设f(x)=2根号3sin(π-x)sinx-(sinx-cosx)²

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设f(x)=2根号3sin(π-x)sinx-(sinx-cosx)²
设函数f(x)=sin(x+π/3)+2sin(x-π/3)-根号3cos(2π/3-x)

因为:f(π/6)=sinπ/2+2sin(-π/6)-根号3cos(π/2)=1-1-0=0所以:f(π/6)f(π/3)=0再问:原式=sin(x+π/3)+√3cos(x+π/3)+2sin(x

数学卷21:设函数f(x)=[(根号2)/2]×cos[2x+(π/4)]+sin²x.

f(x)=√2/2×cos[2x+(π/4)]+sin²x.=√2/2cos2x*√2/2-√2/2*sin2x*√2/2+sin²x=1/2cos2x-1/2sin2x+sin&

设函数f(x)=sin(2x+φ)(-π

你啊,要好好学习了!还没有悬赏分?把对称轴即x=∏/8代入原式子,即sin(∏/4+φ)=1或者-1,再用(-π

已知f(x)=sin(2x+π/3)-根号3sin^2x+sinxcosx+根号3/2

由题意可得:f(x)=sin(2x+π/3)-√3sin^2x+sinxcosx+√3/2=sin(2x+π/3)-√3(1/2-1/2cos2x)+1/2sin2x+√3/2=2sin(2x+π/3

已知f(x)=sin(2x+л/3)+sin(2x-л/3) g(x)=根号三cos2x (1)设

f(x)=sin(2x+π3)+sin(2x-π/3)f(x)=sin(2x)cos(π/3)+co(2x)sin(π/3)+sin(2x)cos(π/3)-cos(2x)sin(π/3)f(x)=2

设f(x)=2cosx.sin(x+π/3)-根号3 sin平方x+sinx.cosx

f(x)=2cosx*sin(x+π/3)-√3sinx^2+sinx*cosx=2cosx*(sinxcosπ/3+cosxsinπ/3))-√3sinx^2+sinx*cosx=sinxcosx+

设函数f(x)=sinx+sin(x+π/3)

1)由三角函数和差化积公式:f(x)=2sin(x+x+π/3)/2cos(x-x-π/3)/2=2sin(x+π/6)cos(π/6)=√3sin(x+π/6)f(x)的最小值为-√3.当x+π/6

已知函数f(x)=2根号3sin平方x-sin(2x-π/3)

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

设函数f(x)sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x) (1)求f(π/6),f(π/

题目是f(x=sin(x+π/3)+2sin(x+π/3)-√3cos(2π/3-x)!再问:设函数f(x)=sin(x+π/3)+2sin(x+π/3)-根号3cos(2π/3-x)再答:f(x)=

设f(x)=根号1-x化简f(sin2)+f(sin(-2))

1、f(sin2)+f(sin(-2))=√(1-sin2)+√[1-sin(-2)]=√(1-sin2)+√(1+sin2)1-sin2=(sin1)^2+(cos1)^2-2sin1cos1=(s

设函数f(x)=(sinθ/3)x^3+((根号3)cosθ/2)x^2+tanθ,则f'(π/4)=

f'(x)=(sinθ)x^2+((根号3)cosθ)x+tanθf'(π/4)=(sinθ)(π^2)/16+[(根号3)*(π/4)*cosθ]+tanθ题目很阴险啊,想让别人把θ和x弄混.

已知函数f(x)=sin^2 x+2根号3sinxcosx+sin(x+π/4)sin(x-π/4),x属于R,求f(x

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

设函数 f(x)=sin(2x+y),(-π

f(x)=sin2(x+y/2)由于sin2x对称轴为π/4+kπ/2;故x+y/2=π/4+kπ/2x=π/4+kπ/2-y/2;将x=x=π/8代入,得y=π/4+kπ,根据y的范围可知:y=-3

设函数f x=SIN(2X+φ)(-π

1)f(x)=sin(2x+φ)一条对称轴是X=π/8则kπ+π/2=2*π/8+φ===>φ=kπ+π/4因为-π

设函数f(x)=sin(2x+ φ)(-π

1.由f(x)=sin(2x+φ)一条对称轴是直线x=π/2可得:在x=π/2时,函数取极值.则2*π/2+φ=kπ+π/2(k∈Z)φ=kπ-π/2又-π

已知函数f(x)=2cosxcos(π/6-x)-根号下3×sin²x+sinxcosx设x∈【-π/3,π/

根据诱导公式进行化简这个要利用和角和倍角公式逆用

设函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,x属于[0,π/2],求f(x)的

f(x)=2cosxsin(x+π/3)-√3sin²x+sinxcosx=2cosx[sinxcos(π/3)+cosxsin(π/3)]-√3sin²x+sinxcosx=2c

已知函数f(x)=-根号3sin^2x+sinxcosx (1)求f((23π)/6) (2)设x属于(0,π),求f(

f(x)=-根号3sin^2x+sinxcosx=-√3/2(1-cos2x)+1/2sin2x=√3/2cos2x+1/2sin2x-√3/2=sin(2x+π/3)-√3/2(1)f((23π)/