△ ABC中,满足acos ²C/2+cos²A/2=（3/2）b ①求证a+c=2b②B= π/4,b=2,求S△ ABC

△ ABC中,满足acos ²C/2+cos²A/2=（3/2）b ①求证a+c=2b②B= π/4,b=2,求S△ ABC
△ ABC中,满足acos ²C/2+cos²A/2=（3/2）b ①求证a+c=2b②B= π/4,b=2,求S△ ABC

△ ABC中,满足acos ²C/2+cos²A/2=（3/2）b ①求证a+c=2b②B= π/4,b=2,求S△ ABC
(1) 你题目抄漏了,cos²(A/2)前面有个c
acos²(C/2)+ccos²(A/2)=(3/2)b
a(1+cosC)/2+c(1+cosA)/2=(3/2)b
a(1+cosC)+c(1+cosA)=3b
由正弦定理得
sinA(1+cosC)+sinC(1+cosA)=3sinB
sinA+sinAcosC+sinC+sinCcosA=3sinB
sinA+sinC+sin(A+C)=3sinB
sinA+sinC+sinB=3sinB
sinA+sinC=2sinB
由正弦定理得
a+c=2b
(2)
由余弦定理得
cosB=(a²+c²-b²)/(2ac)
[(a+c)²-2ac-b²]/(2ac)=cosB
[(2b)²-2ac-b²]/(2ac)=cos(π/4)
(3b²-2ac)/(2ac)=√2/2
b=2代入,整理,得
(2+√2)ac=12
ac=12/(2+√2)=12(2-√2)/2=6(2-√2)
S△ABC=(1/2)acsinB
=(1/2)·6(2-√2)·sin(π/4)
=(1/2)·6(2-√2)·(√2/2)
=3√2-3