440 CHAPTER 6 Trigonometric Functions 73. 22 23 3 6 74. 22 22 2 2 75. 2p 3 2p 3 2 24 4 76. 2p 24 4 p Mixed Practice In Problems 77–80, find the average rate of change of f from 0 to π 2 . 77. ( ) = f x x sin 78. ( ) = f x x cos 79. ( ) = f x x sin 2 80. ( ) ( ) = f x x cos 2 Mixed Practice In Problems 81–84, find ( )( ) f g x and ( )( ) g f x , and graph each of these functions. 81. ( ) ( ) = = f x x g x x sin 4 82. ( ) ( ) = = f x x g x x cos 1 2 83. ( ) ( ) = − = f x x g x x 2 cos 84. ( ) ( ) = − = f x x g x x 3 sin Applications and Extensions 85. Graph π π π ( ) = ≤ < ≤ ≤ ⎧ ⎨ ⎪⎪ ⎪⎪ ⎩ ⎪⎪ ⎪⎪ f x x x x x sin if 0 5 4 cos if 5 4 2 86. Graph π π π { ( ) = ≤ ≤ + < ≤ g x x x x x 2 sin if 0 cos 1 if 2 87. Graph π π = − ≤ ≤ y x x sin , 2 2 . 88. Graph π π = − ≤ ≤ y x x cos , 2 2 . 89. Alternating Current (AC) Circuits The current I, in amperes, flowing through an AC (alternating current) circuit at time t , in seconds, is π ( ) ( ) = ≥ I t t t 220sin 60 0 What is the period? What is the amplitude? Graph this function over two periods. 90. Alternating Current (AC) Circuits The current I, in amperes, flowing through an AC (alternating current) circuit at time t, in seconds, is π ( ) ( ) = ≥ I t t t 120sin 30 0 What is the period? What is the amplitude? Graph this function over two periods. 91. Alternating Current (AC) Generators The voltage V produced by an AC generator is sinusoidal. As a function of time, the voltage V is π ( ) ( ) = V t V ft sin 2 0 where f is the frequency, the number of complete oscillations (cycles) per second. [In the United States and Canada, f is 60 hertz (Hz).] The power P delivered to a resistance R at any time t is defined as ( ) [ ( )] = P t V t R 2 (a) Show that π ( ) ( ) = P t V R ft sin 2 . 0 2 2 (b) The graph of P is shown in the figure. Express P as a sinusoidal function. P V0 2 ––R 1 – 2f 3 – 4f 1 –f 1 – 4f t Power in an AC generator (c) Deduce that π π ( ) ( ) [ ] = − ft ft sin 2 1 2 1 cos 4 2
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