sin^3x=-1 2 x=?

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sin(-x)=sin(x),cos(-x)=-cosx对吗?

sin(-x)=sin(x),cos(-x)=-cosx对吗?不对,反了应sin(-x)=-sin(x),cos(-x)=cosx

若sin x/2-cos x/2=1/3,则sin x=

两边平方得sin²x/2-2sinx/2cosx/2+cos²x/2=1/91-sinx=1/9sinx=8/9

已知函数f(x)=2√3sin²x-sin(2x-π/3)

(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间:π/2+2

三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x

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

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si

化简!f(x)=sin(pai-x)cos(3/2pai+x)+sin(pai+x)sin(3/2pai-x)

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

求导f(x) = cos(3x) * cos(2x) + sin(3x) * sin(2x).

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

已知函数f(x)=2根号3sin平方x-sin(2x-π/3)

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

高中数学:已知函数f(x)=2sin(x+π/2).sin(x+7π/3)-

fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3(cos*2x-sin*2x)=sin2x+根号3cos2x=2sin(2x+派

求极限 ((sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1

由和差化积公式分子=2sin[(x^3+x^2)/2]cos[(x^3+x^2-2x)/2]x→0,则(x^3+x^2)/2→0,sin则(x^3+x^2)/2和(x^3+x^2)/2是等价无穷小而c

s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)

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)

已知函数F(X)=√3sin 2x 2sin^2(x∈r)

(1)F(X)=√3sin2x+2sin²x=√3sin2x+1-cos2x=2(√3/2sin2x-1/2cos2x)+1=2sin(2x-π/6)+1F(x)的最小正周期T=2π/2=π

证明成立:[cos(3x)-sin(3x)]/(cosx+sinx)=1-2sin(2x).

cos(3x)-sin(3x)=cos(2x+x)-sin(2x+x)=cos2xcosx-sin2xsinx-sin2xcosx-cos2xsinx=(cos^2x-sin^2x)cosx-2sin

matlab菜鸟问题>> x=0:.12:1;>> y=(x.^2-3*x+5).*exp(-5*x).*sin(x);

最后一行y1',y2',y3',y4'之间应该用,或空格隔开.plot(x1,[y1'y2'y3'y4'],':',x,y,'o',x1,y0);

cos x sin x=3,求cos x+sin x

因为cosxsinx=3,所以2cosxsinx=6又因为(cosx+sinx)²=cos²x+2cosxsinx+sin²x且cos²x+sin²x

证明sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=3/4

sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=sin^2(x)+cos(x+30)[cos(x+30)+sinx]=sin^2(x)+cos(x+30)(cosxcos30

化简y=sin^2(x)+2sin(x)cos(x)+3cos^2(x)

y=sin²x+2sinxcosx+3cos²xy=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1=1+sin2x+cos2

已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)