作业帮已知函数Y=sin^2x/2++√3/2sinx,求其最小正周期及其值域,单调递增区间
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已知函数Y=sin^2x/2++√3/2sinx,求其最小正周期及其值域,单调递增区间
已知函数Y=sin^2x/2++√3/2sinx,求其最小正周期及其值域,单调递增区间
已知函数Y=sin^2x/2++√3/2sinx,求其最小正周期及其值域,单调递增区间
y=sin^2x/2++√3/2sinx
=(1-cosx)/2+√3/2sinx
=1/2+√3/2sinx-1/2cosx
=1/2+sin(x-π/6)
T=2π
值域为[-1/2,3/2]
单调递增区间为2kπ-π/3
先用半角公式，再化简就行
y=sin^2x/2++√3/2sinx
=(1-cosx)/2+√3/2sinx
=sin(x-π/6)+1
因此函数最小正周期T=2π
值域为[0,2]
单调递增区间为[2kπ-π/3,2kπ+2π/3],k∈Z
Y=sin^2x/2++√3/2sinx
=(1-cosx)/2+√3/2sinx
=1+sin(x-π/6)
∴函数最小正周期为2π
值域为[0,2]
单调递增区间为[2kπ-π/3,2kπ+2π/3],k∈Z
y=sin^2x/2++√3/2sinx
=(1-cosx)/2+√3/2sinx
=1/2+√3/2sinx-1/2cosx
=1/2+sin(x-π/6)
T=2π
值域为[-1/2,,3/2]
单调递增区间为2kπ-π/3
Y=sin^2x/2+√3/2sinx
=sin^2x/2+√3/2sin(2*x/2)
=sin^2x/2+√3/2*2sinx/2cosx/2
=sinx/2(sinx/2+√3cosx/2)
=2sinx/2(1/2sinx/2+√3/2cosx/2)
=2sinx/2sin(x/2+π/3)
=2*【cos(x/2-x/2-π/3)-cos(...
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Y=sin^2x/2+√3/2sinx
=sin^2x/2+√3/2sin(2*x/2)
=sin^2x/2+√3/2*2sinx/2cosx/2
=sinx/2(sinx/2+√3cosx/2)
=2sinx/2(1/2sinx/2+√3/2cosx/2)
=2sinx/2sin(x/2+π/3)
=2*【cos(x/2-x/2-π/3)-cos(x/2+x/2+π/3)】/2
=cos（-π/3）-cos（x+π/3）
=-cos(x+π/3)-1/2
最小正周期为2π
值域cosπ<=cosx<=cos0 cos(π+π/3)<=cos(x+π/3)<=cosπ/3
-cosπ/3<=-cos(x+π/3)<=-cos(π+π/3)
-1/2-cosπ/3<=-1/2-cos(x+π/3)<=-1/2-cos(π+π/3) 即值域-1<=-1/2-cos(x+π/3)<=0
增区间cosπ<=cosx<=cos3π/2 cos(π+π/3)<=cos(x+π/3)<=cos(3π/2+π/3）
-cos11π/3<=-cos(x+π/3)<-cos4π/3
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