f(x)=sin²ωx+√3cosωxXcos(π/2-ωx)(ω>0),且函数y=f(x)的图像相邻两条对称轴之间的距离为π/2.1）求ω的值及f(x)的单调递增区间（2）在△ABC中,a,b,c分别是角A,B,C的对边,若a=√3,b=√2,f(A)=3/2,求角C

1）f(x)=sin²ωx+√3cosωxXcos(π/2-ωx)
=sin²ωx+√3cosωxsinωx
=(1-cos2ωx)/2+√3/2sin2ωx
=1/2-1/2cos2ωx+√3/2sin2ωx
=1/2+sin(2ωx-π/6)
函数y=f(x)的图像相邻两条对称轴之间的距离为π/2
所以y=f(x)的周期为π
则2π/2ω=π 所以ω=1

2kπ- π/2 ≦2x-π/6≦2kπ+π/2
解得kπ- π/6≦x≦kπ+π/3,k为整数
所以f(x)的单调递增区间〔kπ- π/6,kπ+π/3,〕

（2）f(A)=1/2+sin(2A-π/6)=3/2
解得A=π/3
有余弦定理a²=b²+c²-2bccosA
将a=√3，b=√2，A=π/3代入得c=(√2+√6)/2
再根据正弦定理a/sinA=c/sinC
解得sinC=(√2+√6)/4
C=75°

1、显然cos(π/2-ωx)=sinωx,故√3cosωx *cos(π/2-ωx)=√3sinωx *cosωx= 0.5√3 sin2ωx而sin²ωx=0.5*(1-cos2ωx),故f(x)=0.5*(1-cos2ωx) + 0.5√3 sin2ωx=0.5√3 sin2ωx - 0.5cos2ωx +0.5=cos(π/6...