帮下 已知sin(π/4+2a) * sin(π/4-2a) =1/4 ,a∈(π/4,π/2,求2sin^a+tana
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帮下 已知sin(π/4+2a) * sin(π/4-2a) =1/4 ,a∈(π/4,π/2,求2sin^a+tana-cota-1的值.
已知sin(π/4+2a) * sin(π/4-2a) =1/4 ,a∈(π/4,π/2,求2sin^a+tana-cota-1的值.
a∈(π/4,π/2)
已知sin(π/4+2a) * sin(π/4-2a) =1/4 ,a∈(π/4,π/2,求2sin^a+tana-cota-1的值.
a∈(π/4,π/2)
sin(π/4+2a) * sin(π/4-2a) =1/4
则(1/2)[cos4a-cos(π/2)]=1/4
cos4a=1/2
a∈(π/4,π/2)
2a∈(π/2,π)
可见cos2a0
由cos4a=2cos²2a-1=1/2 cos²2a=3/4
cos2a=-√3/2 (1)
又cos4a=1-2sin²2a=1/2 sin²2a=1/4
sin2a=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
则(1/2)[cos4a-cos(π/2)]=1/4
cos4a=1/2
a∈(π/4,π/2)
2a∈(π/2,π)
可见cos2a0
由cos4a=2cos²2a-1=1/2 cos²2a=3/4
cos2a=-√3/2 (1)
又cos4a=1-2sin²2a=1/2 sin²2a=1/4
sin2a=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
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