in the interval \( \left\lbrack {1,{12}}\right\rbrack \), which has a length of 11, we need to think of the data point at \( t = 1 \) as a representative sample of the amount of daylight for every day in January. That is, it represents \( H\left( t\right) \) over the interval \( \left\lbrack {0,1}\right\rbrack \) . Similarly, \( t = 2 \) is a sample of \( H\left( t\right) \) over \( \left\lbrack {1,2}\right\rbrack \), and so forth.\n\n---\n\nmeans \( \omega = \frac{2\pi }{T} = \frac{2\pi }{12} = \frac{\pi }{6} \) . The last quantity to find is the phase \( \phi \) . Unlike the previous example, it is easier in this case to find the phase shift \( - \frac{\phi }{\omega } \) . Since we picked \( A > 0 \), the phase shift corresponds to the first value of \( t \) with \( H\left( t\right) = {12.55} \) (the baseline value). \( {}^{7} \) Here, we choose \( t = 3 \), since its corresponding \( H \) value of 12.4 is closer to 12.55 than the next value, 15.9, which corresponds to \( t = 4 \) . Hence, \( - \frac{\phi }{\omega } = 3 \), so \( \phi = - {3\omega } = - 3\left( \frac{\pi }{6}\right) = - \frac{\pi }{2} \) . We have \( H\left( t\right) = {9.25}\sin \left( {\frac{\pi }{6}t - \frac{\pi }{2}}\right) + {12.55} \) . Below is a graph of our data with the curve \( y = H\left( t\right) \) .\n\n
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Yes
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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 equilibrium position at time \( t \) is given by \( x\left( t\right) = A\sin \left( {{\omega t} + \phi }\right) \) where
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\[ \text{-}\omega = \sqrt{\frac{k}{m}}\text{and}A = \sqrt{{x}_{0}^{2} + {\left( \frac{{v}_{0}}{\omega }\right) }^{2}} \] \[ \text{-}A\sin \left( \phi \right) = {x}_{0}\text{and}{A\omega }\cos \left( \phi \right) = {v}_{0}\text{.} \]
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Yes
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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?
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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 \( {64} = k \cdot 8 \), or \( k = 8\frac{\text{ lbs }}{\text{ ft }} \) . To find \( m \), we use \( w = {mg} \) with \( w = {64} \) lbs. and \( g = {32}\frac{\text{ ft }}{{s}^{2}} \) . We get \( m = 2 \) slugs. We can now proceed to apply Theorem 11.1.\n\n1. With \( k = 8 \) and \( m = 2 \), we get \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{8}{2}} = 2 \) . We are told that the object is released 3 feet below the equilibrium position ’from rest.’ This means \( {x}_{0} = 3 \) and \( {v}_{0} = 0 \) . Therefore, \( A = \sqrt{{x}_{0}^{2} + {\left( \frac{{v}_{0}}{\omega }\right) }^{2}} = \sqrt{{3}^{2} + {0}^{2}} = 3 \) . To determine the phase \( \phi \), we have \( A\sin \left( \phi \right) = {x}_{0} \) , which in this case gives \( 3\sin \left( \phi \right) = 3 \) so \( \sin \left( \phi \right) = 1 \) . Only \( \phi = \frac{\pi }{2} \) and angles coterminal to it\n\n\( {}^{13} \) The sign conventions here are carried over from Physics. If not for the spring, the object would fall towards the ground, which is the 'natural' or 'positive' direction. Since the spring force acts in direct opposition to gravity, any movement upwards is considered 'negative'. satisfy this condition, so we pick \( {}^{14} \) the phase to be \( \phi = \frac{\pi }{2} \) . Hence, the equation of motion is \( x\left( t\right) = 3\sin \left( {{2t} + \frac{\pi }{2}}\right) \) . To find when the object passes through the equilibrium position we solve \( x\left( t\right) = 3\sin \left( {{2t} + \frac{\pi }{2}}\right) = 0 \) . Going through the usual analysis we find \( t = - \frac{\pi }{4} + \frac{\pi }{2}k \) for integers \( k \) . Since we are interested in the first time the object passes through the equilibrium position, we look for the smallest positive \( t \) value which in this case is \( t = \frac{\pi }{4} \approx {0.78} \) seconds after the start of the motion. Common sense suggests that if we release the object below the equilibrium position, the object should be traveling upwards when it first passes through it. To check this answer, we graph one cycle of \( x\left( t\right) \) . Since our applied domain in this situation is \( t \geq 0 \), and the period of \( x\left( t\right) \) is \( T = \frac{2\pi }{\omega } = \frac{2\pi }{2} = \pi \), we graph \( x\left( t\right) \) over the interval \( \left\lbrack {0,\pi }\right\rbrack \) . Remembering that \( x\left( t\right) > 0 \) means the object is below the equilibrium position and \( x\left( t\right) < 0 \) means the object is above the equilibrium position, the fact our graph is crossing through the \( t \) -axis from positive \( x \) to negative \( x \) at \( t = \frac{\pi }{4} \) confirms our answer.
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Yes
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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.
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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 \( a = \sqrt{33} \) units. Now that all three sides of the triangle are known, there are several ways we can find \( \alpha \) using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to \( \alpha \) . According to Theorem 10.4, \( \cos \left( \alpha \right) = \frac{4}{7} \) . Since \( \alpha \) is an acute angle, \( \alpha = \arccos \left( \frac{4}{7}\right) \) radians. Converting to degrees, we find \( \alpha \approx {55.15}^{ \circ } \) . Now that we have the measure of angle \( \alpha \), we could find the measure of angle \( \beta \) using the fact that \( \alpha \) and \( \beta \) are complements so \( \alpha + \beta = {90}^{ \circ } \) . Once again, we opt to use the data given to us in the problem. According to Theorem 10.4, we have that \( \sin \left( \beta \right) = \frac{4}{7} \) so \( \beta = \arcsin \left( \frac{4}{7}\right) \) radians and we have \( \beta \approx {34.85}^{ \circ } \) .
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Yes
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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\[ \frac{\sin \left( \alpha \right) }{a} = \frac{\sin \left( \beta \right) }{b} = \frac{\sin \left( \gamma \right) }{c} \]\n\nor, equivalently,\n\n\[ \frac{a}{\sin \left( \alpha \right) } = \frac{b}{\sin \left( \beta \right) } = \frac{c}{\sin \left( \gamma \right) } \]
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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 from vertex \( B \), we divide the triangle into two right triangles: \( \bigtriangleup {ABQ} \) and \( \bigtriangleup {BCQ} \) . If we call the length of the altitude \( h \) (for height), we get from Theorem 10.4 that \( \sin \left( \alpha \right) = \frac{h}{c} \) and \( \sin \left( \gamma \right) = \frac{h}{a} \) so that \( h = c\sin \left( \alpha \right) = a\sin \left( \gamma \right) \) . After some rearrangement of the last equation, we get \( \frac{\sin \left( \alpha \right) }{a} = \frac{\sin \left( \gamma \right) }{c} \) . If we drop an altitude from vertex \( A \), we can proceed as above using the triangles \( \bigtriangleup {ABQ} \) and \( \bigtriangleup {ACQ} \) to get \( \frac{\sin \left( \beta \right) }{b} = \frac{\sin \left( \gamma \right) }{c} \), completing the proof for this case.
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Yes
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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, \( \beta = {45}^{ \circ } \)
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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 \( b = \frac{7\sin \left( {45}^{ \circ }\right) }{\sin \left( {120}^{ \circ }\right) } = \frac{7\sqrt{6}}{3} \approx {5.72} \) units. Now that we have two angle-side pairs, it is time to find the third. To find \( \gamma \), we use the fact that the sum of the measures of the angles in a triangle is \( {180}^{ \circ } \) . Hence, \( \gamma = {180}^{ \circ } - {120}^{ \circ } - {45}^{ \circ } = {15}^{ \circ } \) . To find \( c \), we have no choice but to used the derived value \( \gamma = {15}^{ \circ } \), yet we can minimize the propagation of error here by using the given angle-side opposite pair \( \left( {\alpha, a}\right) \) . The Law of Sines gives us \( \frac{c}{\sin \left( {15}^{ \circ }\right) } = \frac{7}{\sin \left( {120}^{ \circ }\right) } \) so that \( c = \frac{7\sin \left( {15}^{ \circ }\right) }{\sin \left( {120}^{ \circ }\right) } \approx {2.09} \) units. \( {}^{5} \)
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Yes
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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 \( a = h \), then \( \gamma = {90}^{ \circ } \) so exactly one (right) triangle exists which satisfies the criteria.\n- If \( h < a < c \), then two distinct triangles exist which satisfy the given criteria.\n- If \( a \geq c \), then \( \gamma \) is acute and exactly one triangle exists which satisfies the given criteria
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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 \( \sin \left( \gamma \right) = \frac{c\sin \left( \alpha \right) }{a} > \frac{a}{a} = 1 \), which is impossible. In the figure below, we see geometrically why this is the case.\n\n\n\n\n\nSimply put, if \( a < h \) the side \( a \) is too short to connect to form a triangle. This means if \( a \geq h \) , we are always guaranteed to have at least one triangle, and the remaining parts of the theorem\n\n---
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No
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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 from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point?
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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 that end, we note that the angles \( \gamma \) and \( {45}^{ \circ } \) are supplemental, so that \( \gamma = {180}^{ \circ } - {45}^{ \circ } = {135}^{ \circ } \). We can now find \( \beta = {180}^{ \circ } - {30}^{ \circ } - \gamma = {180}^{ \circ } - {30}^{ \circ } - {135}^{ \circ } = {15}^{ \circ } \). By the Law of Sines, we have \( \frac{d}{\sin \left( {30}^{ \circ }\right) } = \frac{5}{\sin \left( {15}^{ \circ }\right) } \) which gives \( d = \frac{5\sin \left( {30}^{ \circ }\right) }{\sin \left( {15}^{ \circ }\right) } \approx {9.66} \) miles. Next, to find the point on the coast closest to the island, which we’ve labeled as \( C \), we need to find the perpendicular distance from the island to the coast. Let \( x \) denote the distance from the second observation point \( Q \) to the point \( C \) and let \( y \) denote the distance from \( C \) to the island. Using Theorem 10.4, we get \( \sin \left( {45}^{ \circ }\right) = \frac{y}{d} \). After some rearranging, we find \( y = d\sin \left( {45}^{ \circ }\right) \approx {9.66}\left( \frac{\sqrt{2}}{2}\right) \approx {6.83} \) miles. Hence, the island is approximately \( {6.83} \) miles from the coast. To find the distance from \( Q \) to \( C \), we note that \( \beta = {180}^{ \circ } - {90}^{ \circ } - {45}^{ \circ } = {45}^{ \circ } \) so by symmetry, we get \( x = y \approx {6.83} \) miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point.
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Yes
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Find the area of the triangle in Example 11.2.2 number 1.
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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 \( \beta = {45}^{ \circ } \), and we calculated \( c = \frac{7\sin \left( {15}^{ \circ }\right) }{\sin \left( {120}^{ \circ }\right) } \) . Using these values, we find \( A = \frac{1}{2}\left( 7\right) \left( \frac{7\sin \left( {15}^{ \circ }\right) }{\sin \left( {120}^{ \circ }\right) }\right) \sin \left( {45}^{ \circ }\right) = \approx {5.18} \) square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4.
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Yes
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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
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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 \( {b}^{2} = {7}^{2} + {2}^{2} - 2\left( 7\right) \left( 2\right) \cos \left( {50}^{ \circ }\right) \) which yields \( b = \sqrt{{53} - {28}\cos \left( {50}^{ \circ }\right) } \approx {5.92} \) units. In order to determine the measures of the remaining angles \( \alpha \) and \( \gamma \), we are forced to used the derived value for \( b \) . There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair \( \left( {\beta, b}\right) \) we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not\n\n---
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Yes
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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 \( {60}^{ \circ } \), find the width of the pond.
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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 \( {w}^{2} = {950}^{2} + {1000}^{2} - 2\left( {950}\right) \left( {1000}\right) \cos \left( {60}^{ \circ }\right) = {952500} \) from which we get \( w = \sqrt{952500} \approx {976} \) feet.
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Yes
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Find the area enclosed of the triangle in Example 11.3.1 number 2.
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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 \( A = \sqrt{s\left( {s - a}\right) \left( {s - b}\right) \left( {s - c}\right) } = \sqrt{\left( 8\right) \left( 4\right) \left( 1\right) \left( 3\right) } = \sqrt{96} = 4\sqrt{6} \approx {9.80} \) square units.
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Yes
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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 \( r < 0 \) .
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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 \n\nWe now set about finding alternate descriptions \( \left( {r,\theta }\right) \) for the point \( P \) . Since \( P \) is 2 units from the pole, \( r = \pm 2 \) . Next, we choose angles \( \theta \) for each of the \( r \) values. The given representation for \( P \) is \( \left( {2,{240}^{ \circ }}\right) \) so the angle \( \theta \) we choose for the \( r = 2 \) case must be coterminal with \( {240}^{ \circ } \) . (Can you see why?) One such angle is \( \theta = - {120}^{ \circ } \) so one answer for this case is \( \left( {2, - {120}^{ \circ }}\right) \) . For the case \( r = - 2 \), we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate \( \theta = {60}^{ \circ } \) to arrive at location coterminal with \( {240}^{ \circ } \) . Hence, our answer here is \( \left( {-2,{60}^{ \circ }}\right) \) . We check our answers by plotting them.\n\n
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Yes
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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\[ \text{-}{x}^{2} + {y}^{2} = {r}^{2}\text{and}\tan \left( \theta \right) = \frac{y}{x}\left( {\text{provided}x \neq 0}\right) \]
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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 \( \left( {-r,\theta + \pi }\right) \). Since \( \cos \left( {\theta + \pi }\right) = - \cos \left( \theta \right) \) and \( \sin \left( {\theta + \pi }\right) = - \sin \left( \theta \right) \), applying the theorem to \( \left( {-r,\theta + \pi }\right) \) gives \( x = \left( {-r}\right) \cos \left( {\theta + \pi }\right) = \left( {-r}\right) \left( {-\cos \left( \theta \right) }\right) = r\cos \left( \theta \right) \) and \( y = \left( {-r}\right) \sin \left( {\theta + \pi }\right) = \left( {-r}\right) \left( {-\sin \left( \theta \right) }\right) = r\sin \left( \theta \right) \). Moreover, \( {x}^{2} + {y}^{2} = {\left( -r\right) }^{2} = {r}^{2} \), and \( \frac{y}{x} = \tan \left( {\theta + \pi }\right) = \tan \left( \theta \right) \), so the theorem is true in this case, too. The remaining case is \( r = 0 \), in which case \( \left( {r,\theta }\right) = \left( {0,\theta }\right) \) is the pole. Since the pole is identified with the origin \( \left( {0,0}\right) \) in rectangular coordinates, the theorem in this case amounts to checking ’ \( 0 = 0 \) .’
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Yes
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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.
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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 \( {r}^{2} = {x}^{2} + {y}^{2} = {\left( 2\right) }^{2} + {\left( -2\sqrt{3}\right) }^{2} = \) \( 4 + {12} = {16} \) so \( r = \pm 4 \) . Since we are asked for \( r \geq 0 \), we choose \( r = 4 \) . To find \( \theta \), we have that \( \tan \left( \theta \right) = \frac{y}{x} = \frac{-2\sqrt{3}}{2} = - \sqrt{3} \) . This tells us \( \theta \) has a reference angle of \( \frac{\pi }{3} \), and since \( P \) lies in Quadrant IV, we know \( \theta \) is a Quadrant IV angle. We are asked to have \( 0 \leq \theta < {2\pi } \) , so we choose \( \theta = \frac{5\pi }{3} \) . Hence, our answer is \( \left( {4,\frac{5\pi }{3}}\right) \) . To check, we convert \( \left( {r,\theta }\right) = \left( {4,\frac{5\pi }{3}}\right) \) back to rectangular coordinates and we find \( x = r\cos \left( \theta \right) = 4\cos \left( \frac{5\pi }{3}\right) = 4\left( \frac{1}{2}\right) = 2 \) and \( y = r\sin \left( \theta \right) = 4\sin \left( \frac{5\pi }{3}\right) = 4\left( {-\frac{\sqrt{3}}{2}}\right) = - 2\sqrt{3} \), as required.
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Yes
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Graph the following polar equations.\n\n1. \( r = 4 \)
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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 definition of circle, centered at the origin, with a radius of 4 .
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Yes
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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
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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 \( \left\lbrack {0,\frac{\pi }{2}}\right\rbrack ,\left\lbrack {\frac{\pi }{2},\pi }\right\rbrack \) , \( \left\lbrack {\pi ,\frac{3\pi }{2}}\right\rbrack \) and \( \left\lbrack {\frac{3\pi }{2},{2\pi }}\right\rbrack \), and proceed as we did above. As \( \theta \) ranges from 0 to \( \frac{\pi }{2}, r \) decreases from 4 to 2 . This means that the curve in the \( {xy} \) -plane starts 4 units from the origin on the positive \( x \) -axis and gradually pulls in towards the origin as it moves towards the positive \( y \) -axis.\n\n \n\n---\n\n\( {}^{4} \) The graph of \( r = 6\cos \left( \theta \right) \) looks suspiciously like a circle, for good reason. See number 1a in Example 11.4.3.\n\n---\n\n\n\nNext, as \( \theta \) runs from \( \frac{\pi }{2} \) to \( \pi \), we see that \( r \) increases from 2 to 4 . Picking up where we left off, we gradually pull the graph away from the origin until we reach the negative \( x \) -axis.\n\n\n\nOver the interval \( \left\lbrack {\pi ,\frac{3\pi }{2}}\right\rbrack \), we see that \( r \) increases from 4 to 6 . On the \( {xy} \) -plane, the curve sweeps out away from the origin as it travels from the negative \( x \) -axis to the negative \( y \) -axis.\n\n\n\nFinally, as \( \theta \) takes on values from \( \frac{3\pi }{2} \) to \( {2\pi }, r \) decreases from 6 back to 4 . The graph on the \( {xy} \) -plane pulls in from the negative \( y \) -axis to finish where we started.\n\n\n\nWe leave it to the reader to verify that plotting points corresponding to values of \( \theta \) outside the interval \( \left\lbrack {0,{2\pi }}\right\rbrack \) results in retracing portions of the curve, so we are finished.\n\n \( r = 4 - 2\sin \left( \theta \right) \) in the \( {\theta r} \) -plane\n\n\n\n\( r = 4 - 2\sin \left( \theta \right) \) in the \( {xy} \) -plane.\n
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No
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Sketch the region in the \( {xy} \) -plane described by the following sets.\n\n1. \( \left\{ {\left( {r,\theta }\right) \mid 0 \leq r \leq 5\sin \left( {2\theta }\right) ,0 \leq \theta \leq \frac{\pi }{2}}\right\} \)
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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 the graph of \( r = 5\sin \left( {2\theta }\right) \) is a rose. Moreover, we know from our work there that as \( 0 \leq \theta \leq \frac{\pi }{2} \), we are tracing out the ’leaf’ of the rose which lies in the first quadrant. The inequality \( 0 \leq r \leq 5\sin \left( {2\theta }\right) \) means we want to capture all of the points between the origin \( \left( {r = 0}\right) \) and the curve \( r = 5\sin \left( {2\theta }\right) \) as \( \theta \) runs through \( \left\lbrack {0,\frac{\pi }{2}}\right\rbrack \) . Hence, the region we seek is the leaf itself.\n\n\n\n\n\n\[ \left\{ {\left( {r,\theta }\right) \mid 0 \leq r \leq 5\sin \left( {2\theta }\right) ,0 \leq \theta \leq \frac{\pi }{2}}\right\} \]
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Yes
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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 \( P\left( {{x}^{\prime },{y}^{\prime }}\right) \) are related by the following systems of equations
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\[ \left\{ {\begin{array}{l} x = {x}^{\prime }\cos \left( \theta \right) - {y}^{\prime }\sin \left( \theta \right) \\ y = {x}^{\prime }\sin \left( \theta \right) + {y}^{\prime }\cos \left( \theta \right) \end{array}\;\text{ and }\left\{ \begin{array}{l} {x}^{\prime } = x\cos \left( \theta \right) + y\sin \left( \theta \right) \\ {y}^{\prime } = - x\sin \left( \theta \right) + y\cos \left( \theta \right) \end{array}\right. }\right. \]
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Yes
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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.
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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 \( {x}^{\prime } = x\cos \left( \theta \right) + y\sin \left( \theta \right) = 2\cos \left( \frac{\pi }{3}\right) + \left( {-4}\right) \sin \left( \frac{\pi }{3}\right) \) which simplifies to \( {x}^{\prime } = 1 - 2\sqrt{3} \) . Similarly, \( {y}^{\prime } = - x\sin \left( \theta \right) + y\cos \left( \theta \right) = \left( {-2}\right) \sin \left( \frac{\pi }{3}\right) + \left( {-4}\right) \cos \left( \frac{\pi }{3}\right) \) which gives \( {y}^{\prime } = - \sqrt{3} - 2 = - 2 - \sqrt{3} \) . Hence \( P\left( {{x}^{\prime },{y}^{\prime }}\right) = \left( {1 - 2\sqrt{3}, - 2 - \sqrt{3}}\right) \) . To check our answer algebraically, we use the formulas in Theorem 11.9 to convert \( P\left( {{x}^{\prime },{y}^{\prime }}\right) = \left( {1 - 2\sqrt{3}, - 2 - \sqrt{3}}\right) \) back into \( x \) and \( y \) coordinates. We get\n\n\[ x = {x}^{\prime }\cos \left( \theta \right) - {y}^{\prime }\sin \left( \theta \right) \]\n\n\[ = \;\left( {1 - 2\sqrt{3}}\right) \cos \left( \frac{\pi }{3}\right) - \left( {-2 - \sqrt{3}}\right) \sin \left( \frac{\pi }{3}\right) \]\n\n\[ = \left( {\frac{1}{2} - \sqrt{3}}\right) - \left( {-\sqrt{3} - \frac{3}{2}}\right) \]\n\n\[ = 2 \]\n\nSimilarly, using \( y = {x}^{\prime }\sin \left( \theta \right) + {y}^{\prime }\cos \left( \theta \right) \), we obtain \( y = - 4 \) as required. To check our answer graphically, we sketch in the \( {x}^{\prime } \) -axis and \( {y}^{\prime } \) -axis to see if the new coordinates \( P\left( {{x}^{\prime },{y}^{\prime }}\right) = \) \( \left( {1 - 2\sqrt{3}, - 2 - \sqrt{3}}\right) \approx \left( {-{2.46}, - {3.73}}\right) \) seem reasonable. Our graph is below.\n\n
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Yes
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Graph the following equations.\n\n1. \( 5{x}^{2} + {26xy} + 5{y}^{2} - {16x}\sqrt{2} + {16y}\sqrt{2} - {104} = 0 \)
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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 \( \cot \left( {2\theta }\right) = 0 \) which gives \( \theta = \frac{\pi }{4} + \frac{\pi }{2}k \) for integers \( k \) . We choose \( \theta = \frac{\pi }{4} \) so that our rotation equations are \( x = \frac{{x}^{\prime }\sqrt{2}}{2} - \frac{{y}^{\prime }\sqrt{2}}{2} \) and \( y = \frac{{x}^{\prime }\sqrt{2}}{2} + \frac{{y}^{\prime }\sqrt{2}}{2} \) . The reader should verify that\n\n\[ \n{x}^{2} = \frac{{\left( {x}^{\prime }\right) }^{2}}{2} - {x}^{\prime }{y}^{\prime } + \frac{{\left( {y}^{\prime }\right) }^{2}}{2},\;{xy} = \frac{{\left( {x}^{\prime }\right) }^{2}}{2} - \frac{{\left( {y}^{\prime }\right) }^{2}}{2},\;{y}^{2} = \frac{{\left( {x}^{\prime }\right) }^{2}}{2} + {x}^{\prime }{y}^{\prime } + \frac{{\left( {y}^{\prime }\right) }^{2}}{2} \n\] \n\nMaking the other substitutions, we get that \( 5{x}^{2} + {26xy} + 5{y}^{2} - {16x}\sqrt{2} + {16y}\sqrt{2} - {104} = 0 \) reduces to \( {18}{\left( {x}^{\prime }\right) }^{2} - 8{\left( {y}^{\prime }\right) }^{2} + {32}{y}^{\prime } - {104} = 0 \), or \( \frac{{\left( {x}^{\prime }\right) }^{2}}{4} - \frac{{\left( {y}^{\prime } - 2\right) }^{2}}{9} = 1 \) . The latter is the equation of a hyperbola centered at the \( {x}^{\prime }{y}^{\prime } \) -coordinates \( \left( {0,2}\right) \) opening in the \( {x}^{\prime } \) direction with vertices \( \left( {\pm 2,2}\right) \) (in \( {x}^{\prime }{y}^{\prime } \) -coordinates) and asymptotes \( {y}^{\prime } = \pm \frac{3}{2}{x}^{\prime } + 2 \) . We graph it below.
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Yes
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Example 11.6.3. Use Theorem 11.11 to classify the graphs of the following non-degenerate conics.
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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 work from Example 11.6.1 number 2.\n\n2. Here, \( A = 5, B = {26} \) and \( C = 5 \), so \( {B}^{2} - {4AC} = {26}^{2} - 4\left( 5\right) \left( 5\right) = {576} > 0 \) . Theorem 11.11 classifies the graph as a hyperbola, which matches our answer to Example 11.6.2 number 1.\n\n3. Finally, we have \( A = {16}, B = {24} \) and \( C = 9 \) which gives \( {24}^{2} - 4\left( {16}\right) \left( 9\right) = 0 \) . Theorem 11.11 tells us that the graph is a parabola, matching our result from Example 11.6.2 number 2.
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Yes
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Sketch the graphs of the following equations.\n\n1. \( r = \frac{4}{1 - \sin \left( \theta \right) } \)
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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 the vertex is located at the point (in rectangular coordinates) \( \left( {0, - 2}\right) \) and must open upwards. With \( d = 4 \), we have a focal diameter of \( {2d} = 8 \), so the parabola contains the points \( \left( {\pm 4,0}\right) \) . We graph \( r = \frac{4}{1 - \sin \left( \theta \right) } \) below.
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Yes
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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 \( \left( {0,0}\right) \) .
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- 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 \( 0 < e < 1 \), the graph is an ellipse whose major axis has length \( \frac{2ed}{1 - {e}^{2}} \) and whose minor axis has length \( \frac{2ed}{\sqrt{1 - {e}^{2}}} \)\n\n- If \( e = 1 \), the graph is a parabola whose focal diameter is \( {2d} \) .\n\n- If \( e > 1 \), the graph is a hyperbola whose transverse axis has length \( \frac{2ed}{{e}^{2} - 1} \) and whose conjugate axis has length \( \frac{2ed}{\sqrt{{e}^{2} - 1}} \) .
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Yes
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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.
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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 \( \left( {r,\theta }\right) \) with \( r \geq 0 \) for the point \( P\left( {\sqrt{3}, - 1}\right) \) associated with \( z \) . We know \( {r}^{2} = {\left( \sqrt{3}\right) }^{2} + {\left( -1\right) }^{2} = 4 \), so \( r = \pm 2 \) . Since we require \( r \geq 0 \) , we choose \( r = 2 \), so \( \left| z\right| = 2 \) . Next, we find a corresponding angle \( \theta \) . Since \( r > 0 \) and \( P \) lies in Quadrant IV, \( \theta \) is a Quadrant IV angle. We know \( \tan \left( \theta \right) = \frac{-1}{\sqrt{3}} = - \frac{\sqrt{3}}{3} \), so \( \theta = - \frac{\pi }{6} + {2\pi k} \) for integers \( k \) . Hence, \( \arg \left( z\right) = \left\{ {-\frac{\pi }{6} + {2\pi k} \mid k\text{is an integer}}\right\} \) . Of these values, only \( \theta = - \frac{\pi }{6} \) satisfies the requirement that \( - \pi < \theta \leq \pi \), hence \( \operatorname{Arg}\left( z\right) = - \frac{\pi }{6} \) .
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Yes
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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- \( \left| z\right| = \sqrt{\operatorname{Re}{\left( z\right) }^{2} + \operatorname{Im}{\left( z\right) }^{2}} \)\n\n- Product Rule: \( \left| {zw}\right| = \left| z\right| \left| w\right| \)\n\n- Power Rule: \( \left| {z}^{n}\right| = {\left| z\right| }^{n} \) for all natural numbers, \( n \)\n\n- Quotient Rule: \( \left| \frac{z}{w}\right| = \frac{\left| z\right| }{\left| w\right| } \), provided \( w \neq 0 \)
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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 \( {r}^{2} = {a}^{2} + {b}^{2} \) so that \( r = \pm \sqrt{{a}^{2} + {b}^{2}} \) . Since we require \( r \geq 0 \), then it must be that \( r = \sqrt{{a}^{2} + {b}^{2}} \), which means \( \left| z\right| = \sqrt{{a}^{2} + {b}^{2}} \) . Using the distance formula, we find the distance\n\nfrom \( \left( {0,0}\right) \) to \( \left( {a, b}\right) \) is also \( \sqrt{{a}^{2} + {b}^{2}} \), establishing the first property. \( {}^{5} \) For the second property, note that since \( \left| z\right| \) is a distance, \( \left| z\right| \geq 0 \) . Furthermore, \( \left| z\right| = 0 \) if and only if the distance from \( z \) to 0 is 0, and the latter happens if and only if \( z = 0 \), which is what we were asked to show. \( {}^{6} \) For the third property, we note that since \( a = \operatorname{Re}\left( z\right) \) and \( b = \operatorname{Im}\left( z\right), z = \sqrt{{a}^{2} + {b}^{2}} = \sqrt{\operatorname{Re}{\left( z\right) }^{2} + \operatorname{Im}{\left( z\right) }^{2}} \) .\n\nTo prove the product rule, suppose \( z = a + {bi} \) and \( w = c + {di} \) for real numbers \( a, b, c \) and \( d \) . Then \( {zw} = \left( {a + {bi}}\right) \left( {c + {di}}\right) \) . After the usual arithmetic \( {}^{7} \) we get \( {zw} = \left( {{ac} - {bd}}\right) + \left( {{ad} + {bc}}\right) i \) . Therefore,\n\n\[ \left| {zw}\right| = \sqrt{{\left( ac - bd\right) }^{2} + {\left( ad + bc\right) }^{2}} \]\n\n\[ = \sqrt{{a}^{2}{c}^{2} - {2abcd} + {b}^{2}{d}^{2} + {a}^{2}{d}^{2} + {2abcd} + {b}^{2}{c}^{2}}\;\text{Expand} \]\n\n\[ = \sqrt{{a}^{2}{c}^{2} + {a}^{2}{d}^{2} + {b}^{2}{c}^{2} + {b}^{2}{d}^{2}}\;\text{Rearrange terms} \]\n\n\[ = \sqrt{{a}^{2}\left( {{c}^{2} + {d}^{2}}\right) + {b}^{2}\left( {{c}^{2} + {d}^{2}}\right) } \]\nFactor\n\n\[ = \sqrt{\left( {{a}^{2} + {b}^{2}}\right) \left( {{c}^{2} + {d}^{2}}\right) } \]\nFactor\n\n\[ = \sqrt{{a}^{2} + {b}^{2}}\sqrt{{c}^{2} + {d}^{2}} \]\nProduct Rule for Radicals\n\n\[ = \left| z\right| \left| w\right| \]\nDefinition of \( \left| z\right| \) and \( \left| w\right| \]\n\nHence \( \left| {zw}\right| = \left| z\right| \left| w\right| \) as required.
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Yes
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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 \( \operatorname{Re}\left( z\right) = 0 \) and \( \operatorname{Im}\left( z\right) > 0 \), then \( \arg \left( z\right) = \left\{ {\frac{\pi }{2} + {2\pi k} \mid k}\right. \) is an integer \( \} \) .\n\n- If \( \operatorname{Re}\left( z\right) = 0 \) and \( \operatorname{Im}\left( z\right) < 0 \), then \( \arg \left( z\right) = \left\{ {-\frac{\pi }{2} + {2\pi k} \mid k\text{is an integer}}\right\} \) .\n\n- If \( \operatorname{Re}\left( z\right) = \operatorname{Im}\left( z\right) = 0 \), then \( z = 0 \) and \( \arg \left( z\right) = \left( {-\infty ,\infty }\right) \) .
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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 \( \left( {a, b}\right) = \left( {\operatorname{Re}\left( z\right) ,\operatorname{Im}\left( z\right) }\right) \) . From Section 11.4, we know that if \( \left( {r,\theta }\right) \) is a polar representation for \( \left( {\operatorname{Re}\left( z\right) ,\operatorname{Im}\left( z\right) }\right) \), then \( \tan \left( \theta \right) = \frac{\operatorname{Im}\left( z\right) }{\operatorname{Re}\left( z\right) } \), provided \( \operatorname{Re}\left( z\right) \neq 0 \) . If \( \operatorname{Re}\left( z\right) = 0 \) and \( \operatorname{Im}\left( z\right) > 0 \), then \( z \) lies on the positive imaginary axis. Since we take \( r > 0 \), we have that \( \theta \) is coterminal with \( \frac{\pi }{2} \), and the result follows. If \( \operatorname{Re}\left( z\right) = 0 \) and \( \operatorname{Im}\left( z\right) < 0 \), then \( z \) lies on the negative imaginary axis, and a similar argument shows \( \theta \) is coterminal with \( - \frac{\pi }{2} \) . The last property in the theorem was already discussed in the remarks following Definition 11.2.
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Yes
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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: \( {zw} = \left| z\right| \left| w\right| \operatorname{cis}\left( {\alpha + \beta }\right) \)
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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\[ = \left| z\right| \left| w\right| \left\lbrack {\cos \left( \alpha \right) + i\sin \left( \alpha \right) }\right\rbrack \left\lbrack {\cos \left( \beta \right) + i\sin \left( \beta \right) }\right\rbrack \]\n\nWe now focus on the quantity in brackets on the right hand side of the equation.\n\n\[ \left\lbrack {\cos \left( \alpha \right) + i\sin \left( \alpha \right) }\right\rbrack \left\lbrack {\cos \left( \beta \right) + i\sin \left( \beta \right) }\right\rbrack = \cos \left( \alpha \right) \cos \left( \beta \right) + i\cos \left( \alpha \right) \sin \left( \beta \right) \]\n\n\[ + i\sin \left( \alpha \right) \cos \left( \beta \right) + {i}^{2}\sin \left( \alpha \right) \sin \left( \beta \right) \]\n\n\[ = \cos \left( \alpha \right) \cos \left( \beta \right) + {i}^{2}\sin \left( \alpha \right) \sin \left( \beta \right) \]\nRearranging terms\n\n\[ + i\sin \left( \alpha \right) \cos \left( \beta \right) + i\cos \left( \alpha \right) \sin \left( \beta \right) \]\n\n\[ = \left( {\cos \left( \alpha \right) \cos \left( \beta \right) - \sin \left( \alpha \right) \sin \left( \beta \right) }\right) \]\nSince \( {i}^{2} = - 1 \)\n\n\[ + i\left( {\sin \left( \alpha \right) \cos \left( \beta \right) + \cos \left( \alpha \right) \sin \left( \beta \right) }\right) \]\nFactor out \( i \)\n\n\[ = \cos \left( {\alpha + \beta }\right) + i\sin \left( {\alpha + \beta }\right) \]\nSum identities\n\n\[ = \operatorname{cis}\left( {\alpha + \beta }\right) \]\nDefinition of 'cis'\n\nPutting this together with our earlier work, we get \( {zw} = \left| z\right| \left| w\right| \operatorname{cis}\left( {\alpha + \beta }\right) \), as required.
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Yes
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