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已知sinα=1/3,α∈(π/2,π),求sin2α,cos2α,tan2α的值
人气:320 ℃ 时间:2019-12-19 02:52:49
解答
α∈(π/2,π)
cosα = -√(1-1/9) = -2√2/3
sin2α = 2sinα * cosα = 1/3 * -2√2/3 = -2√2/9
cos2α = 2cos²α-1 = 2 * 8/9 - 1 = 7/9
tan2α = sin2α/cos2α = -2√2/7
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