Question

数学这道题怎么做

Answer 1

b/sinB=a/sinA=c/sinC=2R=2
sinA=a/2,sinC=c/2 ,sinB=b/2
a=2sinA ,b=2sinB
2[(sinA)^2-(sinC)^2]=(sinA-sinB)b

2[a^2/4-c^2/4]=(a/2-b/2)b
a^2-c^2=(a-b)b
a^2-c^2=ba-b^2
a^2+b^2-c^2=ab
cosC=(a^2+b^2-c^2)/2ab=ab/2ab=1/2
cosC=1/2
C=π/3

S=1/2absinC=1/2absinπ/3=√3/4ab=√3/4*4*sinA sinB
=√3sinA sinB
=√3sinA sin(2π/3-A)
=√3sinA(√3/2cosA+1/2sinA)

=3/4sin2A+√3/2(sinA)^2
=3/4sin2A+√3/4(1-cos2A)
=√3/2(√3/2sin2A-1/2cos2A)+√3/4
=√3/2sin(2A-π/6)+√3/4
A(0,2π/3)
2A-π/6在(-π/6,7π/6)
sin(2A-π/6)在(-π/6,7π/6)值域为(-1/2,1 ]
S=√3/2sin(2A-π/6)+√3/4在(-π/6,7π/6)值域为:(0,3√3/4]
所以,S的最大值=3√3/4

Answer 2

2(a²/4-c²/4)=(a/2-b/2)/b

追答

a²-c²=ab-b²

所以cosc=1/2

s=1／2*2Sina*2sinb*√3／2

s=√3Sinasinb

s=√3sina（√3cosa+1／2sina）