已知sin(π/4+2a)sin(π/4-2a)=1/4,a∈(π/4,π/2)求2sin方a+tana-cota-1
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已知sin(π/4+2a)sin(π/4-2a)=1/4,a∈(π/4,π/2)求2sin方a+tana-cota-1
![已知sin(π/4+2a)sin(π/4-2a)=1/4,a∈(π/4,π/2)求2sin方a+tana-cota-1](/uploads/image/z/962230-22-0.jpg?t=%E5%B7%B2%E7%9F%A5sin%28%CF%80%2F4%2B2a%29sin%28%CF%80%2F4-2a%29%3D1%2F4%2Ca%E2%88%88%EF%BC%88%CF%80%2F4%2C%CF%80%2F2%29%E6%B1%822sin%E6%96%B9a%2Btana-cota-1)
由和差化积公式:cosa-cosb=-2sin(a+b)/2sin(a-b)/2
可知sin(π/4+2a)sin(π/4-2a)=1/2*(cos4a-cosπ/2)
所以cos4a=1/2因为a∈(π/4,π/2)
所以4a=5/3*π所以2a=5/6*π
2sin方a+tana-cota-1=-cos2a-2tana=(5根号3)/2
可知sin(π/4+2a)sin(π/4-2a)=1/2*(cos4a-cosπ/2)
所以cos4a=1/2因为a∈(π/4,π/2)
所以4a=5/3*π所以2a=5/6*π
2sin方a+tana-cota-1=-cos2a-2tana=(5根号3)/2
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