SECTION 7.6 Double-angle and Half-angle Formulas 525 The following theorem summarizes the Double-angle Formulas . THEOREM Double-angle Formulas • θ θ θ ( ) = sin 2 2 sin cos (1) • θ θ θ ( ) = − cos 2 cos sin 2 2 (2) • θ θ ( ) = − cos 2 1 2 sin2 (3) • θ θ ( ) = − cos 2 2 cos 1 2 (4) 1 Use Double-angle Formulas to Find Exact Values Finding Exact Values Using Double-angle Formulas If θ π θ π = < < sin 3 5 , 2 , find the exact value of: (a) θ ( ) sin 2 (b) θ ( ) cos 2 EXAMPLE 1 2 Use Double-angle Formulas to Establish Identities (a) Because θ θ θ ( ) = sin 2 2 sin cos , and because θ = sin 3 5 is known, begin by finding θ cos . Since θ π θ π = = < < y r sin 3 5 , 2 , let = y 3 and = r 5, and place θ in quadrant II. The point ( ) ( ) = = P x y x , , 3 is on a circle of radius 5, + = x y 25 2 2 . See Figure 38. Then + = = − = = − x y x x 25 25 9 16 4 2 2 2 This means that θ = = − = − x r cos 4 5 4 5 . Now use Double-angle Formula (1) to obtain θ θ θ ( ) ( ) = = ⋅ ⋅ − = − sin 2 2 sin cos 2 3 5 4 5 24 25 (b) Because θ = sin 3 5 is given, it is easiest to use Double-angle Formula (3) to find θ ( ) cos 2 . θ θ ( ) = − = − ⋅ = − = cos 2 1 2 sin 1 2 9 25 1 18 25 7 25 2 Figure 38 θ π θ π = < < sin 3 5 , 2 u y x x2 1 y2 5 25 (x, 3) 25 25 5 5 5 Solution = y 3 <x 0 Now Work PROBLEMS 9(a) AND (b) CAUTION In finding θ ( ) cos 2 in Example 1(b), a version of Double-angle Formula (3) was used. Note that it is not possible to use the Pythagorean Identity θ θ ( ) ( ) =± − cos 2 1 sin2 , 2 with θ ( ) =− sin 2 24 25 , because there is no way of knowing which sign to choose. Since π θ π < < 2 , we have π θ π < < 2 2 , which mean θ2 could lie in quadrant III or IV. j Establishing Identities (a) Develop a formula for θ ( ) tan 2 in terms of θ tan . (b) Develop a formula for θ ( ) sin 3 in terms of θ sin and θ cos . EXAMPLE 2 (continued)
RkJQdWJsaXNoZXIy NjM5ODQ=