cot2x等于1/sin2x么-搜答案-百度作业网

sin2x+1=2tanx/(1+tan^2x)+1=4/5+1=9/5

这样做比较简单：令i=∫[(sinx)^2*cosx/(sinx+cosx)]dxj=∫[sinx*(∴i=-(1/8)(sin2x+cos2x)+(1/4)In|sinx+cosx|+C

假设sin2x*[(根号1-tanx)+(根号1+tanx)]=2sin2x则(根号1-tanx)+(根号1+tanx)=2两边平方得1-tanx+1+tanx+2根号(1-tan^2x)=42根号(

从您的解题过程猜测原题可能是x->0,lim[1/x^2-(cotx)^2]错误的原因是等价无穷小只能替换因式,你上面的替换分子就有问题了,分母的替换是正确的.lim[1/x^2-(cotx)^2]=

sinx=2cosx,sinx^2+(1/2sinx)^2=1,得sinx^2=4/5,sin2x=2sinxcosx=sinx^2=4/5希望采纳

cos^2x+2SINXCOSX+SIN^2X=1(SINX+COSX)^2=1SINX+COSX=1或-1希望对你帮助

y=ln(x/(1+x))-cot2xdy=[(1+x)/x]d(x/(1+x))+(csc2x)^2.d(2x)={(1+x)/[x(1+x)^2]+2(csc2x)^2}dx

cot2x=cos2x/sin2x=(2cos2xcos2x)/2sin2xcos2x=(1+cos4x)/sin4x

x=0代入f(0)=cos0-sin0+2(3sin0cos0+1)=1-0+2(0+1)=3

原式＝(-2cos2x/1+sin2x+cos2x)+1=(-2cos^2x+2sin^2x)/(1+2sinxcosx+cos^2x-sin^2x)+1=[2(sinx+cosx)(sinx-cos

题目条件不充分啊cos2x+sin2x=1cos2x=cos^2x-sin^2xsin2x=2sinxcosxcos2x+sin2x=cos^2x-sin^2x+2sinxcosx

1+sin2x-cos2x=1+2sinxcosx-1+2sinx^2=2sinx(cosx+sinx)1+sin2x+cos2x=1+2sinxcosx+2cosx^2-1=2cosx(cosx+s

tan(π/4-x)=1/2所以tan[2(π/4-x)]=2tan(π/4-x)/{1-[tan(π/4-x)]^2}=2×(1/2)/[1-(1/2)^2]=1/(3/4)=4/3即tan(π/2

sinx/cosx=tanx=2sinx=2cosxsin²x=4cos²x因为sin²x+cos²x=1所以cos²x=1/5sin2x=2sinx

sinx/cosx=tanx所以sin2x/cos2x=tan2x

sin2x＋cos2x＝√2（√2＆＃47；2sin2x＋√2＆＃47；2＊cos2x）＝√2（cospai＆＃47；4sin2x＋sinpai＆＃47；4cos2x）＝√2sin（2x＋pai＆＃4

等式左边=2sinxcosx/[sinx+(cosx-1)][sinx-(cosx-1)]=2sinxcosx/[sin^2x-(cosx-1)^2]=2sinxcosx/(sin^2x-cos^2x

1/2sin4x

等式左边=2sinxcosx/[sinx+(cosx-1)][sinx-(cosx-1)]=2sinxcosx/[sin^2x-(cosx-1)^2]=2sinxcosx/(sin^2x-cos^2x

sinx^2+cosx^2=1,这就相当于一个公式,中间的变数x当然可以换成任何值了!