Question

在△ABC中,内角A,B,C的对边分别为a,b，c，已知a，b，c成等比数列，且cosB=3/4.求cosA\sinA+cosC/sinC

Answer 1

ac=b^2
即(sinA*2R)(sinC*2R)=(sinB*2R)^2
∴sinAsinC=(sinB)^2
cosA/sinA+cosC/sinC
=(cosAsinC+cosCsinA)/sinAsinC
=sin(A+C)/sinAsinC
=sin(π-B)/(sinB)^2
=1/sinB （sinB>0)
=1/√(1-cos^2B)
=1/√(1-9/16)
=4/√7
=4√7/7

Answer 2

b^2=ac，(sinB)^2=sinAsinC。
sinB=√[1-(cosB)^2]=√7/4。
cosA/sinA+cosC/sinC
=(cosAsinC+sinAcosC)/(sinAsinC)
=sin(A+C)/(sinB)^2
=sinB/(sinB)^2
=1/sinB
=4√7/4