由余弦定理:a^2+b^2-c^2-2abcosC=0
正弦定理:a/sinA=b/sinB=c/sinC=2R
转化 1-(cosA)^2+1-(cosB)^2-[1-(cosC)^2]-2sinAsinBcosC=0
即 (cosA)^2+(cosB)^2-(cosC)^2+2sinAsinBcosC-1=0
得 (cosA)^2+(cosB)^2-(cosC)^2+2cosC[cos(C)+cosAcosB]-1=0
(cosA)^2+(cosB)^2+(cosC)^2=1-2cosAcosBcosC