518 CHAPTER 7 Analytic Trigonometry As a result, α β α β α β ( ) ( ) + = + = + = + − ⋅ − − − u v uv u v sin sin cos sin sin cos cos sin 1 1 1 1 2 2 Option 2 Let α = − u sin 1 for π α π − < < 2 2 , and β = − v cos 1 for β π < < 0 . Then α β = − ≤ ≤ = − ≤ ≤ u u v v sin for 1 1 cos for 1 1 Since π α π − < < 2 2 , then α lies in either quadrant I or IV. If > u 0, then α lies in quadrant I; if < u 0, then α lies in quadrant IV. See Figure 35(a). In either case, = − a u 1 2, and α = = − a u cos 1 1 2 . Since β π < < 0 , then β lies in either quadrant I or II. If > v 0, then β lies in quadrant I; if < v 0, then β lies in quadrant II. See Figure 35(b). In either case, = − b v 1 2 , and β = = − b v sin 1 1 2 . As a result, α β α β α β ( ) ( ) + = + = + = + − ⋅ − − − u v uv u v sin sin cos sin sin cos cos sin 1 1 1 1 2 2 Figure 35(b) ( ) = P v b , y x x2 1 y2 5 1 P 5 (v, b); v . 0 P 5 (v, b); v , 0 b b Figure 35(a) ( ) = P a u, y x x2 1 y2 5 1 P 5 (a, u); u . 0 P 5 (a, u); u , 0 a a Solving a Trigonometric Equation Linear in Sine and Cosine Solve the equation: θ θ θ π + = ≤ < sin cos 1, 0 2 EXAMPLE 11 Solution Option 1 Attempts to use available identities do not lead to equations that are easy to solve. (Try it yourself.) So, given the form of this equation, square both sides. θ θ θ θ θ θ θ θ θ θ θ θ ( ) + = + = + + = = = sin cos 1 sin cos 1 sin 2 sin cos cos 1 2 sin cos 0 sin cos 0 2 2 2 Setting each factor equal to zero leads to θ θ = = sin 0 or cos 0 The apparent solutions are 0 2 3 2 θ θ π θ π θ π = = = = Because both sides of the original equation were squared, these apparent solutions must be checked to see whether any are extraneous. θ θ π π π θ π π π θ π π π ( ) = + = + = = + = + − = − = + = + = = + = − + = − 0: sin0 cos0 0 1 1 : sin cos 0 1 1 2 : sin 2 cos 2 1 0 1 3 2 : sin 3 2 cos 3 2 1 0 1 The values θ π = and θ π = 3 2 are extraneous. The solution set is π { } 0, 2 . Now Work PROBLEM 87 4 Solve Trigonometric Equations Linear in Sine and Cosine Sometimes it is necessary to square both sides of an equation to obtain expressions that allow the use of identities. Remember, squaring both sides of an equation may introduce extraneous solutions. As a result, apparent solutions must be checked. Square both sides. Remove parentheses. θ θ + = sin cos 1 2 2 A solution Not a solution A solution Not a solution
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