Question

已知sina+sinb=1，cosa+cosb=0求cos（a+b）的值

Answer 1

cosa+cosb=A,sina+sinb=B,求cos(a+b)
cosa+cosb=A--->(cosa+cosb)^=cos^a+cos^b+2cosacosb=A^
sina+sinb=A--->(sina+sinb)^=sin^a+sin^b+2sinasinb=B^
相加:2+2(cosacosb+sinasinb)=A^+B^--->1+cos(a-b)=(A^+B^)/2
相减:cos2A+cos2B+2cos(a+b)
=2cos(a+b)cos(a-b)+2cos(a+b)
=2cos(a+b)[cos(a-b)+1]
=2cos(a+b)[(A^+B^)/2]
=(A^+B^)cos(a+b)=A^-B^
∴cos(a+b)=(A^-B^)/(A^+B^)