cos(a+β)=3/5,cos(a-β)=12/13,0

分类: 作业答案 • 2021-12-19 18:49:13

问题描述：

数学人气：390 ℃时间：2020-03-26 15:03:13

优质解答

∵0∴0sin(a-β)=√[1-cos^2(a-β)]=5/13.
sin(a+β)=√[1-cos^2(a+β)]=4/5.
cos2a=cos[(a+β)+(a-β)]=cos(a+β)*cos(a-β)-sin(a+β)*sin(a-β)=16/65.

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答

∵0∴0sin(a-β)=√[1-cos^2(a-β)]=5/13.
sin(a+β)=√[1-cos^2(a+β)]=4/5.
cos2a=cos[(a+β)+(a-β)]=cos(a+β)*cos(a-β)-sin(a+β)*sin(a-β)=16/65.