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问题: 高1数学题8-TOY专问

求值域y=sinxcosx+sinx+cosx

解答:

解1.
令t=sinx+cosx--->t=2^.5*sin(x+pi/4)--->-2^.5=<t=<2^.5
--->t^2=1+2sinxcosx--->sinxcosx=(t^2-1)/2
y=(t^2-1)/2+t
=(t^2+2t-1)/2
=[(t+1)^2-2]/2
=0.5(t+1)^2-1
t属于[-2^.5,2^.5]--->ymin=y(-1)=-1 & ymax=y(2^.5)=1/2+2^.5
值域是[-1,1/2+2^.5]
解2.
y=.5*sin2x+2^.5*cos(x-pi/4)
=0.5cos(pi/2-2x)+2^.5*cos(x-pi/4)
=0.5cos[2(x-pi/4)]+2^.5cos(x-pi/4)
=[cos(x-pi/4)]^2+2^.5*cos(x-pi.4)-1/2
=[cos(x-pi/4)+(2^.5)/2]^2-1
-1=<cos(x-pi/4)=<1
--->-1=<y=<1/2+2^.5.