Question

在三角形ABC中,角A,B,C所对的边分别为a,b,c,且c^2=a^2+b^2-ab

Answer 1

(1)由c^2=a^2+b^2-ab可得a^2+b^2-c^2=ab

又cosC=(a^2+b^2-c^2)/2ab=1/2

所以在三角形中C=π/3

tanA-tanB=√3/3(1+tanAtanB)

又tan(A-B)=(tanA-tanB)/(1+tanAtanB)=√3/3

所以A-B=π/6

又A+B=180-C=2π/3

所以B=π/4

(2)m*n=3sinA+cos2A=-2(sinA)^2+3sinA+1

=-(sinA-3/2)^2+1+9/4

因为A的范围为(0,2π/3)

当sinA=1时,m*n有最大值3