三角形ABC中内角ABC对应边abc,求证:(cotA/4 --cscA/2)/(cotB/2+cotC/2)=(b+c-a)/2a

题目

三角形ABC中内角ABC对应边abc,求证:(cotA/4 --cscA/2)/(cotB/2+cotC/2)=(b+c-a)/2a

答案

cotA/4=cosA/4/sinA/4 =2cos²A/4/(2sinA/4cosA/4) =(1+cosA/2)/sinA/2 故cotA/4-cscA/2 =(1+cosA/2)/sinA/2-1/sinA/2 =cosA/2/sinA/2 =(1+cosA)/sinA cotB/2+cotC/2 =cosB/2/sinB/2+cosC/2/sinC/2 =(1+cosB)/sinB+(1+cosC)/sinC =(sinC+sinCcosB+sinB+sinBcosC)/(sinBsinC) =(sinC+sinB+sinA)/(sinBsinC) 故(cotA/4-cscA/2)/(cotB/2+cotC/2) =[(1+cosA)/sinA]/[(sinC+sinB+sinA)/(sinBsinC)] =[(1+cosA)(sinBsinC)]/[(sinC+sinB+sinA)sinA] 由正弦定理,余弦定理 原式={[1+(b²+c²-a²)/(2bc)]bc}/[(a+b+c)a] =[(b²+c²-a²+2bc)/2]/[(a+b+c)a] =[(b+c)²-a²]/[2(a+b+c)a] =(a+b+c)(b+c-a)/[2(a+b+c)a] =(b+c-a)/(2a)