1-cos+sinx

章爬灵1006若(1+sinx)/cosx= - 1/2,则cos/sinx - 1= -
陈春唯19717676145 ______ (1+sinx)/cosx=-1/2 (1+sinx)cosx/(cosx*cosx)=-1/2 (1+sinx)cosx/(cosx)^2=-1/2 (1+sinx)cosx/[1-(sinx)^2]=-1/2 (1+sinx)cosx/[(1-sinx)(1+sinx)]=-1/2 cosx/(1-sinx)=-1/2 cosx/(sinx-1)=1/2

章爬灵1006(sin x+1)/cos x=(sin x - cos x+1)/(sin x+cos x - 1) 证明 -
陈春唯19717676145 ______ 证明:∵ cos²x=1-sin²x ∴ cosx*cosx=(1+sinx)*(1-sinx) ∴ (1+sinx)/cosx=cosx/(1-sinx) 利用等比性质 ∴ (1+sinx)/cosx=(1+sinx-cosx)/[cosx-(1-sinx)] 即 (sinx+1)/cosx=(sinx-cosx+1)/(sinx+cosx-1)

章爬灵1006y=(1 - cosX)/(1+sinx)求导
陈春唯19717676145 ______ y'=[(1+sinx)'*(1-cosx)-(1+sinx)(1-cosx)']/(1-cosx)² =[cosx(1+cosx)+sinx(1+sinx)]/(1-cosx)² =(cosx+cos²x+sinx+sin²x)/(1-cosx)² ==(sinx+cosx+1)/(1-cosx)²

章爬灵10061、已知cos4α=2/3,则(sin^4 α - cos^4 α)^2= 2、(1 - cosx+sinx)/(1+cosx+sinx)= - 2,则tanx= -
陈春唯19717676145 ______ 1. cos4α=cos^22a-sin^22a=2/3(sin^4 α-cos^4 α)^2=(cos^22a+sin^22a)(cos^22a-sin^22a)=cos^22a-sin^22a=2/32. (1-cosx+sinx)/(1+cosx+sinx)=(2sin^2x/2+2sinx/2cosx/2)/(2cos^2x/2+2sinx/2cosx/2)=2sinx/2(sinx/2+cosx/2)/2cosx/2(sinx/2+cosx/2)=tanx/2=-2 tanx=2tanx/2/(1-ta/^2x/2)=-4/(1-4)=-4/3

章爬灵1006若sinx/(1+cos)=1/2,求sinx+cosx的值? -
陈春唯19717676145 ______[答案] (sin(x)/(1+cos(x)))^2 = (1-cos(x)^2) / (1+cos(x))^2 = (1-cos(x)) / (1+cos(x)) = 1/4 => cos(x) = 3/5 => sin(x) = 4/5 => cos(x)+sin(x) = 7/5

章爬灵1006(sinx/1 - cosx)=(1+cosx)/(sinx)
陈春唯19717676145 ______ 证明: 1)交叉相乘,得sin^2x=1-cos^2x 化简得,sin^2x+cos^2x=1 ∴(sinx/1-cosx)=(1+cosx)/(sinx) 2)由题意得,(sinα/cosα)+(cosα/sinα)=1/(sinαcosα) 通分得,(sin^2α+cos^2α)/sinαcosα ∴ tanα+(1/tanα)=1/(sinαcosα)

章爬灵1006求导:cosx(1 - cos(sinx)) -
陈春唯19717676145 ______[答案] [ cosx(1-cos(sinx))]' =-sinx*(1-cos(sinx))+cosx*(cosx*sin(sinx))

章爬灵1006化简:(1 - cos*+sin*)/(1+cos*+sin*)
陈春唯19717676145 ______ 原式=[1-(1-2sin²x/2)+2sinx/2cosx/2)/[1-(2cos²x/2-1)+2sinx/2cosx/2] =(2sin²x/2+2sinx/2cosx/2)/(2cos²x/2+2sinx/2cosx/2) =[2sinx/2(sinx/2+cosx/2)]/[2cosx/2(sinx/2+cosx/2)] =sinx/2/*cosx/2 =tanx/2

章爬灵1006(1+cosX - sinX)/(1 - sinx - cos)+(1 - cosx - sinx)/(1 - sinx+cosx)=? -
陈春唯19717676145 ______[答案] 原式=(1+cosx-sinx)/(1-cosx-sinx)+(1-cosx-sinx)/(1+cosx-sinx) =[(1+cosx-sinx)(1+cosx+sinx)]/[(1-cosx-sinx)(1+cosx+sinx)] +[(1-cosx-sinx)(1-cosx+sinx)]/[(1+cosx-sinx)(1-cosx+sinx)] =(1+2cosx+cos2x)/(-sin2x)+(1-2cosx+cos2x)/(sin2x) =(-4cosx)/sin2...

章爬灵1006已知(1 - cos x+sin x)/(1+cos x+sin x)= - 2,若x不等于kπ+(π/2),则tan x= -
陈春唯19717676145 ______ 1-cos x+sin x=-2-2cos x-2sin x3sinx=-cosx-3两边平方9sin²x=9(1-cos²x)=cos²x+6cosx+910cos²x+6cosx=0x不等于kπ+(π/2)所以cosx=-3/53sinx=-cosx-3sinx=4/5所以tanx=sinx/cosx=-4/3