Question

在三角形ABC中，角A,B,C所对的边为a,b,c,且（b^2-a^2-c^2）/ac=cos(A+C)/sinAcosA，若sinB/cosC>根号2，

求C的范围

Answer 1

由余弦定理得，
b²-a²-c²=-2ac·cosB
∴(b²-a²-c²)/(ac)=-2cosB
∴-2cosB=cos(A+C)/(sinAcosA)
由于A+B+C=π
∴cosB=-cos(A+C)
∴-2cosB=2cos(A+C)
∴2cos(A+C)=cos(A+C)/(sinAcosA)
∴1/2sin(2A)=1/2
即sin(2A)=1
∴A=π/4
∴ B+C=π-A=3π/4
∴ B=3π/4-C
sinB/cosC>√2
sin(3π/4-C)/cosC>√2
sin(3π/4-C)>√2 cosC
sin(C+π/4)>√2 cosC
√2 sin(C+π/4)>2 cosC
sinC+cosC>2cosC
sinC>cosC
tanC>1
所以
π/4<C<π/2

Answer 2

(b^2-a^2-c^2)/ac=-2cosB
cos(A+C)/sinAcosA=-cosB/(sinAcosA)
2sinAcosA=1
sin2A=1
A=π/4
sinB/cosC=sin(A+C)/cosC=sinA+cosAsinC/cosC>√2
sinC/cosC>1
C的范围是(π/4,π/2)