Question

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

Answer 1

(1)化简得y=sin(2ωx-π/6)+1/2,由y=f(x)的图像相邻两条对称轴之间的距离为π/2.
则T=π,ω=1,单调递增，2kπ-π/2<=2x-π/6<=2kπ+π/2
kπ-π/6<=x<=kπ+π/3
(2)f(A)=3/2,即sin(2A-π/6)=1,A=π/3
sinA=(√3)/2
又正弦定理，即a>b
sinb=(√2)/2
B=π/4
C=π-A-B=5π/12