f(x)=sin(x^2)在x=0处有极限么
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①原式=f(x)=2cos2x+sinx^2=2cos2x+1-cos2x/2=3/2cos2x+1/2故f(π/3)=3/2*cos2π/3+1/2=-3/4+1/2=-1/4②依f(x)=3/2c
f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T
1、由于函数g(x)=sin(2(x-a)+π/3)为偶函数,所以g(x)的图像关于y轴对称,即函数g(x)当x=0时取得最值,所以g(0)=±1,解得sin(π/3-2a)=±1,sin(2a-π/
f(x)=sin2x-2sin^2x=sin2x+cos2x-1=√2sin(2x+π/4)-1.(1)T=2π/2=π.(2).当2x+π/4=2kπ+π/2,k∈Z,即x=kπ+π/8,k∈Z时,
因为f(x)是定义在R上的偶函数,且对任意实数x满足f(x+2)=f(x)所以f(x)的周期为2f(-3)=f(3)f(-2)=f(2)f(0)=f(2)f(1)=f(3)f(x)在(-3,-2)上单
周期是4π,减区间为[π/2,π]
设x0所以f(-x)=sin2(-x)+cos(-x)=-sin2x+cosx因为f(x)为奇函数,所以f(-x)=-f(x)得f(x)=-f(-x)=sin2x-cosx(x
(1)f(x)=cos(-x/2)+sin(π-x/2)=cos(x/2)+sin(x/2)=√2[(√2/2)cos(x/2)+(√2/2)sin(x/2)]=√2[sin(π/4)cos(x/2)
π2x+π/6属于[π/2+2kπ,3π/2+2kπ]时为减区间,所以x属于[π/6+kπ,2π/3+kπ],k属于Z列表:三行2x+π/60π/2π3π/22πx(根据上面一行的值求出x对应的值)f
f(x)=sin2x+cos2x-1=√2sin(2x+π/4)-1.1、最小正周期是π,最大值时2x+π/4=2kπ+π/2,即x=kπ+π/4,k是整数.再问:已知函数f(x)=2sin(∏-X)
cosx=sinx是取得最大值,m=1再问:能给我详细过程吗?再答:f(x)=2(sin^4x+cos^4x)+m(sinx+cosx)^4=2(sin^2x+cos^2x)^2-4sin^2xcos
f(x)=sin(π-x)cos(3π/2+x)+sin(π+x)sin(3π/2-x)=(sinx)(sinx)+(-sinx)(-cosx)=sinx(sinx+cosx)f'(x)=cosx(s
这道题我会,您稍等,我这就给您写答案,但是请把采纳留给我,不然我白忙活了,我会很伤心的,呜呜~再答:再问:亲,过程能在详细一点吗再答:
f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx
(根号2+根号6)÷4再问:如何做的??????????谢谢
f(x)=cos(-x/2)+sin(π-x/2)=cos(x/2)+sin(x/2)=√2/2sin(x/2+π/4)周期T=2π*2=4ππ/2+2kπ≤x/2+π/4≤3π/2+2kπ解得π/2
令t=sin^2x,则sinx=√t和-√t.若sinx=√t,即x=arcsin√t所以f(t)=arcsin√t/√t.若sinx=-√t,x=-arcsin√t.f(t)=arcsin√t/√t
f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x
f(x)=sin^2x+asin^2(x/2)=sin^2x+a(1-cosx)=1-cos^2x+a-acosx1=-(cos^2x+acosx)+a+1=-(cos^2x+acosx+a^2/4)
f(x)=√2sin(2x+π/4)=√2sin2(x+π/8),x在(-3π/8,π/8)是增函数,在(π/8,5π/8)时减函数;f(x)在[0,π/2]上的最大值为x=π/8时,此时f(x)=√