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

来源：百度知道 编辑：UC知道 时间：2024/07/08 02:39:23

a b都是锐角
即0<a<π/2 0<b<π/2
-π/2<a-b<π/2
0<a+b<π
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

∵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