sin^6x cos^6x

1、(1)、y=√3/2sin2ωx-1/2cos2ωx+1=sin(2ωx-π/6)+1,T=2π/|2ω|=π,故|ω|=1,又当x=π/6时,函数有最小值,所以ω=-1.∴y=1-sin(2x+

那个前半括号里面相加等于一

（1）f（x）=3cos2ωx+sinωxcosωx=32cos2ωx+12sin2ωx+32…2分=sin（2ωx+π3）+32，…4分∵2ω×π6+π3=π2，…6分∴ω=12…7分（2）∵f（x

cos^4x+sin^2xcos^2x+sin^2x=cos^4x+(1-cos²x)cos²x+sin²x=cos^4x+cos²x-cos^4x+sin&#

∵[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

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

原式=∫4dx/(2sinxcosx)²=4∫dx/sin²2x=2∫csc²2xd2x=-2cot2x+C

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

利用半角公式如图降次计算.经济数学团队帮你解答,请及时采纳.

f(x)=a(sin²x+cos²x)(sin^4x-sin²xcos²x+cos^4x)+b(sin^4x+cos^4x)+6sin^2xcos^2x=a(s

（1）原式f（x)=1/2sin2xsinφ+cos^xcosφ-1/2cosφ=1/2sin2xsinφ+cosφ（cos^x-1/2）=1/2sin2xsinφ+1/2cos2xcosφ=1/2c

∫e^sinx(xcosx-sinx/cosx^2)dx=∫e^xsinx*xcosxdx-∫e^sinxsinxdx/(cosx)^2=∫xe^sinxdsinx-∫e^sinxd(1/cosx)=

∫(1/sin²xcos²x)dx=∫(sin2x+cos2x/sin²xcos²x)dx=∫(1/sin²x+1/cos²x)dx=-co

∫arctan(1/x)dx=∫(x)'arctan(1/x)dx=xarctan(1/x)-∫x*{1/[1+x^(-2)]}*[-1/x^2]dx=xarctan(1/x)+∫1/(x+1/x)d

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

1/[(sinx)^3(cosx)^3]=[sinx/(cosx)^3]+(2/sinxcosx)+[cosx/(sinx)^3]∫(1/sin³xcos³x)dx=[(1/2)/

∫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