limx2-9 sin(x-3)-搜答案-拍题作业网

1和3等价无穷小替代,sinxx,答案为2/5和w,第二题用洛必达法则,答案cosa,过程应该会写吧,我用手机回答的,输入不方便,请谅解

因为:sin(45-3x)=sin[90-(45+3x)]=cos(45+3x)cos(60-3x)=cos[90-(30+3x)]=sin(30+3x).则有cos(45+3x)*sin(30+3x

用公式a³+b³=(a+b)(a²-ab+b²)cos^6x+sin^6x=(cos²x)³+(sin²x)³=(cos

第一步,二倍角公式cos2@=1-2sin²@2-（1/2）（cos6x+cos10x+cos14x+cos18x）=2得cos6x+cos18x+cos10x+cos14x=0第二步,和差

∵cosπ/6=√3/2sinπ/6=1/2cosπ/6sinx+sinπ/6cosx=sin(x+π/6)∴√3sinX+cosX=2(√3/2sinX+1/2cosX)=2(cosπ/6sinx+

f(x)=sin(π-x)cos(3π/2+x)+sin(π+x)sin(3π/2-x)=(sinx)(sinx)+(-sinx)(-cosx)=sinx(sinx+cosx)f'(x)=cosx(s

lim(x²-9)/(x²-2x-3)x趋近于3=lim(x+3)(x-3)/(x+1)(x-3)x趋近于3=lim(x+3)/(x+1)x趋近于3=(3+3)/(3+1)=3/2

x趋向无穷的时候,lim【x2/(x-a)(x+b)】^x=lim1/(1-a/x)^x(1+b/x)^x=lim1/(e^-a)(e^b)=e^(a-b)

我知道这个题是个定积分题,请追问我给出积分限.我按我以前做过的同一题给你做吧,积分限是0→π∫[0→π]√(sin^3x-sin^5x)dx=∫[0→π]√[sin³x(1-sin²

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

sin(45-3x)*cos(60-3x)-cos(30+3x)*sin(45+3x)=sin(45-3x)*cos(60-3x)-sin[90-(30+3x)]*cos[90-(45+3x)]=si

先积化和差sin3xsin5x=0.5（cos2x-cos8x）∫sin3xsin5xdx=∫0.5（cos2x-cos8x）dx=0.25sin2x-0.0625sin8x+c

x=0:0.1:2*pi;s=2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x);plot(x,s)

lim(x趋向于3)sin(x-3)/(x*x-9)=lim(x趋向于3)sin（x-3）/[(x-3)(x+3)]=[lim(x趋向于3)sin（x-3）/(x-3)]*[lim(x趋向于3)1/(

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

3/2cosx-3/2(sinx)^2

原式=(-sinx*sinx)/(cos(1.5π-x)*sin(4.5π+x))=sinx/cosx=tanx=-3/4

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

再答：