数学必修五解三角形提问会的来,

1:在三角形ABC中求证COSB/COSC=（c-bCOSA）/（b-cCOSA） 证：根据余弦定理,得 COSA=(b^2+c^2-a^2)/(2bc)COSB=(a^2+c^2-b^2)/(2ac)COSC=(a^2+b^2-c^2)/(2ab)c-bCOSA=c-b*(b^2+c^2-a^2)/(2bc)=(a^2+c^2-b^2)/(2c)b-cCOSA =b-c*(b^2+c^2-a^2)/(2bc)=(a^2+b^2-c^2)/(2b)(c-bCOSA)/(b-cCOSA )=[(a^2+c^2-b^2)/(2c)]/[(a^2+b^2-c^2)/(2b)]=[b*(a^2+c^2-b^2)]/[c*(a^2+b^2-c^2)]COSB/COSC=[(a^2+c^2-b^2)/(2ac)]/[(a^2+b^2-c^2)/(2ab)]=[b*(a^2+c^2-b^2)]/[c*(a^2+b^2-c^2)]故cosB/cosC=(c-b*cosA)/(b-c*cosA )2:在三角形ABC中求证（a^2-b^2）/c^2=sin(A-B)/sinC 证：根据正弦定理,得ab*sinC/2=bc*sinA/2a/c=sinA/sinCa^2/c^2=sin^2A/sin^2C（a^2-b^2）/c^2=(sin^2A-sin^2B)/sin^2C∵cos2A=1-2sin^2Asin^2A=(1-cos2A)/2cos2B-cos2A=-2sin(A+B)*sin(A-B)sinC=sin[180°-(A+B)]=sin(A+B)∴sin(A+B)/sinC=1上方程两边*sin(A-B)/sinC,得2sin(A+B)*sin(A-B)/(2sin^2C)=sin(A-B)/sinC-(cos2A-cos2B)/(2sin^2C)=sin(A-B)/sinC(sin^2A-sin^2B)/sin^2C=sin(A-B)/sinC（a^2-b^2）/c^2=sin(A-B)/sinC