y=2cos²*2分之x 1 求最小正周期....-搜答案-智慧作业帮

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

x1

y=sin²x+cos²x+2sinxcosx+2cos²x-1+1=1+sin2x+cos2x+1=√2sin(2x+π/4)+2-1

对不起,

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