y=2cos²*2分之x 1 求最小正周期....
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把平方打开,y=根号下(sina)^2+(cosa)^2+2(sina-cosa)+2=根号下2(sina-cosa)+3=根号下(2根号2×sin(a-π/4)+3)所以,最小值为:根号下3-2根号
可以根据:asinα+bcosα=√(a^2+b^2)*sin(α+θ),其中θ由a,b的符号和tanθ=b/a确定具体到这题,就是:y=√[(√3+2)/2]^2+(1/2)^2*sin(2x+θ)
首先,因为x∈[π/6,2π/3],所以(x-π/8)∈[π/24,13π/24]由余弦函数的曲线图可知在(x-π/8)=13π/24时y取最小值所以y=cos(13π/24)因为13π/24=97.
sinx+siny=1/3sinx=1/3-siny(siny)^2+(cosy)^2=1(cosy)^2=1-(siny)^2u=sinx-cos^2yu=(1/3-siny)-[1-(siny)^
X1+X2=-B/A=2X1*X2=C/A=1/2求得X1=1+根号2或者X1=1-根号2从而求出X2的值X1/X2+X2/X1=(X1*X1+X2*X2)/(X1X2)=6
y=sin²x+cos²x+2sinxcosx+2cos²x-1+1=1+sin2x+cos2x+1=√2sin(2x+π/4)+2-1
sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3(1/2-cos^2y)+sin^2y=1.5-3cos^2y)+sin^2y又有sin^2y+c
-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k
y=cos^2x-sin^2-√3cos(3π/2+2x)+1=cos2x-√3sin2x+1=2cos(2x+π/3)+1当2kπ+π/2≤2x+π/3≤2kπ+3π/2时,函数单调递减2kπ+π/
Sinx-siny=2/3cosx-cosy=1/2分别平方得(Sinx-siny)^2=(2/3)^2(cosx-cosy)^2=(1/2)^2展开相加得-2cos(x-y)+2=4/9+1/4-2
最小正周期π最大值2最小值0
y=2cos^2[(x+1)/2]=1+cos(x+1)T=2π最大值2最小值0
y=(sinχ+cosχ)²+2cos²χ求导得:y‘=2(sinx+cosx)(cosx-sinx)-4cosxsinx=2(cos²x-sin²x)-2si
y=cosx*(-2)sin[(x+x+2/3)/2]sin[(-2/3)/2]=2cosxsin(x+1/3)sin(1/3)=2sin(1/3)cosxsin(x+1/3)=2sin(1/3)(1
x2/x1²+x1/x2²=(x1³+x2³)/x1²x2²=(x1+x2)(x1²-x1x2+x2²)/(x1x2)&
y=(sinx)^2+3(cosx)^2+2sinxcosx=2+2(cosx)^2+2sinxcosx=2+cos2x+1+sin2x=3+sin2x+cos2x=√2sin(2x+π/4)+3最大
y=2asinx-cos²x+a²+2=2asinx+sin²x+a²+1=(sinx+a)²+1当a>=0时最小值为f(x)=(a-1)²