在三角形ABC中,已知A,B,C成等差数列,且b=根号3 ,（1）若sinA+cosA=根号2,求a；（2）求三角形ABC面积的最大值

(1)
A,B,C成等差数列,A + C = 2B
A+ B + C = B + 2B = 180˚
B = 60²
sinA+cosA = √2
(sinA + cosA)² = 2
sin²A + cos²A + 2sinAcosA = 2
1 + sin(2A) = 2
sin(2A) = 1
2A = 90˚,A = 45˚
a/sinA = b/sinB
a = bsinA/sinB = √3sin45˚/sin60˚ = √3(√2/2)/(√3/2) = √2
(2)
a/sinA = b/sinB = csinC
a = bsinA/sinB
c = bsinC/sinB
S = (1/2)ac*sinB = (1/2)(sinB)(bsinA/sinB)(bsinC/sinB)
= (1/2)b²sinAsinC/sinB
= [(1/2)*3/(√3/2)]sinAsinC
= √3sinAsinC
= (√3/2)[cos(A - C) - cos(A + C)]
= (√3/2)[cos(A - C) -cos120˚]
= (√3/2)[cos(A - C) + 1/2]
A - C = 0,S最大
S = (√3/2)(cos0˚ + 1/2)
= (√3/2)(1 + 1/2)
= 3√3/4