510 CHAPTER 7 Analytic Trigonometry 77. sec 1 sin 1 sin cos3 θ θ θ θ − = + 78. 1 sin 1 sin sec tan 2 θ θ θ θ ( ) + − = + 79. v v v v v v sec tan 1 csc sec tan 2 tan 2 ( ) ( ) − + − = 80. v v v v v v sec tan tan sec sin cos 2 2 − + = + 81. sin cos cos sin cos sin sec csc θ θ θ θ θ θ θ θ + − − = 82. sin cos sin cos sin cos sec csc θ θ θ θ θ θ θ θ + − − = 83. sin cos sin cos 1 sin cos 3 3 θ θ θ θ θ θ + + = − 84. sin cos 1 2 cos sec sin tan 1 3 3 2 θ θ θ θ θ θ + − = − − 85. cos sin 1 tan cos 2 2 2 2 θ θ θ θ − − = 86. cos sin sin sin cot cos 3 2 θ θ θ θ θ θ + − = + 87. 2 cos 1 cos sin 1 2 sin 2 2 4 4 2 θ θ θ θ ( ) − − = − 88. 1 2 cos sin cos tan cot 2 θ θ θ θ θ − = − 89. 1 sin cos 1 sin cos 1 cos sin θ θ θ θ θ θ + + + − = + 90. 1 cos sin 1 cos sin sec tan θ θ θ θ θ θ + + + − = + 91. a b a b a b sin cos cos sin 2 2 2 2 θ θ θ θ ( ) ( ) + + − = + 92. a a a 2 sin cos cos sin 2 2 2 2 2 2 θ θ θ θ ( ) ( ) + − = 93. tan tan cot cot tan tan α β α β α β + + = 94. tan tan 1 cot cot cot cot 1 tan tan 0 α β α β α β α β ( )( ) ( )( ) + − + + − = 95. sin cos cos sin cos sin 2 cos sin cos 2 α β β α β α β α β ( ) ( )( ) ( ) + + + − = + 96. sin cos cos sin cos sin 2 cos sin cos 2 α β β α β α β α β ( ) ( )( ) ( ) − + + − = − − 97. ln sec ln cos θ θ = − 98. ln tan ln sin ln cos θ θ θ = − 99. ln 1 cos ln 1 cos 2 ln sin θ θ θ + + − = 100. ln sec tan ln sec tan 0 θ θ θ θ + + − = In Problems 101–104, show that the functions f and g are identically equal. 101. f x x x g x x x sin tan sec cos ( ) ( ) = ⋅ = − 102. f x x x g x x x cos cot csc sin ( ) ( ) = ⋅ = − 103. f g 1 sin cos cos 1 sin 0 θ θ θ θ θ θ ( ) ( ) = − − + = 104. f g tan sec cos 1 sin θ θ θ θ θ θ ( ) ( ) = + = − 105. Show that 16 6 tan 4 sec if 2 2 . 2 θ θ π θ π + = − < < 106. Show that 9 sec 9 3tan if 3 2 . 2 θ θ π θ π − = ≤ < Applications and Extensions 107. Searchlights A searchlight at the grand opening of a new car dealership casts a spot of light on a wall located 75 meters from the searchlight. The acceleration r of the spot of light is found to be r 1200 sec 2 sec 1 . 2 θ θ ( ) = − Show that this is equivalent to θ θ ( ) = + r 1200 1 sin cos . 2 3 Source: Adapted from Hibbeler, Engineering Mechanics: Dynamics, 13th ed., Pearson © 2013. 108. Optical Measurement Optical methods of measurement often rely on the interference of two light waves. If two light waves, identical except for a phase lag, are mixed together, the resulting intensity, or irradiance, is given by I A4 csc 1 sec tan csc sec . t 2 θ θ θ θ θ ( )( ) = − + Show that this is equivalent to I A2 cos . t 2 θ ( ) = Source: Experimental Techniques, July/August 2002 109. Challenge Problem Prove: x x sin sin 1 1 ( ) − = − − − 110. Challenge Problem Prove: x x cot tan 1 1 1( ) = − − Explaining Concepts 111. Write a few paragraphs outlining your strategy for establishing identities. 112. Write down the three Pythagorean Identities. 113. Why do you think it is usually preferable to start with the side containing the more complicated expression when establishing an identity? 114. Make up an identity that is not a basic identity. Retain Your Knowledge Problems 115–124 are based on previously learned material. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for subsequent sections, a final exam, or later courses such as calculus. 115. Determine whether f x x x 3 120 50 2 ( ) = − + + has a maximum or a minimum value, and then find the value. 116. If f x x x 1 2 ( ) = + − and g x x3 4, ( ) = − find f g.
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