SECTION 7.5 Sum and Difference Formulas 519 Figure 36 y x x2 1 y2 5 1 21 21 1 1 p– 4 Ax, B –– 2 2 3p ––– 4 A2x, B –– 2 2 Option 2 Start with the equation θ θ + = sin cos 1 and divide both sides by 2. Then θ θ + = 1 2 sin 1 2 cos 1 2 The left side now resembles the formula for the sine of the sum of two angles, one of which is θ. The other angle is unknown (call it φ.) Then θ φ θ φ θ φ ( ) + = + = sin sin cos cos sin 1 2 (8) Comparing (8) to θ θ + = 1 2 sin 1 2 cos 1 2 , we see that cos 1 2 2 2 sin 1 2 2 2 0 2 φ φ φ π = = = = ≤ < The angle φ is therefore π 4 . As a result, equation (8) becomes θ π ( ) + = sin 4 2 2 In the interval π [ ) 0, 2 , there are two angles whose sine is 2 2 : π 4 and π3 4 . See Figure 36. As a result, θ π π θ π π θ θ π + = + = = = 4 4 or 4 3 4 0 or 2 The solution set is π { } 0, 2 . Figure 37 x y f P = (a, b) cos f = sin f = a2 1 b2 a2 1 b2 b a2 1 b2 a x2 1 y2 5 a2 1 b2 The second option can be used to solve any linear equation of the form θ θ + = a b c sin cos , where a, b, and c are nonzero constants, by dividing both sides of the equation by + a b . 2 2 Solving a Trigonometric Equation Linear in θ sin and θ cos Solve: θ θ + = a b c sin cos (9) where a, b, and c are constants and either ≠ a 0 or ≠ b 0. EXAMPLE 12 Solution Divide both sides of equation (9) by + a b . 2 2 Then θ θ + + + = + a a b b a b c a b sin cos 2 2 2 2 2 2 (10) There is a unique angle φ φ π ≤ < , 0 2 , for which φ φ = + = + a a b b a b cos and sin 2 2 2 2 (11) Figure 37 shows the angle φ for > a 0 and > b 0. (continued)
RkJQdWJsaXNoZXIy NjM5ODQ=