帮下 已知sin(π/4+2a) * sin(π/4-2a) =1/4 ,a∈(π/4,π/2,求2

帮下 已知sin(π/4+2a) * sin(π/4-2a) =1/4 ,a∈(π/4,π/2,求2

sin(π/4+2a) * sin(π/4-2a) =1/4则(1/2)[cos4a-cos(π/2)]=1/4cos4a=1/2a∈(π/4,π/2)2a∈(π/2,π)可见cos2a0由cos4a=2cos²2a-1=1/2 cos²2a=3/4cos2a=-√3/2 (1)又cos4a=1-2sin²2a=1/2 sin²2a=1/4sin2a=1/2 (2)2sin^a+tana-cota-1=sina/cosa+cosa/sina-(1-2sin²a)=(sin²a+cos²a)/(sinacosa)-cos2a=2/sin2a-cos2a(1)(2)代入得原式=2/(1/2)-(-√3/2)=4+√3/2======以下答案可供参考======供参考答案1:sin(π/4+2a) * sin(π/4-2a) =（1/2)[cos4a-cos(π/2)]=1/4,∴cos4a=1/2,a∈(π/4,π/2),4a∈(π,2π),∴4a=5π/3,a=5π/12,2(sina)^2=1-cos2a=1-cos(5π/6)=1+(√3）/2，tana-cota=[(sina)^2-(cosa)^2]/(sinacosa)=-2cos2a/sin2a=-2cot2a=-2cot(5π/6)=2√3，∴2sin^a+tana-cota-1=（5√3）/2.供参考答案2:∵sin（π/4+2a)sin(π/4-2a)=[(√2/2)cos2a+(√2/2)sin2a][(√2/2)cos2a-(√2/2)sin2a]=(√2/2)²[cos²2a-sin²2a]=(1/2)cos4a=1/4.∴cos4a=1/2.又a∈(π/4,π/2),∴4a∈(π，2π）。于是4a=π+π/3=4π/3,即a=π/3或4a=2π-π/3=5π/3,即a=5π/12.当a=π/3=60°时，sina=√3/2；tana=√3；cota=-√3/3,sin²a=3/4,所以2sin²a+tana-cota-1=2*3/4+√3-√3/3-1=1/2+2√3/3。当a=5π/12=75°时，sina=(√6+√2)/4；tana=2+√3；cota=2-√3。2sin²a+tana-cota-1=2*[(√6+√2)/4]²+(2+√3)-(2-√3)-1=5√3/2。供参考答案3:sin(π/4+2a) * sin(π/4-2a) =（1/2)[cos4a-cos(π/2)]=1/4,∴cos4a=1/2,a∈(π/4,π/2),4a∈(π,2π),∴4a=5π/3,a=5π/12,2(sina)^2=1-cos2a=1-cos(5π/6)=1+(√3）/2，tana-cota=[(sina)^2-(cosa)^2]/(sinacosa)=-2cos2a/sin2a=-2cot2a=-2cot(5π/6)=2√3，∴2sin^a+tana-cota-1=（5√3）/2. (∩_∩)答案:5√3）/2

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