Question

已知sina＋sinb＝1,cosa＋cosb＝0,求cos2a＋cos2b的值

Answer 1

cosa+cosb=0……(1)

sina+sinb=1……(2)

(1)^2+(2)^2:1+2(cosacosb+sinasinb)+1=1

cos(a-b)=-1/2.

(1)2sin[(a+b)/2]*cos[(a-b)/2]=0……(3)

(2)2cos[(a+b)/2]*cos[(a-b)/2]=1……(4)

(3)/(4):tan[(a+b)/2]=0,cos[(a+b)/2]=0

cos(a+b)=1-2{cos[(a+b)/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