Question

三角形ABC面积s=c^2-(a-b)^2,求c/2的正切值

Answer 1

∵c^2=a^2+b^2-2abcosC

∴S=[c^2-(a-b)^2]

=(2ab-2abcosC)

=2ab(1-cosC)

=2ab(1-cosC)

又S=absinC/2

2ab(1-cosC)=absinC/2

4ab(1-cosC)=absinC

4(1-cosC)=sinC

∴sinC=4(1-cosC)

∴sinC/(1-cosC)=4

∴(1-cosC)/sinC=1/4

即tan(C/2)=1/4

注：tan(A/2)=(1-cosA)/sinA=sinA/(1+cosA);