已知sin(π/4+2α）sin(π/4-2α)=1/4,α属于（π/4，π/2），求2sin平方α+tanα-cotα-1

已知sin(π/4+2α）sin(π/4-2α)=1/4,α属于（π/4，π/2），求2sin平方α+tanα-cotα-1

sin(π/4+2α)sin(π/4-2α) 
=1/2*(cos(π/4+2α-π/4+2α)-cos(π/4+2α+π/4-2α)) 
=1/2*(cos(4α)-cos(π/2)) 
=1/2*cos(4α) 
=1/4 
所以cos(4α)=1/2
因为α属于(π/4,π/2)
所以π<4α<2π
所以π/2<2α<π
所以cos(2α)=-√3/2,sin(2α)=1/2
2(sinα)^2+tanα-cotα-1
=sinα/cosα-cosα/sinα-(1-2(sinα)^2)
=((sinα)^2-(cosα)^2)/(sinαcosα)-cos(2α)
=(-cos(2α))/(1/2*sin(2α))-cos(2α
=-2cos(2α)/sin(2α)-cos(2α)
=-2*(-√3/2)/(1/2)-(-√3/2)
=2√3+√3/2
=5√3/2
懂了吗？
希望能帮到你 O(∩_∩)O~

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