已知函数f(x)=2sin^2x-sin(π+2x),x属于R.(1)求f(x)的最小正周期 (2)求函数f(x)的值域

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已知函数f(x)=2sin^2x-sin(π+2x),x属于R.(1)求f(x)的最小正周期 (2)求函数f(x)的值域
已知函数f(x)=2sin^2x-sin(π+2x),x属于R.(1)求f(x)的最小正周期 (2)求函数f(x)的值域

已知函数f(x)=2sin^2x-sin(π+2x),x属于R.(1)求f(x)的最小正周期 (2)求函数f(x)的值域
解f(x)=2sin^2x-sin(π+2x),
=2sin^2x+sin(2x)
=2(1-cos2x)/2+sin(2x)
=1-cos2x+sin(2x)
=sin(2x)-cos2x+1
=√2sin(2x-π/4)+1
即周期T=2π/2=π
由-1≤sin(2x-π/4)≤1
即-√2≤√2sin(2x-π/4)≤√2
即-√2+1≤√2sin(2x-π/4)+1≤√2+1
即-√2+1≤f(x)≤√2+1
即值域为[-√2+1,√2+1]

f(x)=2sin^2x-sin(π+2x),
=1-cos2x+sin2x,
=√2sin(2x-π/4)-1
T=2π/2=π
f(x)的值域:[-1-√2,√2-1]