420 CHAPTER 6 Trigonometric Functions Finding Exact Values Given One Value and the Sign of Another Given that θ = sin 1 3 and θ < cos 0, find the exact value of each of the remaining five trigonometric functions. Solution Option 1 Using a Circle EXAMPLE 5 Suppose that ( ) = P x y , is the point on a circle that corresponds to θ. Since θ = > sin 1 3 0 and θ < cos 0, the point ( ) = P x y , is in quadrant II. Because θ = = y r sin 1 3 , we let = y 1 and = r 3. The point ( ) ( ) = = P x y x , , 1 that corresponds to θ lies on the circle of radius 3 centered at the origin: + = x y 9. 2 2 See Figure 44. To find x, we use the fact that + = = x y y 9, 1, 2 2 and P is in quadrant II ( ) < x so 0 . Figure 44 y x 3 u P 5 (x, 1) x2 1 y2 5 9 23 23 3 O + = + = = = − x y x x x 9 1 9 8 2 2 2 2 2 2 2 = y 1 <x 0 First, solve identity (5), θ θ + = sin cos 1 2 2 , for θ cos . θ θ θ θ θ θ + = = − = ± − sin cos 1 cos 1 sin cos 1 sin 2 2 2 2 2 Because θ < cos 0, choose the minus sign and use the fact that θ = sin 1 3 . θ θ =−− =−−=− =− cos 1 sin 1 1 9 8 9 2 2 3 2 ↑ θ = sin 1 3 Now, use θ = sin 1 3 and θ = − cos 2 2 3 and fundamental identities. θ θ θ θ θ θ θ θ θ = = − = − = − = = − = = − = − = − = = = tan sin cos 1 3 2 2 3 1 2 2 2 4 cot 1 tan 2 2 sec 1 cos 1 2 2 3 3 2 2 3 2 4 csc 1 sin 1 1 3 3 Since = − = x y 2 2, 1, and = r 3, we find that θ θ θ θ θ = = − = = − = − = = = = = − = − = = − = − x r y x r y r x x y cos 2 2 3 tan 1 2 2 2 4 csc 3 1 3 sec 3 2 2 3 2 4 cot 2 2 1 2 2 Option 2 Using Identities
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