tan(派z)的积分-搜答案-智慧作业帮

设z－4=r(cos60°＋isin60°),则z=4＋(1/2)r＋[(√3/2)r]i,利用|z|=√21,代入计算出r=1或r=－5(舍去),从而z=9/2＋√3/2i.

根号（cosx-cosx^3）dx=根号cosx(1-cosx^2）dx=(根号cosx)(-sinx)dx(-Pi/2,0)+(根号cosx)(sinx)dx(0,Pi/2)=2/3(cosx)^(

tan(α+π/4)=tan(α+π/4+β-β)=tan【(α+β)-(β-π/4)】=【tan(α+β)-tan(β-π/4)】/【1+tan(α+β)tan(β-π/4)】=(2/5-1/4)/

x属于[-派/3,派/4],则tanx属于[-根号3,1]原式：y=(tanx+1)^2-3tanx=-1时有最小值-3tanx=1时有最大值1.则值域为[-3,1]应该能看明白吧

tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=4tanβ；故得tanα+tanβ=4tanβ-4tanαtan²β；(1+4tan²β)tanα=3tanβ故

原式=∫(0,π/2)cosxdx-∫(π/2,π)cosxdx=(sinx)│(0,π/2)-(sinx)│(π/2,π)=(1-0)-(0-1)=2

=-cosa=-1/3

sin(2派-x)tan(派+x)cot(-x-派)/tan(3派-x)cos(派-x)=-sinxtanxcot(-x)/tan(-x)(-cosx)=sinxtanxcotx/tanxcosx=s

z²+2z+4=0的根为：[-2±√(4-16)]/2=-1±i√3这两个点均不在单位圆内,因此被积函数在单位圆内解析,所以本题积分结果为0希望可以帮到你,如果解决了问题,请点下面的"选为满

=sin(-23派/6)[cos(-23派/6)tan(-23派/6)]=sin²(-23派/6)=sin²(-4派+派/6)=sin²(派/6)=(1/2)²

=sin(4π+5π/6)+cos(π/3-10π)+tan(3π/4-7π)=sin5π/6+cosπ/3+tan3π/4=1/2+1/2-1=0

1.f(π/9)=tan(π/3+π/4)=(tanπ/3+tanπ/4)/(1-tanπ/3tanπ/4)=(√3+1)/(1-√3)=-√3-2

令x=π/2-y,dx=-dy当x=0,y=π/2；当x=π/2,y=0L=∫[0,π/2]dx/[1+(tanx)^2010]=-∫[π/2,0]dy/[1+(tan(π/2-y))^2010]=∫

tan(派+a)=-2.tana=-2sina=2v5/5或者-2v5/5cosa=-v5/5或者v5/5sin(5派/2-a)=sin(π/2-a)=cosasin(3派-a)=sin(π-a)=s

和角公式tan(A+B)=[tanA+tanB]/(1-tanAtanB)tan(α+π/4)=-3[tanα+tanπ/4]/[1-tanαtanπ/4]=-3(1+tanα)/(1-tanα)=-

z=a(i+1)+i=(a+1)i+a,a＞0所以z的轨迹可以表示成y=x+1(x＞0)再问：小弟不才。。没有看懂第一步是怎么来的？a是设的吗？派/4呢？再答：π/4是角度，你还没理解虚数的定义这么说

sin(25π/6)+cos(25π/3)+tan(-25π/4)=sin(25π/6-4π)+cos(25π/3-8π)+tan(-25π/4+7π)=sin(π/6)+cos(π/3)+tan(3