530 CHAPTER 7 Analytic Trigonometry (c) Because α 2 lies in quadrant II, α < tan 2 0, so use the − sign in formula (12) to get α α α ( ) ( ) = − − + = − − − + − = − = − = − tan 2 1 cos 1 cos 1 3 5 1 3 5 8 5 2 5 4 2 Another way to solve Example 7(c) is to use the results of parts (a) and (b). α α α = = − = − tan 2 sin 2 cos 2 2 5 5 5 5 2 Now Work PROBLEMS 9(c), (d), AND (f) There is a formula for α tan 2 that does not contain + and − signs, making it more useful than formula (12). To derive it, use the formulas α α − = 1 cos 2 sin 2 2 Formula (9) and α α α α ( ) = ⋅ = sin sin 2 2 2 sin 2 cos 2 Double–angle Formula Then α α α α α α α α − = = = 1 cos sin 2 sin 2 2 sin 2 cos 2 sin 2 cos 2 tan 2 2 Because it also can be shown that α α α α − = + 1 cos sin sin 1 cos this results in the following two Half-angle Formulas: Half-angle Formulas for tan α 2 α α α α α = − = + tan 2 1 cos sin sin 1 cos (13) With this formula, the solution to Example 7(c) can be obtained as follows: α π α π α α =− < < =−− =−− =− =− cos 3 5 3 2 sin 1 cos 1 9 25 16 25 4 5 2 Then, by equation (13), α α α ( ) = − = − − − = − = − tan 2 1 cos sin 1 3 5 4 5 8 5 4 5 2
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