比较sin^3x cos^4x x^2sin^3x - cos^4x
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/18 09:51:16
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
(1)∵f(x)=4sinωx(cosωxcosπ3−sinωxsinπ3)+3,------(1分)=2sinωxcosωx−23sin2ωx+3=sin2ωx+3cos2ωx---(3分)=2si
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
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^
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
∫sin^3xcos^2xdx=-∫sin^2xcos^2xdcosx=-∫(1-cos^2x)*cos^2xdcosx=-∫(cos^2x-cos^4x)dcosx=(1/5)*cos^5x-(1/
y=sin⁴3xcos³4xdy/dx=cos³4x*d(sin⁴3x)/dx+sin⁴3x*d(cos³4x)/dx=cos
sin^2(a+π)Xcos(π+a)Xcot(-a-2π)/tan(π+a)Xcos^3(-a-π)=sin^2a(-cosa)(-cota)/tana(-cos^3a)=-sinacos^2a/s
看:(对不起,第一条的变数全部都是t,刚才做的时候不小心把t打错作x了)
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
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
这题方法有很多,你可以把cos^2x换成1-sin^2x4sin^2xcos^2x=4(sin^2x-sin^4x)sin^2x和sin^4x积分是有公式的.但是一般人估计也记不得,所以方法二:为了方
2sinxcosx=sin2x那么sin^2xcos^2x=sin^22x/4另外sinx等价于x,所以sin^2x等价于x^2,也即x^2sin^2x变成了x^4不知您是否明白,若有不明还可问(⊙o
证明:因为左边=sin²X(sin²X+cos²X)+cos²X=sin²X+cos²X=1=右边,所以:(sinX)^4+sin²
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