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函数F(x)=sin2x+2sinx在[π,2π]上的最小值为?

f(x)=sin2x+2sinx f'(x)=2cos2x+2cosx =2(2cos²x-1)+2cosx =2(2cos²x+cosx-1) 驻点cosx=-1 cosx=½ x=2kπ±π x=2kπ±π/3 经判断x=2kπ±π不是极值点 x=2kπ-π/3是极小值点 x=2kπ+π/3是极大值点 ∴单调递增区间为x∈(2kπ-π/3,2kπ+π/3)

f(x)=sin²x+√3sinxsin(x+½π) =(1-cos2x)/2+√3sinxcosx =-1/2cos2x+√3/2sin2x+1/2 =sin(2x-π/6)+1/2 最小正周期T=2π/2=π f(x)在区间【0,½π】上的取值范围 因为x∈[0,π/2] 则2x∈[0,π] 则2x-π/6∈[-π/6,5π/6] 所以根据三角函数的性...

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

(1)保证tanx及tan(x/2)有意义,有x≠kπ+π/2,x/2≠kπ+π/2,联立解得x≠kπ+π/2且x≠2kπ+π (k∈Z) (2)y=sin2xtanx+sinxtan(x/2) =2sinxcosxsinx/cosx+sinx[(1-cos)/sinx] =2sin2x+1-cos =2(1-cos2x)+1-cosx = -2cos2x-cosx+3 =(25/8)-2(cosx+1/4)2 当co...

f(x)=(sin2x-2sin²x)/sinx =(2sinxcosⅹ-2sin²x)/sinx =2(cosx-sinx) =2√2[cosxcos(π/4)-sinxsin(π/4)] =2√2cos(x+π/4) 显然,定义域为x∈R,即(-∞,+∞). cos(x+π/4)=1, 即x=2kπ-π/4时, 最大值f(x)|max=2√2. 0

f(x)=2sinxcosx+2√3sin²x-√3 =2sinxcosx+√3sin²x+√3(sin²x-1) =2sinxcosx+√3sin²x-√3cos²x =2sinxcosx+√3(sin²x-cos²x) =sin2x-√3cos2x =2(1/2 sin2x-√3/2 cos2x) =2sin(2x-π/3) sinx的单调递减区间为[π/2+2...

2x+π/3∈(π/3,4π/3) 2x+π/3看成一个整体a.取值a∈(π/3,4π/3) sin a (-√3/2,1]就是个sin的函数. 所以sin(2x+π/3)∈(-√3/2,1]

函数f(x)=sinxcosx-cos2x=12sin2x-1+cos2x2=12(sin2x?cos2x)?12=22sin(2x?π4)?12.∴T=π.∴命题①错误;由?π2+2kπ≤2x?π4≤π2+2kπ,k∈Z.解得:?π8+kπ≤x≤3π8+kπ,k∈Z.取k=0,得?π8≤x≤3π8.∴f(x)在区间(0,π8)上为增函数.∴命题②正确;取x=3π8...

(Ⅰ)∵f(x)=sinxcosx+sin2x,∴f(π4)=sinπ4cosπ4+sin2π4,…(1分)=(22)2+(22)2 …(4分)=1.…(6分)(Ⅱ)f(x)=sinxcosx+sin2x=12sin2x+1?cos2x2,…(8分)=12(sin2x?cos2x)+12=22sin(2x?π4)+12,…(9分)由x∈[0,π2]得 2x?π4∈[?π4,3π4],...

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