Question

在三角形ABC中，若a:b:c=2:3:4,求2sinA-sinB/sin2C的值

Answer 1

sinA:sinB:sinC=2:3:4
sinC=2sinA
由余弦定理:cosC=(a^2+b^2-c^2)/2ab=(2^2+3^2-4^2)/(2*2*3)=-1/4
（2sinA-sinB)/sin2C
=(2sinA-sinB)/(-sinC/2)
=2(2sinA-sinB)/(-sinC)
=2(2*2-3)/(-4)
=-1/2

Answer 2

解：可设a=2t,b=3t,c=4t,(t>0)
由余弦定理可知，cosC=(4t²+9t²-16t²)/(12t²)=-1/4.
===>1/cosC=-4.
∴(2sinA-sinB)/sin2C=[(2sinA-sinB)/(2sinC)]×(1/cosC)=-1/2.