数学正余弦定理题
问题描述：
数学正余弦定理题
在三角形ABC中角A,B,C所对的边分别为a,b,c,BC=根号5,AC=3,sinC=2sinA,⑴求AB的值⑵求sin（2A-π/4）的值
答
(1)由正弦定理,sinC/sinA = c/a = 2所以AB = c = 2a = 2根号5(2)sin（2A-π/4）= sin2Acosπ/4 - cos2Asinπ/4由余弦定理cosA = (b^2 + c^2 - a^2)/2bc = 2/根号5sinA = 1/根号5所以sin2A = 2sinA...