△ABC中,(2a+c)cosB+bcosC=0,求y=sin²A+sin²C的范围
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![△ABC中,(2a+c)cosB+bcosC=0,求y=sin²A+sin²C的范围](/uploads/image/z/1600459-43-9.jpg?t=%E2%96%B3ABC%E4%B8%AD%2C%282a%2Bc%29cosB%2BbcosC%3D0%2C%E6%B1%82y%3Dsin%26%23178%3BA%2Bsin%26%23178%3BC%E7%9A%84%E8%8C%83%E5%9B%B4)
△ABC中,(2a+c)cosB+bcosC=0,求y=sin²A+sin²C的范围
△ABC中,(2a+c)cosB+bcosC=0,求y=sin²A+sin²C的范围
△ABC中,(2a+c)cosB+bcosC=0,求y=sin²A+sin²C的范围
(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
。。。
应该是【1/2,3/4)