作业帮已知cosB=cosαsinA,cosC=sinαsinA,求证sinAsinA+sinBsinB+sinCsinC=2
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/07/15 05:02:51
已知cosB=cosαsinA,cosC=sinαsinA,求证sinAsinA+sinBsinB+sinCsinC=2?
cosB=cosαsinA 平方一下
cos^2B=cos^2asin^aA
1-sin^2B=cos^2asin^aA 1
cosC=sinαsinA 平方一下
cos^2C=sin^2asin^2A
1-sin^2C=sin^2asin^2A 2
1式+2式得
2-sin^2B-sin^2C=cos^2asin^2A+sin^2asin^2A=sin^2A(sin^2a+cos^2a)=sin^2A
所以
sin^2A+sin^2B+sin^2C=2
即sinAsinA+sinBsinB+sinCsinC=2
cos^2B=cos^2asin^aA
1-sin^2B=cos^2asin^aA 1
cosC=sinαsinA 平方一下
cos^2C=sin^2asin^2A
1-sin^2C=sin^2asin^2A 2
1式+2式得
2-sin^2B-sin^2C=cos^2asin^2A+sin^2asin^2A=sin^2A(sin^2a+cos^2a)=sin^2A
所以
sin^2A+sin^2B+sin^2C=2
即sinAsinA+sinBsinB+sinCsinC=2
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