sin2x+根号三cos2x
答:=2[sin(2x)cos(π3)+cos(2x)sin(π/3)]=2sin(2x+π/3)当sin(2x+π/3)=1时,y有最大值ymax=2 当sin(2x+π/3)=-1时,y有最小值ymin=-2 函数的最大值为2,最小值为-2。
答:原式:y=sin2x+√3cos2x/y=2[(1/2)sin2x+(√3/2)cos2x]=2[sin2xcos(π/3)+cos2xsin(π/3)].=2sin(2x+π/3).最小正周期T=2π/2=π.当2x+π/3=2kπ+π/2,即x=kπ+π/12,k∈Z时,sin(2x+π/3)=1,函数y具有最大值,ymax=2;...
答:f(x)=sin2x+根号3cos2xf(x)=2[(1/2)sin2x+(√3/2)cos2x]f(x)=2[cos60°sin2x+sin60°cos2x] 根据两角和的正弦公式 sin(A+B) = sinAcosB + cosAsinBf(x)=2sin(2x+60°)sin(X)的对阵中心坐标为(kπ,0),k属于整...
答:y=sin2x+√3*cos2x =2(1/2*sin2x+√3/2*cos2x)=2sin(2x+π/3)函数的周期为:T=2kπ/2=kπ,k为整数 ∵-1≤sin(2x+π/3)≤1 ∴函数的值域为:[-2,2]
答:sin2x+根号3cos2x =2(1/2sin2x+√3/2cos2x)=2(sin2xcosπ/3+cos2xsinπ/3)=2sin(2x+π/3)再把图像向左移动5/6π 解析式为:2sin[(2(x+5/6π)+π/3)=2sin(2x+2π)=2sin2x
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答:f(x)=2[cos60°sin2x+sin60°cos2x]根据两角和的正弦公式 sin(A+B) = sinAcosB + cosAsinB f(x)=2sin(2x+60°)sin(X)的对阵中心坐标为(kπ,0),k属于整数 X=kπ 所以,用2x+60°取代X,即2x+(π/3)=kπ 解得x=(kπ/2)-(π/6)所以,f(x)对称中心坐标为 ...
答:f(x)=sin2x+(根号下3)cos2x =2[(1/2)*sin2x+(√3/2)*cos2x]=2(cos60°*sin2x+sin60°*cos2x)=2sin(2x+60°)f(x)的最小正周期为∏ 。 (∏,pi,圆周率)
答:y=sin2x-√3cos2x =2sin(2x-π/3).而-1≤sin(2x-π/3)≤1.∴sin(2x-π/3)=1,即2x-π/3=2kπ+π/2 →x=kπ+5π/12时,所求最大值为:y|max=2.∴sin(2x-π/3)=-1 即2x-π/3=2kπ+3π/2 →x=kπ+11π/12时,所求最小值:y|min=-2。
答:y=sin2x-√3 cos2x =2(1/2sin2x-√3/2cos2x)=2(sin2x*cos60°-cos2x*sin60°)=2sin(2x-60°)值域:[-2,2]
网友评论:
刁弘19142071265:
函数y=sin2x+根号3cos2x的单调区间 -
27405贺景
:[答案] y=sin2x+√3cos2x =2sin(2x+π/3) 单调增区间: 令2kπ-π/2<2x+π/3<2kπ+π/2,k∈Z 得kπ-5π/12
刁弘19142071265:
求函数y=sin2x+根号3cos2x的最小正周期和最值求过程,非常感谢 -
27405贺景
:[答案] 原式:y=sin2x+√3cos2x/y=2[(1/2)sin2x+(√3/2)cos2x]=2[sin2xcos(π/3)+cos2xsin(π/3)].=2sin(2x+π/3).最小正周期T=2π/2=π.当2x+π/3=2kπ+π/2,即x=kπ+π/12,k∈Z时,sin(2x+π/3)=1,函数y具有最大值,ymax=2;...
刁弘19142071265:
函数y=sin2x+根号3cos2x的最大值是多少 -
27405贺景
:[答案] y=sin2x+根号3cos2x =2(1/2sin2x+√3/2cos2x) =2(sin2xcos60+cos2xsin60) =2(sin(2x+60)) ∵sin(2x+60)的最大值是1 ∴y的最大值是2
刁弘19142071265:
三角函数化解y=sin2x+根号3cos2x如何化解...帮忙解答..主要是带上公式... -
27405贺景
:[答案] asinx+bsinx=√(a^2+b^2)sin(x+z) 其中tanz=b/a 所以此处y=√[1^2+(√3)^2]sin(2x+z) =2sin(2x+z) 其中tanz=√3/1=√3 z=π/3 所以y=2sin(2x+π/3)
刁弘19142071265:
求y=sin2x+根号3cos2x的周期和值域 -
27405贺景
: y=sin2x+√3*cos2x=2(1/2*sin2x+√3/2*cos2x)=2sin(2x+π/3) 函数的周期为:T=2kπ/2=kπ,k为整数 ∵-1≤sin(2x+π/3)≤1 ∴函数的值域为:[-2,2]
刁弘19142071265:
已知y=sin2x+根号3cos2x求出该函数的周期 最大值及递增区间 -
27405贺景
: y=sin2x+根号3cos2x=2sin(2x+π/3) 最小正周期为 2π/2=π 最大值为 2 单调增区间: 2x+π/3∈[2kπ-π/2,2kπ+π/2] x∈[kπ-5π/12,kπ+π/12] k∈z
刁弘19142071265:
已知f(x)=sin2x+根号3cos2x -
27405贺景
: 解:f(x)=sin2x+√3cos2x=2(1/2sin2x+√3/2cos2x)=2sin(2x+π/3) f(x)的最大值=2,最小值=-2 f(x)的单调区间:∵2kπ-π/2≤2x+π/3≤2kπ+π/2 ∴单调递增区间为kπ-5π/12≤x≤kπ+π/12,即[kπ-5π/12,kπ+π/12] ∵2kπ+π/2≤2x+π/3≤2kπ+3π/2 ∴单调递减区间为kπ+π/12≤x≤kπ+11π/12,即[kπ+π/12,kπ+11π/12]
刁弘19142071265:
一道高中三角函数的题化简sin2x + 根号3 cos2x 答案是2sin(2x+π/3) 是如何化简来的? -
27405贺景
:[答案] sin2x +√3 cos2x =2(1/2sin2x+√3/2cos2x)=2(sin2xcosπ/3+sinπ/3cos2x)=2sin(2x+π/3)
刁弘19142071265:
函数y=sin2x+根号3cos2x的最大值是多少 -
27405贺景
: y=sin2x+√3cos2x=2sin(2x+π/3) max y =2
刁弘19142071265:
求y=sin2x+根号3cos2x的周期和值域 -
27405贺景
:[答案] y=sin2x+√3*cos2x =2(1/2*sin2x+√3/2*cos2x) =2sin(2x+π/3) 函数的周期为:T=2kπ/2=kπ,k为整数 ∵-1≤sin(2x+π/3)≤1 ∴函数的值域为:[-2,2]