若sin^8x cos^8x=41 128,x∈﹙0,pai 2),则x=
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/03 22:02:59
f(x)=sin^4x+cos^4x+sin^2xcos^2x/2-2sinxcosx=(sin²x+cos²x)²-3sin²xcos²x/2-2s
先化简.f(x)=(sin^4x+cos^4x+sin^2xcos^2x)/(2-2sinxcosx)-1/2sinxcosx+1/4cos^2x=【(sin²x+cos²x)&s
cos^4x+sin^2xcos^2x+sin^2x=cos^4x+(1-cos²x)cos²x+sin²x=cos^4x+cos²x-cos^4x+sin
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^
y=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=sin^2(2x)/2=(1-cos4x)/4周期显然是pi/2
y=sin^4x+cos^4x+4sin^2xcos^2x-1=(sin^2x+cos^2x)^2+2sin^2xcos^2x-1=1+2sin^2xcos^2x-1=2sin^2xcos^2x=si
∵[xcos(x+y)+sin(x+y)]dx+xcos(x+y)dy=0==>xcos(x+y)dx+xcos(x+y)dy+sin(x+y)dx=0==>xcos(x+y)(dx+dy)+sin(
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
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
x^2-6xcosθ-4y+9cos^2θ+8sinθ=0(θ为参数),配方:(x^2-6xcosθ+9cos^2θ)=4y-8sinθ(x-3cosθ)^2=4(y-2sinθ)曲线是一条抛物线,焦
y=sin⁴3xcos³4xdy/dx=cos³4x*d(sin⁴3x)/dx+sin⁴3x*d(cos³4x)/dx=cos
f(x)=[(sin^2x+cos^2x)^2-sin^2xcos^2x]/(2-2sinxcosx)=(1-sinxcosx)(1+sinxcosx)/2(1-sinxcosx)=1/2sinxco
(sinx)^4+(cosx)^4=(sinx)^4+(cosx)^4+2sin²xcos²x-2sin²xcos²x=(sin²x+cos²
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
sin(x-y)=sinxcosy-cosxsiny,sin(x+y)=sinxcosy+cosxsinysin(x-y)sin(x+y)=sin²xcos²y-cos²
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=
sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x
∫sin^2xcos^3xdx=∫sin^2x(1-sin^2x)dsinx=∫sin^2x-sin^4xdx=(1/3)sin^3x-(1/5)sin^5x+C不是让你求助我吗.再问:∫sin^2x