Question

在△ABC中，内角A、B、C的对边分别为a，b，c，已知b^2=ac,且cosB=3/4.

（1）若△ABC面积S△ABC=√7/4，求a+c的值；（2）求cosA/sinA+cosC/sinC的值

Answer 1

由cosB=3/4，可得cos(B/2)=√14/4.
由b^2=ac=BC*BA=3/2, 可得b=√6/2.
由(sinB)^2 = sinAsinC=1/2 * (cos(A-C)-cos(A+C))=1/2 * (cos(A-C)+cosB), 解得cos(A-C)=1/8,
进而可得cos((A-C)/2)=3/4.
由正弦定理可得a+c=b*(sinA + sinC)/sinB=2bsin((A+C)/2)cos((A-C)/2)/sinB
=2bcos(B/2)cos((A-C)/2)/sinB
=2*√6/2 * √14/4 * 3/4 /(√7/4)=3√3/2. cotA + cotC = (cosAsinC + cosCsinA)/(sinAsinC)=sin(A+C)/(sinB)^2 = 1/sinB = 4√7/7.