все тригонометрические формулы

Ответ от Pavel Dokshin[гуру]
sin⁡α±sin⁡β=2sin⁡〖1/2 (α±β)〗 cos⁡〖1/2 (α∓β)〗
cos⁡α+cos⁡β=2cos⁡〖1/2 (α+β)〗 cos⁡〖1/2 (α-β)〗
sin^2 x + cos^2 x = 1
tg x = sin x/ cos x
ctg x = cosx/sinx
tg x * ctgx = 1
tgx = 1/ctgx
1+ctg^2x = 1/sin2x
1+tg^2x = 1/cos2x
sin2x=2sinxcosx
cos2x = cos^2-sin^2x=2cos^2x – 1= 1 – 2sin^2x
tg2x = 2tgx/(1-tg2x)
sin^2(x/2) = (1-cosx)/2
cos^2(x/2) = (1+cosx)/2
tg^2(x/2) = (1- cosx)/(1+cosx)
sinx = 2tg(x/2)/(1+tg^2(x/2))
cosx = (1-tg2(x/2))/ (1+tg^2(x/2))
tgx = 2tg(x/2)/ (1-tg2(x/2))
ctgx = (1-tg^2(x/2))/2tg^2(x/2)
cos(x+α)= cosx*cosα – sinx*sin α
cos(x-α)= cosx*cosα + sinx*sin α
sin(X+α)=sinx*cos α + cosxsin α
sin(X-α)=sinx*cos α - cosxsin α
tg(x+α)=(tgx+tgα)/(1-tgxtgα)
tg(x-α)=(tgx-tgα)/(1+tgxtgα)
cosx-cos α=-2sin((x+ α)/2)cos((x- α)/2)
tgx±tg α=sin(x± α)/cosxcos α
sinxsin α=1/2(cos(x- α)-cos(x+ α))
cosxcos α=1/2(cos(x+ α)+cos(x- α))
sinxcos α=1/2(sin(x+ α)+sin(x- α))