已知sin(a-β)=4/5,sin(a+β)=-12/13,a-β∈(π/2,π),a+β∈(3π/2,2π),求si
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已知sin(a-β)=4/5,sin(a+β)=-12/13,a-β∈(π/2,π),a+β∈(3π/2,2π),求sin2α,cos2β.
![已知sin(a-β)=4/5,sin(a+β)=-12/13,a-β∈(π/2,π),a+β∈(3π/2,2π),求si](/uploads/image/z/17216952-24-2.jpg?t=%E5%B7%B2%E7%9F%A5sin%28a-%CE%B2%29%3D4%2F5%2Csin%28a%2B%CE%B2%29%3D-12%2F13%2Ca-%CE%B2%E2%88%88%28%CF%80%2F2%2C%CF%80%29%2Ca%2B%CE%B2%E2%88%88%283%CF%80%2F2%2C2%CF%80%29%2C%E6%B1%82si)
α-β∈(π/2,π) cos(α-β)0
cos(α+β)=√[1-sin²(α+β)]=√[1-(-12/13)²]=5/13
cos(2α)=cos[(α+β)+(α-β)]
=cos(α+β)cos(α-β)-sin(α+β)sin(α-β)
=(5/13)(-3/5)-(-12/13)(4/5)
=33/65
cos(2β)=cos[(α+β)-(α-β)]
=cos(α+β)cos(α-β)+sin(α+β)sin(α-β)
=(5/13)(-3/5)+(-12/13)(4/5)
=-63/65
再问: �����sin2����cos2��
再答: sin(2��)=sin[(��+��)+(��-��)] =sin(��+��)cos(��-��)+cos(��+��)sin(��-��) =(-12/13)(-3/5)+(5/13)(4/5) =56/65
cos(α+β)=√[1-sin²(α+β)]=√[1-(-12/13)²]=5/13
cos(2α)=cos[(α+β)+(α-β)]
=cos(α+β)cos(α-β)-sin(α+β)sin(α-β)
=(5/13)(-3/5)-(-12/13)(4/5)
=33/65
cos(2β)=cos[(α+β)-(α-β)]
=cos(α+β)cos(α-β)+sin(α+β)sin(α-β)
=(5/13)(-3/5)+(-12/13)(4/5)
=-63/65
再问: �����sin2����cos2��
再答: sin(2��)=sin[(��+��)+(��-��)] =sin(��+��)cos(��-��)+cos(��+��)sin(��-��) =(-12/13)(-3/5)+(5/13)(4/5) =56/65
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