已知cos(a-B/2)=-1/9,sin(a/2-B)=2/3,且π/2
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![已知cos(a-B/2)=-1/9,sin(a/2-B)=2/3,且π/2](/uploads/image/z/1786678-70-8.jpg?t=%E5%B7%B2%E7%9F%A5cos%28a-B%2F2%29%3D-1%2F9%2Csin%28a%2F2-B%29%3D2%2F3%2C%E4%B8%94%CF%80%2F2)
已知cos(a-B/2)=-1/9,sin(a/2-B)=2/3,且π/2 已知cos(a-B/2)=-1/9,sin(a/2-B)=2/3,且π/2
已知cos(a-B/2)=-1/9,sin(a/2-B)=2/3,且π/2
由于a/2+B/2=(a-B/2)-(a/2-B),故
cos(a/2+B/2)=cos[(a-B/2)-(a/2-B)]
=cos(a-B/2)cos(a/2-B)+sin(a-B/2)sin(a/2-B)
由π/2又cos(a-B/2)=-1/9,sin(a/2-B)=2/3,故sin(a-B/2)=(4倍根号5)/9,cos(a/2-B)=(根号5)/3
因此cos(a/2+B/2)=(7倍根号5)/27