Stitz-Zeager_College_Algebra_e-book

What bearing would the tour helicopter need to follow

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Unformatted text preview: , b1 = 11π , b−1 = − π , b2 = 17π and 3 3 3 3 3 π b−2 = − 73 . Hence, in terms of the a’s and b’s, our domain is . . . (a−2 , b−2 ) ∪ (b−2 , a−1 ) ∪ (a−1 , b−1 ) ∪ (b−1 , a0 ) ∪ (a0 , b0 ) ∪ (b0 , a1 ) ∪ (a1 , b1 ) ∪ . . . If we group these intervals in pairs, (a−2 , b−2 ) ∪ (b−2 , a−1 ), (a−1 , b−1 ) ∪ (b−1 , a0 ), (a0 , b0 ) ∪ (b0 , a1 ) and so forth, we see a pattern emerge of the form (ak , bk ) ∪ (bk , ak + 1 ) for integers k so that our domain can be written as ∞ ∞ (ak , bk ) ∪ (bk , ak + 1 ) = k=−∞ k=−∞ (6k + 1)π (6k + 5)π , 3 3 ∪ (6k + 5)π (6k + 7)π , 3 3 A second approach to the problem exploits the periodic nature of f . It is based on the same premise on which our equation solving technique is based – for a periodic function, if we understand what happens on a fundamental period, we can know what is happening everywhere by adding integer multiples of the period. Since cos(x) and sin(x) have period 2π , it’s not too difficult to show the function f is perio...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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