2cosx*sin(x π除以3)-根号3-sin^2x sinx*cosx
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由(sinx+cosx)^2=1/25得2sinxcosx=-24/25,(sinx-cosx)^2=48/25得sinx-cosx=-4√3/5,故sin^3x-cos^3x=(sinx-cosx)
f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x=[sinx+(√3)cosx+sinx]cosx-(√3)(sinx)^2=2sinxcosx+(√3)[(cosx)^2-(
cos-x自己画个坐标就一目了然了
令t=x-π/3,则当x->π/3时,t->0∴原式=lim(t->0){[1-2cos(t+π/3)]/sint}=lim(t->0){[1-2(cost/2-√3sint/2)]/sint}=li
1.f(x)=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx=2cosx*sin(x+π/3)-2sinx*[(√3/2)sinx-(1/2)cosx]=2cosx*sin(x
∵f(x)=2sin(π-x)cosx=2sinxcosx=sin2x1、最小正周期T=2π/2=π.2、∵-π/6≤x≤π/2∴-π/3≤2x≤π,∴-√3/2≤f(x)≤1,∴最大值1,最小值-√
f(x)=2cosx*sin(x+π/3)-√3sinx^2+sinx*cosx=2cosx*(sinxcosπ/3+cosxsinπ/3))-√3sinx^2+sinx*cosx=sinxcosx+
f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si
(sinx+cosx)/(sinx-cosx)=3sinx+cosx=3sinx-3cosxsinx=2cosxtanx=sinx/cosx=2sinx=2cosx带入恒等式sin²x+co
是(sinx+cosx)/(sinx-cosx)=2和[2cos(π-x)-3sin(π+x)]/[4cos(-x)+sin(2π-x)]吗?如果是的话:(sinx+cosx)/(sinx-cosx)
=-1/2cos2x+根号3)/2sin2x+3/2=sin(2x-£/6)+3/2
这个简单:f(x)=2cosx(sinxcos(pi/3)+cosxsin(pi/3))-根号33sin^2x+sinx*cosx=2sinxcosx+根号3cos2x=2sin(x+pi/3)所以:
(cosx)'=[sin(π/2-x)]‘=cos(π/2-x)*(π/2-x)'=sinx*(-1)=-sinx
y=2sinx*cos(3π/2+x)+√3cosx*sin(π+x)+sin(π/2+x)*cosx=2sinx*sinx-√3cosx*sinx+cosx*cosx=1+(sinx)^2-√3co
(1)由三角函数的公式化简可得f(x)=2cosx(12sinx+32cosx)-3sin2x+sinx•cosx=2sinxcosx+3cos2x−3sin2x=sin2x+3cos2x=2sin(
sinx+cosx/2sinx-3cosx=3sinx+cosx=3(2sinx-3cosx)=6sinx-9cosx5sinx=10cosxsinx=2cosxsin²x=4cos&sup
f(x)=2cosx(12sinx+32cosx)−3sin2x+sinx•cosx=2sinxcosx+3(cos2x−sin2x)…3′=sin2x+3cos2x=2sin(2x+π3)…5′(I
(1)根据题意,可得f(x)=2cosx•sin(x+π3)+sinx•(cosx−3sinx)=2cosx(12sinx+32cosx)+sinx•cosx−3sin2x=sin2x+3cos2x=
f(x)=[2(sinx*1/2+cosx*√3/2)+sinx]cosx-√3sin²x=(2sinx+√3cosx)cosx-√3sin²x=2sinxcosx+√3(cos&
(1)∵f(x)=2sinxcos(π2-x)-3sin(π+x)cosx+sin(π2+x)cosx=2sin2x+3sinxcosx+cos2x=32sin2x-12cos2x+32=sin(2x