Question

在三角形ABC中,求证：cos2A/(a*a)-cos2B/(b*b)=1/(a*a)-1/(b*b).

Answer 1

记为等价符号

cos2A/(a*a)-cos2B/(b*b)=1/(a*a)-1/(b*b)

(2*cosA*cosA-1)/(a*a)-(2*cosB*cosB-1)/(b*b)=1/(a*a)-1/(b*b)

(cosA*cosA-1)/(a*a)=(cosB*cosB-1)/(b*b)

sinA*sinA/(a*a)=sinB*sinB/(b*b)

sinA/a=sinB/b

正弦定理