数学人才们,为什么要那么做!拍照发送的答案更关注!

^2-a^2-c^2/c^2-a^2-b^2=-2accosB/-2abcosC=sinCcosB/sinBcosC
sincsinbcosc=(2sina-sinc)sinccosb
sinbcosc=2sinacosb-sinccosb
sinbcosc+sinccosb=2sinacosb
sina=2sinacosb
cosb=1/2
B=60°
（2）T=1-cos2A/2+3/4+1-cos2C/2
=1+3/4-cos2A/2-COS(240°-2A)/2
=7/4-(cos2A/2-(cos2A/2-根号3/2*sin2A)/2
=7/4-COS2A/4+根号3/4*sin2A
=7/4+1/2sin(2A-30°）
1/2 ＞1 / 2sin(2A-30°） ＞-1/4
9/4 ＞ T＞3/2

详解过程如下图```

采纳发第二问详解过程·····

(1)
c^2sinC-a^2sinC-b^2sinC=2b^2sinA-2a^2sinA-2c^2sinA-b^2sinC+a^2sinC+c^2sinC
sinC(-a^2)=(b^2-a^2-c^2)sinA
因为a/sinA=b/sinB=c/sinC
所以sinC=csinB/b
sinA=asinB/b
(csinB/b)(-a^2)=(...