在△ABC中,求证(a²-b²)/c²=sin(A-B)/sinC
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由正弦定理:
a/sinA=c/sinC
a/c=sinA/sinC,两边同时乘以2cosB,左边分子分母侍乱同乘以c.得:
2ac*cosB/c²=2sinAcosB/sinC.
由余弦定理简谈念a²+c²-b²=2ac*cosB得:
(a²+c²-b²)/拦困c²=2sinAcosB/sinC
两边同时减去1,可得:
(a²-b²)/c²=(2sinAcosB-sinC)/sinC
且有2sinAcosB-sinC=2sinAcosB-sin(A+B)
=2sinAcosB-(sinAcosB+cosAsinB)
=sinAcosB-cosAsinB
=sin(A-B)
则原式得证.
a/sinA=c/sinC
a/c=sinA/sinC,两边同时乘以2cosB,左边分子分母侍乱同乘以c.得:
2ac*cosB/c²=2sinAcosB/sinC.
由余弦定理简谈念a²+c²-b²=2ac*cosB得:
(a²+c²-b²)/拦困c²=2sinAcosB/sinC
两边同时减去1,可得:
(a²-b²)/c²=(2sinAcosB-sinC)/sinC
且有2sinAcosB-sinC=2sinAcosB-sin(A+B)
=2sinAcosB-(sinAcosB+cosAsinB)
=sinAcosB-cosAsinB
=sin(A-B)
则原式得证.
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