sin2x等于-1/2-搜答案-拍题作业网

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

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(

(1)根据图信息可知sina=4/5cosa=3/5∴(sina^2+sin2a)/(cosa^2+cos2a)=(sina^2+2sinacosa)/(cosa^2+cosa^2-sina^2)=(

第一个式子等於（1+cos2x）/2sin2x=1-cosx是错的应该是sin2x=2sinxcosx

证明:（cosxsinx）^2=sinx^2cosx^22sinx·cosx=12sinx·cosx=1sin2x

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

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

tanx=3则：sinx/cosx=3sinx=3cosxsin2x/cos²x=2sinxcosx/cos²x=6cos²x/cos²x=6

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

∵cos2x=2cos²x-1,∴cos²x=1/2(cos2x+1)∴原式=1/(1/2cos2x+sin2x+1/2)=1/[根号5/2sin(2x+ψ)+1/2](tanψ=

=(2sin2x)/(1－cos2x)²=(4sinxcosx)/(2sin²x)²=cosx/sin³x

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

sin²2x＋sin2x＋cos2x=1sn2x＋cos2x=1－sin²2x=cos²2xsin2x=cos²2x－cos2x----------------

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!

∫cos2xdx上限是π/4下限是π/6=1/2*sin2x上限是π/4下限是π/6=1/2sinπ/2-1/2sinπ/3=1/2-根号3/4我是数学百事通,数学问题想不通,快上数学百事通!