f(x)=2sin^2(x)-cos^2(x)+2sinxcosx-1
=2sin^2(x)-1-cos^2(x)+1/2+2sinxcosx-1/2
=1/2cos2x+sin2x-1/2
=√5/2[1/√5cos2x+2/√5sin2x]-1/2
=√5/2cos(2x+t)-1/2
其中sint=1/√5,cost=2/√5
因此f(x)的值域是[-√5/2-1/2,√5/2-1/2]
求值域 f(x)=2sin^2(x)-cos^2(x)+2sinxcosx-1
1个回答
相关问题
-
f(x)=sin^4(x)-sinxcosx+cos^4,求f(x)的值域.化解得 1-1/2Sin^2(x)-Sinx
-
求函数f(x)=(sinx+cos^4x+sin^2xcos^2x)/2-2sinxcosx-1/2sinxcosx+1
-
已知函数f(x)=cos^4x-2sinxcosx-sin^4x,求f(x)的值域
-
y=sin^2x+3sinxcosx+5cos^2x值域
-
f(x)=2sinxcosx+cos^2x-sin^2x
-
y=sin'2x+根号3sinxcosx+2cos'2x求值域和周期
-
已知f(x)=2sinxcosx+cos2x,x∈【0,π/2]求值域
-
函数f(x)=sin^x+2倍根号3sinxcosx+3cos^2x若x属于[0,π/2]求值域?
-
1.函数f(x)=[sin(x)-1]/√[3-2cos(x)-2sin(x)]求值域
-
求三角函数的值域 f(x)=4cos²x-2sinxcosx