Question

在三角形abc中,角a,b,c,所对的边分别为a,b,c,已知asin^2b/2+bsin^2a

2=c/2.求证：a,c,b成等差数列；

Answer 1

证明：
(1)
已知asin²（B/2）+bsin²（A/2）=c/2

∵cosB = 1-2 sin²（B/2）

∴sin²（B/2）=(1- cosB)/2

同理,

cosA = 1-2 sin²（A/2）

∴sin²（A/2 =(1- cosA)/2

∴asin²（B/2）+bsin²（A/2）= c/2

→a×(1- cosB)/2＋b×(1- cosA)/2 = c/2

→a－a×cosB ＋b－b×cosA = c

a＋b－c = a×cosB+ b×cosA

根据余弦定理,可得

cosB = (a2 + c2 - b2) / (2·a·c)

cosA = (c2 + b2 - a2) / (2bc)

代入到a＋b－c = a×cosB+ b×cosA,得

a＋b－c = a×[(a2 + c2 - b2) / (2·a·c)]+ b×[(c2 + b2 - a2) / (2bc)]

=( a2 + c2 - b2＋c2 + b2 - a2)/2c

= 2 c2/2c

=c
a＋b－c = c → a＋b = 2c ∴ a,b,c成等差数列