在△ABC中,已知sinA/sinC=sin(A-B)/sin(B-C),求证,a*2,b*2,c*2成等差数列
在△ABC中,已知sinA/sinC=sin(A-B)/sin(B-C),求证,a*2,b*2,c*2成等差数列...
在△ABC中,已知sinA/sinC=sin(A-B)/sin(B-C),求证,a*2,b*2,c*2成等差数列
展开
1个回答
展开全部
sinA/sinC=sin(A-B)/sin(B-C)
sinA/sinC=(sinAcosB-cosAsinB)/(sinBcosC-cosBsinC)
a/sinA=b/sinB=c/sinC
cosA=(b²+c²-a²)/(2bc)
cosB=(c²+a²-b²)/(2ac)
cosC=(a²+b²-c²)/(2ab)
a/c=[a(c²+a²-b²)/(2ac)-b(b²+c²-a²)/(2bc)]/[b(a²+b²-c²)/(2ab) -c(a²+c²-b²)/(2ac)]
a/c=[(c²+a²-b²)/(2c)-(b²+c²-a²)/(2c)]/[(a²+b²-c²)/(2a)-(a²+c²-b²)/(2a)]
a/c=[(c²+a²-b²-b²-c²+a²)/(2c)]/[(a²+b²-c²-a²-c²+b²)/(2a)]
a/c=[(a²-b²)/c][(b²-c²)/a]
a/c=(a/c)[(a²-b²)/(b²-c²)]
(a²-b²)/(b²-c²)=1
a²-b²=b²-c²
b²-a²=c²-b²
a²、b²、c²成等差数列。
sinA/sinC=(sinAcosB-cosAsinB)/(sinBcosC-cosBsinC)
a/sinA=b/sinB=c/sinC
cosA=(b²+c²-a²)/(2bc)
cosB=(c²+a²-b²)/(2ac)
cosC=(a²+b²-c²)/(2ab)
a/c=[a(c²+a²-b²)/(2ac)-b(b²+c²-a²)/(2bc)]/[b(a²+b²-c²)/(2ab) -c(a²+c²-b²)/(2ac)]
a/c=[(c²+a²-b²)/(2c)-(b²+c²-a²)/(2c)]/[(a²+b²-c²)/(2a)-(a²+c²-b²)/(2a)]
a/c=[(c²+a²-b²-b²-c²+a²)/(2c)]/[(a²+b²-c²-a²-c²+b²)/(2a)]
a/c=[(a²-b²)/c][(b²-c²)/a]
a/c=(a/c)[(a²-b²)/(b²-c²)]
(a²-b²)/(b²-c²)=1
a²-b²=b²-c²
b²-a²=c²-b²
a²、b²、c²成等差数列。
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询