设函数f(x)=m(cosx+sinx)^2 +1-2sin^2 x ,x属于R 且y=f(x)

显示全部楼层 · 2013-1-9 20:21:46

1,直接代点进去，有m（√2/2+√2/2)^2+1-2×(√2/2)^2=2,所以m=12.f（x）=1×(cos^2 x+sin^2 x+2sinx×cosx)+1-2sin^2x=sin2x+2-2sin^2 =sin2x+2(1-sin^2x)=sin2x+2cos^2 x=sin2x+cos2x+1=√2(√2/2×sin2x+√2/2×cos2x)+1=√2sin(2x+Π/4)+1.显然f（x）的最小值为1-√2.当2x+Π/4=-Π/2+2kΠ（k属于Z)时取得最小值，可得x的集合为x=-3Π/8+kΠ（k属于Z).附加公式cos2x=2cos^2 x-1,sin^2 x+cos^2 x=1.如觉得有用，...

千问 · 2013-1-9 20:21:46

f(x)=m(sin^2x+cos^2x)+msin2x+cos2x=m+msin2x+cos2xf(π/4)=2=m+m,( sinπ/2=1,cosπ/2=0)m=1f(x)=1+sin2x+cos2x=1+√2sin(2x+π/4)最小值为1-√2，此时2x+π/4=2kπ+π2x=2kπ+3π/4x=kπ+3π/8, ...