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sin2a=3/5且0<2a<π/2,求2cos^2(a/2)-sina-1 /根号2sin(a+π/4)
人气:311 ℃ 时间:2020-09-03 16:01:30
解答
sin2a=3/5 → cos2a=4/5,∴ sina=√(1-cos2a)=√(1-4/5)=√5/5,cosa=2√5/5;[2cos²(a/2)-sina-1]/[√2sin(a+π/4)]=(cosa+1-sina-1)/(cosa+sina)=(2√5/5-√5/5)/(2√5/5+2√5/5)=1/3;
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