设三角形ABC的内角A、B、C对边长分别为a、b、c 且满足cosB/sinB+cosC/sinC=1/sinA

题目

设三角形ABC的内角A、B、C对边长分别为a、b、c 且满足cosB/sinB+cosC/sinC=1/sinA
（1）求证a^2=bc
（2）若cos(B-C)+cosA=1,求角A的大小

答案

cosB/sinB + cosC/sinC = 1/sinA
通分
(cosBsinC + sinBcosC)/sinBsinC = 1/sinA
sin(B+C)/sinBsinC = 1/sinA
sinA /sinBsinC = 1/sinA
所以
(sinA)^2 = sinBsinC
用正弦定理 a/sinA = b/sinB = c/sinC = u
两边同乘u^2
即得
a^2 = bc
若cos(B-C)+cosA=1
cos(B-C)+cosA
= cos(B-C) -cos(B+C)
= 2sinBsinC
所以2sinBsinC =1
由（1）知(sinA)^2 = sinBsinC
所以(sinA)^2 = 1/2
sinA = √2/2
因为 a^2 = bc,a为等比中项,不可能为最长边.A不能为钝角
所以A =45°