Question

在△ABC中,sinA:sinB:sinC=4:3:2,则cosA的值为 A2/3 B 1/4 C-2/3 D -1/4

Answer 1

简单
sina/(sinb+sinc)+sinb/(sinc+sina)+sinc/(sina+sinb)
>sina/(sina+sinb+sinc)+sinb/(sinb+sinc+sina)+sinc/(sinc+sina+sinb)=1
下面证
sina<(sinb+sinc)
sin（b+c）=sinbcosc+sinccosb<(sinb+sinc)
即sinb（cosc-1）+sinc（cosb-1）<0
所以sina/(sinb+sinc)<1
下面用一个结论a>0
b>0
a>b
如果t>0
那么(a/b)<(a+t)/(b+t)展开既有
所以sina/(sinb+sinc)+sinb/(sinc+sina)+sinc/(sina+sinb)
<（sina+sina）/(sina+sinb+sinc)+（sinb+sinb)/(sinb+sinc+sina)+（sinc+sinc）/(sinc+sina+sinb)=2得证

Answer 2

2bc
=[(3/:3:3;2)*c*c]
=-1/:2
∴a=2c:sinB,b=(3/:b∵sinA:2
根据正弦定理
∴a:sinC=4;[2*(3/2)c
根据余弦定理
cosA=(b^2+c^2-a^2)/2c)^2+c^2-(2c)^2]/:c=4