Question

三角形ABC中,面积sinA*cosB-sinB=sinC-sinA*cosC且周长为12,则其面积最大为

Answer 1

sinAcosB-sinB=sinC-sinAcosC
sinAcosB-sin(A+C)=sin(A+B)-sinAcosC
sinAcosB-sinAcosC-cosAsinC=sinAcosB+cosAsinB-sinAcosC
-cosAsinC=cosAsinB
cosA(sinB+sinC)=0
cosA=0或
sinB+sinC=0(因为B,C属于[0～180度]
所以 sinB>0,sinC>0,所以舍)
所以cosA=0,所以A=90度
所以△ABC为Rt△ABC
所以S△ABC=6

Answer 2

S = a*c*sinB/2 = b*c*sinA/2 = b*a*sinC/2 的最大值？
2S = a*c*sinB = b*c*sinA = b*a*sinC

sinA*cosB-sinB=sinC-sinA*cosC= -(sinAcosC - sinC),
a+b+c = 12,

sin(B+C)*cosB -sinB = sinC - sin(B+C)*cosC,

[sinBcosC+cosBsinC]*cosB - sinB = sinC -[sinBcosC+cosBsinC]*cosC,

sinBcosBcosC +cosBcosBsinC - sinB = sinC - sinBcosCcosC - cosBsinCcosC,

sinB*cosB*a/b - sinB = sinC - sinC*cosC*a/c,

2S*cosB - bc*sinB = bc*sinC -2S*cosC,

2S = bc*(sinB+sinC)/(cosB+cosC),

留个记号，等答案

上面人证明了A = PI/2, so
b*b+c*c = a*a,
a+b+c = 12,

勾三，股四，弦5是一个解，

2bc = (12-a)^2 - a*a = 144 - 24a,
so S = 36 - 6a , so a = 6 - S/6,

b*b + c*c = (6-S/6)^2 = 36 + S*S/36 - 2S,
6-S/6+b+c = 12, so c = 6+S/6-b,

so b*b + (6+S/6-b)^2 = 36-2S + S*S/36,

so b*b + 36 + S*S/36 + b*b + 2S - 12b - b*S/3 = 36 -2S + S*S/36,

2*b*b +4S - 12b - b*S/3 = 0,

(4-b/3)*S = 12b - 2*b*b,

S = (12b-2*b*b)/(4-b/3)

S' = (12 - 4*b)/(4-b/3) -(12b-2*b*b)*(-1/3)/[(4-b/3)^2]
= [(12-4*b)*(4-b/3)-(12b-2*b*b)*(-1/3)]/[(4-b/3)^2]
=[48-4b-16b+4*b*b/3 +4b-2*b*b/3]/[(4-b/3)^2]
=[2*b*b/3 -16b + 48]/[(4-b/3)^2]
= (2/3)*[b*b - 24b + 72]/[(4-b/3)^2]
let S'=0, so (b-12)^2 -144+72=0, so (b-12)^2 = 72, so b= 12+ sqrt(72), or b= 12- sqrt(72),

so b = 12- 6sqrt(2) = 3.514719,

so S = [12(12-6sqrt(2) - 2*(12- 6sqrt(2))*(12- 6sqrt(2))]/[4-(12- 6sqrt(2))/3]
= [ 144-72sqrt(2) -288 + 288sqrt(2) -144] / [2sqrt(2)]
= [ 216sqrt(2) -288] /[2sqrt(2)]
= 108-72sqrt(2) ,

c = 3.514719 = b, 三角形为等腰三角形面积最大。