求函数f x =log1 2(cosx 3 π 4)的单调区间
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fx=-√3cos2x-sin2x=-2sin(2x+π/3)所以最小正周期为πf'x=-4cos(2x+π/3),f'x>0时递增x在(π/12,π/3)上递增f'x=0,x=π/12.极小值f(π
f(x)=cos2x+根号3sin2x=2sin(2x+π/2)所以周期为π对称轴2x+π/2=π/2+kπ(k是整数)即x=kπ/2k是整数单调区间-π/2+2kπ
f(x)=sin²x+√3sinxcosx+2cos²x,=√3sinxcosx+cos²x+1=√3/2sin2x+1/2(1+cos2x)+1=√3/2sin2x+1
令t=sinx则f=(1-t^2)+2t=-t^2+2t+1=-(t-1)^2+2因为|t|
1)f(x)=sin(x/2)cos(x/2)+√3cos²(x/2)=(sinx)/2+(√3cosx)/2-1/2令cos(π/3)=1/2sin(π/3)=√3/2∴f(x)=sin(
fx=2cos^2x+2根号3sinxcosx-1=2cos^2x-1+2根号3sinxcosx根据倍角公式,sin2α=2sinαcosαcos2α=2cos^2(α)-1fx=cos2x+根号3s
f(x)=cos(2x-4π/3)+2cos^2x=cos(2x-4π/3)+cos2x+1=2cos(2x-2π/3)cos2π/3+1=1-√3cos(2x-2π/3)1.当cos(2x-2π/3
f(x)=[2cos^2(x/2)-1]+sinx=cosx+sinx=√2sin(x+π/4)∵x∈R∴x+π/4∈R∵f(x)=sinx∈(-1,1)∴f(x)=√2sin(x+π/4)∈(-√2
F(X)=cos(√3x+t)F'(X)=-√3sin(√3x+t)F(X)+F'(X)=cos(√3x+t)-√3sin(√3x+t)是奇函数所以F(0)+F'(0)=0即cost-√3sint=0
做一道题给你示范下吧,后面的相信你可以举一反三.第一题:a=ln27/ln12(化对同底数对数,一般以e为底)=3ln3/(2ln2+ln3)(分解成质数)于是得ln2/ln3=(3-a)/(2a)再
f(x)=cos(2x-π/3)-cos2x=1/2cos2x+√3/2sin2x-cos2x=√3/2sin2x-1/2cos2x=sin(2x-π/6)最小正周期T=2π/2=π(2)0
f(x)=cos²x+sinxcosx=(cos2x+1)/2+1/2sin2x=(1/2cos2x+1/2sin2x)+1/2=√2/2*(√2/2cos2x+√2/2sin2x)+1/2
由x−1>02−x≥0,解得1<x≤2,∴函数f(x)的定义域为(1,2].又∵函数y1=log12(x-1)和y2=2−x在(1,2]上都是减函数,∴当x=2时,f(x)有最小值,f(2)=log1
f(x)=2cos²(x/2)-√3sinxf(x)=2cos²(x/2)-2√3sin(x/2)cos(x/2)f(x)=2cos(x/2)[cos(x/2)-√3sin(x/2
设函数fx=2cos^2(π/4-x)+sin(2x+π/3)-1=cos(PI/2-2x)+sin(2x+PI/3)=sin(2x)+sin(2x)/2+cos(2x)*sqrt(3)/2=sqrt
解f(x)=2cos^2x+2√3sinxcosx-1=√3sin2x+cos2x=2sin(2x+π/6)∴最小正周期为:2π/2=π再答:不懂追问再问:在三角形ABC中,角ABC所对的边分别是ab
令u=|x-3|,则在(-∞,3)上u为x的减函数,在(3,+∞)上u为x的增函数.又∵0<12<1,y=log12u是减函数∴在区间(3,+∞)上,y为x的减函数.故答案为:(3,+∞)
(1)、f(x)=2cos²x-(sinx-cosx)²=2cos²x-(1-sin2x)=cos2x+sin2x运用一下化一公式得f(x)=√2sin(2x+π/4),