Question

已知三角形ABC的三个内角A,B,C的对边分别为a,b,c 设角A的对边长a=1,当cosA+2cos(B+C/2)取到

Answer 1

cosA+2cos[(B+C)/2]=cosA+2cos[(pi-A)/2]=cosA+2sin(A/2)

=-2(sin(A/2)-1/2)^2+3/2 sin(A/2)=1/2 时取得最大值 A=60°

S=bc/2*sinA=√3bc/4

1+bc=a^2+2bccosA=b^2+c^2>=2bc

bc