y=2(sinx)^2+sin2x=sin2x-cos2x+1=√2sin(2x-π/4)+1
最小正周期为T=2π/2=π,选C.sin2x-cos2x+1=√2sin(2x-π/4)+1这一步怎么转化的啊？求解？好持久的追问sin2x-cos2x+1=√2[(√2/2)sin2x-(√2/2)cos2x]+1=√2(sin2xcosπ/4-cos2xsinπ/4)+1=√2sin(2x-π/4)+1π\4怎么来的？√2/2=sinπ/4=cosπ/4