△ABC中,(2a+c)cosB+bcosC=0,求y=sin²A+sin²C的范围
网友回答
(2a+c)cosB+bcosC=0
由正弦定理,我们有：
a=2RsinA b=2RsinB
代入上式,得
(2sinA+sinC)cosB+sinBcosC=0
2sinAcosB=-(sinCcosB+sinBcosC)=sin(B+C)=-sinA
cosB=-1/2
故B=2π/3
y=sin²A+sin²C
=1-1/2(cos2A+cos2C)
=1-cos(A+C)cos(A-C)
=1-(1/2)cos(A-C)
A-C=π/3-2C
======以下答案可供参考======
供参考答案1:
。。。供参考答案2:
应该是【1/2，3/4)