2017-10-25
368
sin2x+cos2x=1
tg2x+1= 
ctg2x+1= 
tgx= 
tgx*ctgx=1
| tgx= 
sinx= 
cosx= 
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| cos(-α)=cos α sin(-α)=-sin α tg(-α)=-tg α ctg(-α)=-ctg α | ||||
sin2x=2 sinx cosx
sin3x=2sinx-4sin3x
sin4x=2sin2x–cos2x
cos2x=cos2x-sin2x cos2x=2cos2x-1
cos2x=1-2sin2x
cos3x=4cos3x-3cosx
tg2x= 
tg3x= 
| cos(π/2-x)=sin x cos(π±x)=-cos x cos(π/2+x)=-sin x cos(3π/2-x)=-sin x cos(3π/2+x)=sin x | sin(π/2±x)=cos x sin(π-x)=sin x sin(π+x)=-sin x sin(3π/2±x)=-cos x | |||||
| tg(π/2+x)=-ctg x tg(π/2-x)=ctg x ctg(π/2+x)=-tg x ctg(π/2-x)=tg x |  
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| )0))) | sin(x+y)=sinx·cosy+cosx·siny
sin(x-y)=sinx·cosy-cosx·siny
cos(x+y)=cosx·cosy–sinx·siny
cos(x-y)=cosx·cosy+sinx·siny
tg(x+y)=  tg(x-y)= 
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sinx+siny=2 sin( )·cos( )
sinx-siny=2 sin()·cos()
cosx+cosy=2 sin()·sin()
cosx+cosy=-2 sin()·sin()
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| х2 - у2 = (х - у) (х+у) (х + у)2 = х2 + 2ху + у2 (х - у)2 = х2 – 2ху + у2 (х + у)3 = х3 + 3х2у + 3ху2 + у3 (х - у)3 = х3 – 3х2у + 3ху2 - у3 х3 + у3 = (х + у) (х2 - ху + у2) х3 - у3 = (х - у) (х2 + ху + у2) | alogab = b loga 1 = 0 loga a = 1 | logax= 
logax= 
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loga(x · y) = logax + logay
loga xy = logax - logay
loga xp = p logax
logak x =  loga x, при k ≠ 0
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(f(x)+g(x))’=f ’(x)+g’(x)
(k•(f(x))’ = k•f ’(x)
(f(x)•g(x))’=f ’(x)·g(x)+f(x)·g’(x)
(f(x)/g(x))’= 
(f(g(x))’=f ‘(g(x))•g’(x)
lim f(x)g(x)= 
lim ex = ∞, x→+∞; 0, x→-∞
| при  sin x ̴x
arccsin x ̴x
tg x ̴x
arctg x ̴x 1-cos x ̴ x2/2
| ax ̴1+x lna ex-1 ̴x ln (1+x) ̴x loga(1+x) ̴x/lna (1+x)m-1 ̴mx | ||||||||||||||||
| lim (1+x)1/x=e | ||||||||||||||||||
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