y=1-1 2cosπ 3x的最值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/08 09:35:54
y=cosX+cos(X-π/3)=cosX+cosX*1/2+sinX*(√3)/2=cosX*3/2+sinX*(√3)/2=3sin(X+β)∴最小值:-3最大值:3
y=sec²x+2tanx+1=tan²x+2tanx+2=(tanx+1)²+1x∈[-π/3,π/4],则tanx∈[-√3,1]当tanx=-1,即x=-π/4时,
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)
x=-pi/6,y=-1,最小,x=pi/6,y=5,最大再问:什么意思看不懂再答:pi就是圆周率的那个π再问:没过程a,为啥要取这两个值我不懂
y=2sin²x+2sinx-1=2(sinx+1/2)^2-3/2-1≤sinx≤1-1/2≤sinx+1/2≤3/20≤(sinx+1/2)²≤9/4-3/2≤(sinx+1/
y=-2cos^2x+2sinx+3/2=-2+2sin^2x+2sinx+3/2=-2+2(sin^2x+sinx+1/4)+1=2(sinx+1/2)^2-1当sinx=-1/2即x=-π/6时,
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
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/
=cos²x+sinx=1-sin²x+sinx=-sin²x+sinx-1/4+1/4+1=-(sinx-(1/2))²+(5/4)当|x|≤π/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=sin^2(x)+2sin(x)cos(x)+3cos^2(x)=1+2cos^2(x)+sin2x=2+sin2x+cos2x构造向量a=(sin2x,cos2x),b=(1,1)a+b=(si
和差化积得y=2cos(x-π/6)cosπ/6=√3cos(x-π/6).故ymin=-√3,ymax=√3.
y=(1/cosx)+2tanx+1=secx+2tanx+1=tanx+2tanx+2=(tanx+1)+1由:x∈[-π/3,π/4]→1-√3≤tanx+1≤2故:当x=-π/4时,ymin=0
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
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
X∈(0,3/π)x-π/6∈[-π/6,π/6]y=cos[2(x-π/6)]+2sin(x-π/6)=1-2sin^2(x-π/6)+2sin(x-π/6)令sin(x-π/6)=t(-1/2≤t
这能有什么过程啊,这是最基础的东西,都需要一眼看出定义域:R值域:[-2,4]单调增区间(-2π/3+kπ,-π/6+kπ),k∈Z单调减区间(-π/6+kπ,π/3+kπ)奇偶性:非奇非偶函数最小正
y=-2cos(1/3x+π/6)-1定义域:x∈R-2≤-2cos(1/3x+π/6)≤2-3≤-2cos(1/3x+π/6)-1≤1值域:【-3,1】2π/(1/3)=6π最小正周期:6πf(-x
①对称轴过最高/低点∴cos(1/2x-π/6)=±1∴x/2-π/6=kπ∴x=2kπ+π/3(k∈z)∴对称轴x=2kπ+π/3(k∈z)对称中心在x轴上∴cos(1/2x-π/6)=0∴x/2-