f(x)=sin^4x cos^4x
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 00:13:21
![f(x)=sin^4x cos^4x](/uploads/image/f/575379-27-9.jpg?t=f%28x%29%3Dsin%5E4x+cos%5E4x)
f(x)=根号3cos²ωx+sinωxcosωx=根号3/2(cos²2ωx)+1/2(sin2ωx)=sin(2ωx+π/3);y轴右侧的第一个最高点的横坐标为π/6,π/3ω
最小正周期:T=圆周率(pi),最大值=13/8,最小值=-3/8.
f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)=(1-cos2ωx)/2+(√3/2)sin2ωx=sin(2ωx-π/6)+1/2∵函数y=f(x)的图像相邻两条对称轴之间
先化简.f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-2sinxcosx)-1/2sinxcosx+1/4cos^2x=【(sin²x+cos²x)&s
sin^4x+cos^4x+sin^2x*cos^2x=sin^4x+cos^4x+2sin^2x*cos^2x-sin^2x*cos^2x=(sin^2x+cos^2x)^2-sin^2x*cos^
f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-sin2x)=(sin^4x+cos^4x+2sin^2xcos^2x-sin^2xcos^2x)/(2-sin2x)=[(s
f(x)=sin2ωx+√3cos2ωx=2sin(2ωx+π/3),两对称轴之间的最小值为π/2即半个周期,则周期为π=2π/2ω,所以w=1,所以f(x)=2sin(2x+π/3),f(α)=2s
f(-1)+f(1)=sin(-1)cos(-1)+sin1cos1=-sin1cos1+sin1cos1=0
合并同类项么,很简单的只要你愿意去做左边=cos*x(cos*y+sin*y)+sin*x(cos*y+sin*y)=cos*x+sin*x=1=右边
由题意得f(x)=2sinωxcosωx+23sin2ωx−3=sin2ωx−3cos2ωx=2sin(2ωx−π3)…(2分)由周期为π,得ω=1.得f(x)=2sin(2x−π3)…(4分)由正弦
f'(x)=x'cos3x+x*(cos3x)'=cos3x+x(-sin3x)*(3x)'=cos3x-3xsin3x
题目有点怪!会不会打错了啊?提示你哦:sin^2xcos^2x=sin^2(2x)/42sinxcosx=sin2x1/2sinxcosx=sin2x/41/4cos2x=1/4-sin^2x/2再问
f(x)=[(sin^2x+cos^2x)^2-sin^2xcos^2x]/(2-2sinxcosx)=(1-sinxcosx)(1+sinxcosx)/2(1-sinxcosx)=1/2sinxco
f(x)=a(sin²x+cos²x)(sin^4x-sin²xcos²x+cos^4x)+b(sin^4x+cos^4x)+6sin^2xcos^2x=a(s
t=sinx+cosx=√2sin(x+π/4)-√2=再问:上面那个颠倒的V是什么再答:那是根号呀,√2表示根号2.再问:sin^2x这个颠倒的^也是根号?再答:这个是次方符号呀,sin^2x表示的
f(x)=sin2wx,所以值域为[-1,1].T=4π=2π/|w|,w=1/2.
证明:因为左边=sin²X(sin²X+cos²X)+cos²X=sin²X+cos²X=1=右边,所以:(sinX)^4+sin²
1)f(x)=a(cos^2x+sinxcosx)+b=a/2(1+cos2x+sin2x)+b=a/2根号2sin(2x+π/4)+a/2+b2kπ-π/2=
f(x)={[(sinx)^2+(cosx)^2]^2-(sinxcosx)^2}/(2-sin2x)=[1-(sinxcosx)^2]/(2-2sinxcosx)=(1+sinxcosx)(1-si