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已知F x 2sinx sin2x

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f(x)=sin2x-2sin²x =sin2x-(1-cos2x) =sin2x+cos2x-1 =√2sin(2x+π/4)-1 T=2π/2=π 当2x+π/4=2kπ+π/2时函数取到最大值 此时x=kπ+π/8 f(x)max=√2-1

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

二倍角公式化简f(x)后使用等价无穷小替换即可

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

1、 f(x)=sinx(sinxcosπ/6+cosxsinπ/6) =√3/2*sin²x+1/2*sinxcosx =√3/2*(1-cos2x)/2+1/2*(sin2x)/2 =√3/4+1/2*(1/2*sin2x-√3/2*cos2x) =√3/4+(sin2xcosπ/3-cos2xsinπ/3) =√3/4+sin(2x-π/3) 所以T=2π/2=π 2、 0≤x≤π/2 -π/3≤2x-π/3≤2π/3 -√3...

解:先用降幂公式把函数化为:f(x)=√3/2sin2x-1/2cos2x-1=sin(2x-π/6)-1 (1)最小值为-2,最小正周期为π (2)由f(C)=0知sin(2C-π/6)=1,从而可得C=π/3,再由余弦定理知:c^2=a^2+b^2-2abcosC 3=a^2+4a^2-2a*2acosπ/3,解得a=1,故b=2

f(x)=sin²x+sinxcosx+cos2x =2分之(1-cos2x)+(2分之1)sin2x+cos2x =2分之1+(2分之1)(sin2x+cos2x) =(2分之√2)sin(2x+45°)+(2分之1) 最小正周期 = 2π÷2 = π

(I)由2sinx≠0,得x≠kπ,(k∈Z),所以f(x)的定义域为{x|x≠kπ,k∈Z}.(II)当x≠kπ,(k∈Z)时f(x)=sin2x?cos2x+12sinx=2sinxcosx+2sin2x2sinx=sinx+cosx=2sin(x+π4),所以f(x)的值域为{y|?2≤y≤2,且y≠±1}.(III)因为α是锐角,且tanα2...

f(x)的最小正周期T=2Pai/2=Pai,初相是:Pai/4 g1(x)=sin(2(x-Pai/8)+Pai/4)=sin2x g(x)=sinx 3.h(x)=|sinx-1/2|sinx =(sinx)^2-1/2sinx,(sinx>=1/2) =(sinx-1/4)^2-1/16 故h(x)max=h(1)=1/2,h(x)min=h(1/2)=0 h(x)=1/2sinx-(sinx)^2,(-1

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