已知cos(a+B)=4/5,cos(a-B)=-4/5
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a+B∈(7/4π,2π),所以sin(a+B)<0,
所以sin(a+B)=-√(1-cos²(a+B))=-3/5
a-B∈(3/4π,π),所以sin(a-B)>0
所以sin(a-B)=-√(1-cos²(a-B))=3/5
cos2a=cos(a+B+a-B)
=cos(a+B)cos(a-B)-sin(a+B)sin(a-B)
=4/5*(-4/5)-(-3/5)*(3/5)
=-7/25
cos2B=cos(a+B-(a-B))
=cos(a+B)cos(a-B)+sin(a+B)sin(a-B)
=4/5*(-4/5)+(-3/5)*(3/5)
=-1
所以sin(a+B)=-√(1-cos²(a+B))=-3/5
a-B∈(3/4π,π),所以sin(a-B)>0
所以sin(a-B)=-√(1-cos²(a-B))=3/5
cos2a=cos(a+B+a-B)
=cos(a+B)cos(a-B)-sin(a+B)sin(a-B)
=4/5*(-4/5)-(-3/5)*(3/5)
=-7/25
cos2B=cos(a+B-(a-B))
=cos(a+B)cos(a-B)+sin(a+B)sin(a-B)
=4/5*(-4/5)+(-3/5)*(3/5)
=-1
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