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/2
a/c=sinA/sinC
a^2/c^2=sin^2A/sin^2C
（a^2-b^2）/c^2=(sin^2A-sin^2B)/sin^2C
∵cos2A=1-2sin^2A
sin^2A=(1-cos2A)/2
cos2B-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