已知sinα+sinβ+siny=0,cosα+cosβ+cosy=0,求证:cos(α-y)=-1/2急.急.急.快点帮帮忙已知α∈(π/4,3π/4),β∈(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13,求:cos(α+β)
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/14 04:53:22
![已知sinα+sinβ+siny=0,cosα+cosβ+cosy=0,求证:cos(α-y)=-1/2急.急.急.快点帮帮忙已知α∈(π/4,3π/4),β∈(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13,求:cos(α+β)](/uploads/image/z/7173711-63-1.jpg?t=%E5%B7%B2%E7%9F%A5sin%CE%B1%2Bsin%CE%B2%2Bsiny%3D0%2Ccos%CE%B1%2Bcos%CE%B2%2Bcosy%3D0%2C%E6%B1%82%E8%AF%81%EF%BC%9Acos%EF%BC%88%CE%B1-y%EF%BC%89%3D-1%2F2%E6%80%A5.%E6%80%A5.%E6%80%A5.%E5%BF%AB%E7%82%B9%E5%B8%AE%E5%B8%AE%E5%BF%99%E5%B7%B2%E7%9F%A5%CE%B1%E2%88%88%EF%BC%88%CF%80%2F4%EF%BC%8C3%CF%80%2F4%EF%BC%89%EF%BC%8C%CE%B2%E2%88%88%EF%BC%880%EF%BC%8C%CF%80%2F4%EF%BC%89%EF%BC%8C%E4%B8%94cos%EF%BC%88%CF%80%2F4-%CE%B1%EF%BC%89%3D3%2F5%EF%BC%8Csin%EF%BC%885%CF%80%2F4%2B%CE%B2%EF%BC%89%3D-12%2F13%EF%BC%8C%E6%B1%82%EF%BC%9Acos%EF%BC%88%CE%B1%2B%CE%B2%EF%BC%89)
已知sinα+sinβ+siny=0,cosα+cosβ+cosy=0,求证:cos(α-y)=-1/2急.急.急.快点帮帮忙已知α∈(π/4,3π/4),β∈(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13,求:cos(α+β)
已知sinα+sinβ+siny=0,cosα+cosβ+cosy=0,求证:cos(α-y)=-1/2
急.急.急.
快点帮帮忙
已知α∈(π/4,3π/4),β∈(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13,求:cos(α+β)
已知sinα+sinβ+siny=0,cosα+cosβ+cosy=0,求证:cos(α-y)=-1/2急.急.急.快点帮帮忙已知α∈(π/4,3π/4),β∈(0,π/4),且cos(π/4-α)=3/5,sin(5π/4+β)=-12/13,求:cos(α+β)
证明:因为sinα+sinβ+siny=0,cosα+cosβ+cosy=0
所以 sinβ=-(sinα+siny) cosβ=-(cosα+cosy)
根据 sin^2β+cos^2β=1 将上式代入展开得:1+1+2(sinαsiny+cosαcosy)=1
整理得:sinαsiny+cosαcosy=-1/2 即cos(α-y)=-1/2
后面的问题:
因为α∈(π/4,3π/4),β∈(0,π/4)所以π/4-α ∈(-π/2,0),
5π/4+β∈(5π/4,3π/2) 所以 sin(π/4-α)<0 等于-4/5;cos(5π/4+β)<0 等于-5/13;
原式=-cos(α+β+π)=-cos((5π/4+β)-(π/4-α))
然后展开将上面求得的值代入就行了呵呵