Question

在三角形ABC，设A,B,C的对边分别是a,b,c，∠C=60°,c=√3。（1）(a+2√3cosA)sinB的值。（2）若sinA=√3/3

求三角形ABC的面积

求速度

Answer 1

c/sinC=a/sinA=b/sinB,a/sinA=√3/sin60
a=2sinA,b=2sinB
(a+2√3cosA)/sinB=(2sinA+2√3cosA)/sinB
=2(sinA+√3cosA)/sin(120-A)
=2(sinA+√3cosA)/(sin120cosA-cos120sinA)
=2(sinA+√3cosA)/(√3/2cosA+1/2sinA)
=4
S=1/2cbsinA=1/2*√3*√3/3*2sinB=sinB
sinB=sin(A+C)=sinAcosc+cosAsinc
sinA=√3/3,cosA=2√2/3,sinc=sin60=√3/2,cosc=cos60=1/2
sinB=sin(A+C)=sinAcosc+cosAsinc=√3/3*1/2+2√2/3*√3/2
=(√3+2√6)/6
S=1/2cbsinA=1/2*√3*√3/3*2sinB=sinB=(√3+2√6)/6

Answer 2

解：∵a=c.sinA\sinC ∴a=2sinA ∵C=60º∴A+B=120º
∴原式=(2sinA+2√3cosA)sin(120º-A)
=2√3sinAcosA+sin²A+3cos²A
=√3sin2A+2cos²A+cos²A+sin²A
=√3sin2A+2cos²A+1
=√3sin2A+2cos²A-1+2
=√3sin2A+cos2A+2
=2sin(2A+π\6)+2
就化间到这里了 我觉得你的题目貌似没写全 要是求最大最小值就分别是4和0

Answer 3

2