sina +sin b =1,cos a +cos b =0,求cos2a +cos2b

sina +sin b =1,cos a +cos b =0,求cos2a +cos2b
sina +sin b =1,cos a +cos b =0,求cos2a +cos2b

sina +sin b =1,cos a +cos b =0,求cos2a +cos2b
cosa+cosb=0 (1)
sina+sinb=1 (2)
(1)²+(2)²:1+2(cosacosb+sinasinb)+1=1
得出：cos(a-b)=-1/2.
由(1)式得出：2sin[(a+b)/2]*cos[(a-b)/2]=0 (3)
由(2)式得出：2cos[(a+b)/2]*cos[(a-b)/2]=1 (4)
(3)÷(4)得:tan[(a+b)/2]=0
推出：cos[(a+b)/2]=0
得：cos(a+b)=1-2^2=1-0=1
cos(a+b)cos(a-b)=-1/2
1/2*(cos2a+cos2b)=-1/2
cos2a+cos2b=-1
所以cos2A+cos2B=-1