Question

在三角形ABC,求证:1:sinA+sinB+sinC=4cos(A/2).cos(B/2).cos(C/2) 2:……

在三角形ABC,求证:1:sinA+sinB+sinC=4cos(A/2).cos(B/2).cos(C/2) 2:cosA+cosB+cosC=1+4sin（A/2）.sin（B/2）.sin（C/2） 3:tanA+tanB+tanC=tanA.tanB.tanC

Answer 1

sinA+sinB+sinC=sinBcosC+sinCcosB+sinB+sinC
=sinB(cosC+1)+sinC(cosB+1)
=2sinB(cos(C/2))^2+2sinC(cos(B/2))^2
=4[sin(B/2)(cosC/2)+sin(C/2)cos(B/2)]cos(B/2)cos(C/2)
=4cos(A/2)cos(B/2)cos(C/2)

cosA+cosB+cosC=-cos(B+C)+cosB+cosC
=sinBsinC-cosBcosC+1-2sin(B/2)^2+1-2sin(C/2)^2
=4sin(B/2)cos(B/2)sin(C/2)cos(C/2)-[1-2(sin(B/2))^2][1-2(sin(C/2))^2+2-2(sin(B/2))^2-2(sin(C/2))^2
=1+4sin(B/2)cos(B/2)sin(C/2)cos(C/2)-4sin(B/2)^2sin(C/2)^2
=1+4sin(B/2)sin(C/2)[cos(C/2)cos(B/2)-sin(C/2)sin(B/2)
=1+4sin(B/2)sin(C/2)sin(A/2)

tanA+tanB+tanC=tan(A+B)(1-tanAtanB)+tanC
=tanC(tanAtanB-1)+tanC
=tanA tan B tanC