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.
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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.](/uploads/image/z/987293-29-3.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%82sin2%CE%B1%2Ccos2%CE%B2.)
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sin2α,cos2β.
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