求函数f(⊙)=sin⊙-1 cos⊙-2的最大值和最小值
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f(x)=sin²x=(-1/2)(1-2sin²x)+1/2=-(1/2)cos2x+1/2所以f(x)周期是:π
此题如果去化归的话,可能会比较复杂所以用分析法(1)在定义域内,sin平方x的周期显然是π,根号3sinxcosx=根号1.5*sin2x,所以周期也是π;综上,函数f(x)的最小周期为π(2)(si
由条件知sin²x≤1/4,-1/2≤sinx≤1/2由图像知,-π/6+2kπ≤x≤π/6+2kπ所以f(sin²x)的定义域为[-π/6+2kπ,π/6+2kπ],k∈Z
2kπ-π/2≤2x+π/3≤2kπ+π/2得:kπ-5π/12≤x≤kπ+π/12增区间是:[kπ-5π/12,kπ+π/12],其中k∈Zx∈[-π/6,π/6],则:2x+π/3∈[0,2π/3
f(x)=[2sin(x+π/3)+sinx]cosx-√3sin^2x=[sinx+√3cosx+sinx]cosx-√3sin^2x=2sinxcosx+√3cos^2x-√3sin^2x=sin
f(x)=sin(π-x)sin(π/2-x)+cos²xf(x)=sinxcosx+cos²x=1/2sin2x+1/2+1/2cos2x=1/2(sin2x+cos2x)+1/
f(x)={ln[sin(3x+1)]+C}'=1/sin(3x+1)*cos(3x+1)*3=3cot(3x+1)
因为f(x)=根号3sin(2x-π/6)+2sin的平方(x-π/12)=根号3sin(2x-π/6)-(1-2sin的平方(x-π/12))+1=根号3sin(2x-π/6)-cos(2x-π/6
f(x)=cos(x-π/3)-sin(π/2-x)=(1/2)cosx+(√3/2)sinx-cosx=(√3/2)sinx-(1/2)cosx=sin(x-π/6),它的最小值=-1.
f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a=sinx*cos(π/6)+cosx*sin(π/6)+sinx*cos(π/6)-cosx*sin(π/6)+cosx+a=(√3
1.f(x)=2cosx*sin(x+π/3)-√3﹙sinx﹚^2+sinx*cosx=2cosx*﹙sinxcosπ/3+cosxsinπ/3﹚-√3﹙sinx﹚^2+sinx*cosx=cosx
f(x)=(1+cotx)sin^2(x)-2sin(x+∏/4)sin(x-∏/4)=sin^2(x)+sinxcosx+cos2x=1/2(1-cos2x)+sinxcosx+cos2x=1/2(
f(x)=根号2sin(π/2-x)sin(x+4/π)-1/2=√2cosx(√2/2×sinx+√2/2×cosx)-1/2=sinxcosx+cos²x-1/2=1/2sin2x+1/
(3)若函数f(sinα)的最大值为8.求a、b的值(1)取sinα=1,cosβ=-1代入条件中分别得到:f(1)>=0,f(1)=3令cosβ=1得到f(3)=3(3)∵f(sinα)≥0,f(2
设t=sinx,又cos2x=1-2(sinx)^2则f(t)=1-2(sinx)^2+1=2(1-t^2)所以f(cosx)=2(1-t^2)=2(1-(cosx)^2)=2(sinx)^2
f(x)=sin^2x+sinxcosx-sin^2x+cos^2x=sinxcosx+cos^2x=sin2x/2+(1+cos2x)/2=sin2x/2+cos2x/2+1/2(1)f(a)=si
函数f(x)=负根号3sin^2x+sinxcosx应该没^这个符号的吧?如果是没有的话f(x)=负根号3sin2x+sinxcosx=负根号3sin2x+sin2x=(1/2-3^(1/2))sin
,而sin^2a+cos^2a=1,得sin^2a=4/5f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).=sinx(cosx+sinx)+2sin(x+π/4)
f(x)=cos2x/sin(π/4-x)=(cos²x-sin²x)/[√2/2(-sinx+cosx)]=(cosx+sinx)(cosx-sinx)/[√2/2(cosx-s
求导基本格式①求函数的增量Δy=f(x0+Δx)-f(x0)②求平均变化率③取极限,得导数.Δx趋于0lim(f(x+Δx)-f(x))/Δxlim(sin(x+Δx)-sinx)/Δx和差化积后:=