两道打圈的题(9的第一问和10)求详解!还有10~
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/06 12:13:55
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两道打圈的题(9的第一问和10)求详解!
还有10~
![两道打圈的题(9的第一问和10)求详解!还有10~](/uploads/image/z/5467-67-7.jpg?t=%E4%B8%A4%E9%81%93%E6%89%93%E5%9C%88%E7%9A%84%E9%A2%98%EF%BC%889%E7%9A%84%E7%AC%AC%E4%B8%80%E9%97%AE%E5%92%8C10%EF%BC%89%E6%B1%82%E8%AF%A6%E8%A7%A3%21%E8%BF%98%E6%9C%8910%7E)
9:
cos(A-B) = cos(π - B-C-B) =cos[π-(2B+C)] = -cos(2B+C)] = 2sin(B+C/2)^2 -1 = 7/25
10:有点麻烦,如下:
易知:F(θ)= 1/2 [ cos2θ+1 +cos2(θ+α)+1+cos2(θ+β)+1]
=3/2 +1/2(cos2θ + cos2θcos2α-sin2θsin2α +cos2θcos2β-sin2θsin2β)
= 3/2 +1/2[cos2θ(1+cos2α+cos2β) - sin2θ(sin2α+sin2β)]
要满足题意,1+cos2α+cos2β=0 且sin2α+sin2β=0,
而当2α+2β=2π时,sin2α+sin2β=0满足,此时代入前式可得:2cos2α=-1/2
所以2α =2π/3 ,2β =4π/3.
因此当α =π/3 ,β =2π/3时,F(θ)= 3/2 恒成立
再问: 谢谢第九题的解释!问下第十题怎么做?
cos(A-B) = cos(π - B-C-B) =cos[π-(2B+C)] = -cos(2B+C)] = 2sin(B+C/2)^2 -1 = 7/25
10:有点麻烦,如下:
易知:F(θ)= 1/2 [ cos2θ+1 +cos2(θ+α)+1+cos2(θ+β)+1]
=3/2 +1/2(cos2θ + cos2θcos2α-sin2θsin2α +cos2θcos2β-sin2θsin2β)
= 3/2 +1/2[cos2θ(1+cos2α+cos2β) - sin2θ(sin2α+sin2β)]
要满足题意,1+cos2α+cos2β=0 且sin2α+sin2β=0,
而当2α+2β=2π时,sin2α+sin2β=0满足,此时代入前式可得:2cos2α=-1/2
所以2α =2π/3 ,2β =4π/3.
因此当α =π/3 ,β =2π/3时,F(θ)= 3/2 恒成立
再问: 谢谢第九题的解释!问下第十题怎么做?