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F(x)=sinx^4+2倍根号3 sinxCosx%Cosx^4 求(1)函...

max=3/2,min=1 解: f(x) =sin²x+√3sinxcosx =(1-cos2x)/2+(√3/2)sin2x =sin(2x-π/6)+1/2 ∵x∈[π/4,π/2] ∴2x-π/6∈[π/3,5π/6] ∴ f(x)_max=1+1/2=3/2 f(x)_min=1/2+1/2=1

y=sin^4(x)+2√3(sinxcosx)-cos^4(x) =sin^4(x)-cos^4(x)+2√3sinxcosx =(sin^2x+cos^2x)(sin^2x-cos^2x)+2√3(sinxcosx) =2√3(sinxcosx)-cos2x =√3sin2x-cos2x =2sin(2x-π/6) 最小正周期T=2π/2=π 最小值是:2*(-1)=-2 单调增区间是:-Pai/2+2kPai

f(x)=3cosx+4sinx =5×(3/5×cosx+4/5×sinx) =5×(sina×cosx+cosa×sinx),sina=3/5,cosa=4/5 =5sin(x+a) ∵-1≤sin(x+a)≤1 ∴-5≤sin(x+a)≤5 故,函数的值域为[-5,5]。 答题不易,望采纳~~~

-1/4由题得:4sin(a+π/6)+4cosa-根号3=0展开后化简得:(2根号3)sina+6cosa=根号3即:sina+(根号3)*cosa=1/2而sin(a+4π/3)=sina*cos4π/3+cosa*sin4π/3=(-1/2)*[s...

f(x)=4cos∧2x+4√3sinxcosx-2 =2cos2x+2√3sin2x =4sin(2x+π/6), 它的最小正周期=π,最大值=4,此时,2x+π/6=(2k+1/2)π,k∈Z,解得x=(k+1/6)π。 其增区间是[(k-1/3)π,(k+1/5)π],对称轴由2x+π/6=(k+1/2)确定,即x=(k/2+1/6)π。 变式1 f(x)=(5/2)si...

、先将f(x)=m*n化成一个函数式子,f(x)=m*n=√3sinxcosx/16+cosx/4,根据二倍角公式,等于三十二分之根号三倍的sin2x加上八分之cos2x加上1,。并进行提取公因式得到十六分之一倍的括号里二分之根号三倍的sin2x加上二分之cos2x括号结束再加上一,...

f(x)=根号3sin2x+cos2x=2sin(2x+π/6) 1) T=2π/2=π x∈[0,π/2] 2x+π/6[π/6,7π/6] f(x)小=2sin(π/6)=1 f(x)大=2sin(π/2)=2 2) sin(2x0+π/6)=3/5 co2x0=cos(2x0+π/6-π/6)=(3-4根号3)/10 字限制

f(x)=根号3*sinx/4*cosx/4+cos^2*x/4+1/2 =((根号3)/2)sinx/2+(cosx/2+1)/2+1/2 =((根号3)/2)sinx/2+(1/2)cosx/2+1 =cos30°sinx/2+sin30°cosx/2+1 =sin(x/2+30°)+1 周期为4π 对称中心为(-π/3, 1)

解: (1) cos2x-2√3sinxcosx =-√3sin2x+cos2x =-2sin(2x-π/6) T=2π/2=π (2) sinx=3/4 x=2kπ+arcsin(3/4)或x=2kπ+π-arcsin(3/4) 不是特殊值 。

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