方向向量变化题设向量OP的方向角为A,B,C则(cscA)^2+4(cscB)^2+16(cscC)^2的最小值?
1个回答
由已知 (cos A)^2+(cos B)^2+(cos C)^2=1,
因为 (csc A)^2=(1/sin A)^2=1/(1-(cos A)^2)；
4(csc B)^2=4/(1-(cos B)^2)
=2/(1-(cos B)^2)+2/(1-(cos B)^2)；
16(csc C)^2=16/(1-(cos C)^2)
=4/(1-(cos C)^2)+4/(1-(cos C)^2)+4/(1-(cos C)^2)+4/(1-(cos C)^2)；
(csc A)^2+4(csc B)^2+16(csc C)^2
=1/(1-(cos A)^2)+ [2/(1-(cos B)^2)+2/(1-(cos B)^2)] + [4/(1-(cos C)^2)+4/(1-(cos C)^2)+4/(1-(cos C)^2)+4/(1-(cos C)^2)]
≥7^2/[ (1-(cos A)^2) + (1-(cos B)^2)/2 + (1-(cos B)^2)/2 + (1-(cos C)^2)/4 + (1-(cos C)^2)/4 + (1-(cos C)^2)/4 + (1-(cos C)^2)/4 ] （算术平均A_n≥调和平均H_n）
=49/[1+1+1-( (cos A)^2+(cos B)^2+(cos C)^2)]
=49/2,
当且仅当1/(1-(cos A)^2)=2/(1-(cos B)^2)=4/(1-(cos C)^2)时等号成立.
此时(1-(cos A)^2)=(1-(cos B)^2)/2=(1-(cos C)^2)/4.
即最小值为49/2.
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