v_fox
幼苗
共回答了13个问题采纳率:92.3% 举报
y=2sin²x+2sinxcosx=(1-cos2x)+sin2x=√2sin(2x-π/4)+1
故y的最小周期是π
x属于[-π/2,π/2] 2x-π/4属于[-5π/4,3π/4]
故当2x+π/4=-π/2时,y取最小值=1-√2
当2x+π/4=π/2时,y取最大值=1+√2
y=sin²x+2sinxcosx+cos²x+2cos²x=1+sin2x+(1+cos2x)=2+sin2xcos2x=2+√2sin(2x+π/4)
1年前
4