- 高一数学题,有点难已知函数y=(1/2)cos^2x+(√3/2
- 已知y=(1/2)cos^2x+(√3/2)sinxcosx+1, x为全体实数
⒈当函数y取得最小值时,求自变量的集合
⒉该函数的图象可由y=sinx的图象经过怎样的平移和伸缩变换得到的
- 1.
y=(1/2)(cx)^2+(√3/2)sinxcosx+1
=(1/4)[2(cosx)^2+(2√3)sinxcosx]+1
=(1/4)[1+cos2x+(√3)sin2x]+1
=(1/4)[cos2x+(√3)sin2x]+(5/4)
=(1/2)sin[2x+(π/6)]+(5/4)
既:y=(1/2)sin[2x+(π/6)]+(5/4)
当sin[2x+(π/6)]=-1时,ymin=3/4,此时:
2x+(π/6)=2kπ-(π/2)
既:x=kπ-(π/6),k∈Z
2.
对y=sinx做下变换:
纵坐标缩短(1/2),y=sinx ==> y=(1/2)sinx
横坐标缩短(1/2),y=(1/2)sinx ==> y=(1/2)sin2x
向右平移π/12,y=(1/2)sin2x ==>y=(1/2)sin2[x+(π/12)]
==>y=(1/2)sin[2x+(π/6)]
整体上移(3/4), ==> y=(1/2)sin[2x+(π/6)]+(3/4)