Question

三角形ABC中,若b的平方=a*c,则cos(A-C)+cosB+cos2B的值是?

Answer 1

由公式c=Sqrt(a^2+b^2+2cosC*a*b),

a*c=a^2+c^2+2*cosB*a*c

cos(A-C)+cosB+cos2B=(cosA*cosC-sinA*sinC)+2*cos^2B+cosB-1=(b^2+c^2-a^2)/(2*b*c)*(a^2+b^2-c^2)/(2*a*b)-Sqrt[1-((b^2+c^2-a^2)/(2*b*c))^2]*Sqrt[1-((b^2+a^2-c^2)/(2*b*a))^2]+2*((a^2+c^2-b^2)/(2*a*c))^2+(a^2+c^2-b^2)/(2*a*c)-1=(a/c+c/a)*(1/2*(a/c+c/a)-1)