若cos(α-β/2)=-1/9,sin(α/2-β)=2/3,并且π/2<α<π,0<β<π/2,求cos(α+β)/2
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![若cos(α-β/2)=-1/9,sin(α/2-β)=2/3,并且π/2<α<π,0<β<π/2,求cos(α+β)/2](/uploads/image/z/2666560-40-0.jpg?t=%E8%8B%A5cos%28%CE%B1-%CE%B2%2F2%29%3D-1%2F9%2Csin%28%CE%B1%2F2-%CE%B2%29%3D2%2F3%2C%E5%B9%B6%E4%B8%94%CF%80%2F2%EF%BC%9C%CE%B1%EF%BC%9C%CF%80%2C0%EF%BC%9C%CE%B2%EF%BC%9C%CF%80%2F2%2C%E6%B1%82cos%28%CE%B1%2B%CE%B2%29%2F2)
若cos(α-β/2)=-1/9,sin(α/2-β)=2/3,并且π/2<α<π,0<β<π/2,求cos(α+β)/2
若cos(α-β/2)=-1/9,sin(α/2-β)=2/3,并且π/2<α<π,0<β<π/2,求cos(α+β)/2
若cos(α-β/2)=-1/9,sin(α/2-β)=2/3,并且π/2<α<π,0<β<π/2,求cos(α+β)/2
cos[(α+β)/2]
=cos(α/2+β/2)
=cos[(α-β/2)-(α/2-β)]
=cos(α-β/2)cos(α/2-β)+sin(α-β/2)sin(α/2-β)
∵π/2<α<π,0<β<π/2,cos(α-β/2)<0,sin(α/2-β)>0
∴π/2<α-β/2<π,0<α/2-β<π/2
sin(α-β/2)=4√5/9,cos(α/2-β)=√5/3
cos(α-β/2)cos(α/2-β)+sin(α-β/2)sin(α/2-β)
=-1/9*√5/3+4√5/9*2/3
=7√5/27