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sin(a-b)=3/5,cos(a+b)=5/13且a b都是锐角求cos2a和cos2b

分类: 作业答案 • 2021-12-18 16:25:28
问题描述：

sin(a-b)=3/5,cos(a+b)=5/13且a b都是锐角求cos2a和cos2b

答

∵a b都是锐角
sin(a-b)=3/5
cos(a-b)=4/5
cos(a+b)=5/13
sin(a+b)=12/13
cos2a=cos(a+b+a-b)
=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=5/13*4/5-12/13*3/5
=-16/65
cos2b=cos(a+b-(a-b))
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=5/13*4/5+12/13*3/5
=56/65

答

a b都是锐角
即0-π/20所以cos(a-b)>0 sin(a+b)>0
cos(a-b)=√[1-(sin(a-b))^2]=4/5
sin(a+b)=√[1-(cos(a+b))^2]=12/13
cos2a=cos(a+b+a-b)
=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=5/13*4/5-12/13*3/5
=-16/65
cos2b=cos(a+b-(a-b))
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=5/13*4/5+12/13*3/5
=56/65

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