Question

求大神来解~谢谢

Answer 1

y=cos2x+sinx=1-2sin^2x+sinx= -2(sinx-1/4)^2+9/8
-1<=sinx<=1,
y=cos2x+sinx值域为[-2,0]
y=cosx(cosx+sinx)=cos^2x+sinx*cosx
=1/2(2cos^2x-1)+1/2*2sinx*cosx+1/2
=1/2cos2x+1/2sin2x+1/2
=√2/2sin(2x+π/4)+1/2
y=cosx(cosx+sinx)值域为[(1-√2)/2,(1+√2)/2]
y=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)
=cos(2x-π/3)-1/2*2[cos(x-π/4+x+π/4)-cos(x-π/4-x-π/4)]
=cos(2x-π/3)-cos2x+cosπ/2
=1/2cos2x+√3/2sin2x-cos2x
=√3/2sin2x-1/2cos2x
=sin(2x-π/6)
y=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)值域为[-1,1]

Answer 2

y=cos2x+sinx=1-2sin^2x+sinx= -2(sinx-1/4)^2+9/8
-1<=sinx<=1,
y=cos2x+sinx值域为[-2,0]
y=cosx(cosx+sinx)=cos^2x+sinx*cosx
=1/2(2cos^2x-1)+1/2*2sinx*cosx+1/2
=1/2cos2x+1/2sin2x+1/2
=√2/2sin(2x+π...

Answer 3

还需要不，是不是考试