TÓM tắt CÔNG THỨC TOÁN 11 THỦ KHOA

lcoNG THuc cAN NHd LdP GROUP: 0 THỦ KHOA 111 Cac cdng thuc hiqng giac co ban: * sin a + cos a = 1[ ') * + tan - a = )' , a -:1:-+kJr,k E Z cos- a * ') l s-cot" a=- ' sm-a ') 10 Cong thuc bien dfii tich tfing: * cos a cos b = -[ cos (a+ b) + cos ( a - b) J * sin a sin b = _ _!_ [ cos (a+ b )- cos( a - b) a r k m.k e Z * sin a cos b = _!_ [ sin (a+ b) + sin ( a - b) 1[ * tana.cota = 1, a* k-,k E Z 2 Gia tr] hiqng giac cac cung do'i nhau: * cos(-a) =cos a * sin (-a)= -sin a * * tan(-a) = -tan a cot(-a) = -cot a Gia tr] hiqng giac ciia cac cung bu nhau: * sin ( Jr - a) = sin a * cos(Jr-a)=-cosa * * tan(Jr-a) = -tan a cos(Jr-a) = -cota Gia tr] hiqng giac cua cac cung hon kern Jr : * sin(a+Jr)=-sina * tan(a+Jr) = tana * cos(a+Jr)=-cosa * cot(a+Jr) = cota Gia tr] luong giac ciia cac cung phu nhau: • sin (; - a) = cos a • tan(; -a)= cota * * cos (; - a) = sin a cot-a)= tan a 6.Gia tr] lu'qng giac ciia cac cung hon sin( a+ ;) = cos a tan( a+;)= -cot a kem; cos( a + = - sin a co{ a + ; ) = - tan a * cos( a+ b) = cos a cosb- sin a sinb * sin(a-b)=sina cosb-cosa sinb * sin (a+ b) = sin a cos b + cos a sin b tan a± tan b + b) = tan ( a_ + tan a tan b Cong thuc nhan doi va nhan ha: =sin'La = 2sina.cosa =cos 2a = cos ' a-sin2 a 2tana = 2cos2 a-1 *tan2a= l= tan" a = 1-2sin2 a * * Cos3a = -lcos'a - 3cosa * Sin 3a = 3sina - -lsirr'a 9.Cong thuc bac: ') +cos2a *COS- a= -2 ') 1-cos2a *sin- a= -2 J 11 Cong thuc bien dfii t6ng tich: u+v u-v * cosu+ cosv = 2cos cos-2 u+v u-v * cosu-cosv=-2sm sm-2 u+v u-v * smu+smv=2sm cos-2 u+v u-v * smu-sm v = 2cos sm-2 12 Vai ti so' hrong giac thong dung: O(rad) Cung sin cos tang oo 1[ 1[ 1[ 1[ - - - - 30° 45° 60° - J2 - 2 Jj J2 - - 2 Jj Jj 2 90° - fj II - ;) Cong thuc d)ng: * cos ( a - b) = cos a cos b + sin a sin b J cotg II fj Jj - 13.Phu'ong trmh lu'qng giac co ban : • sinx = a (1) neu a la nghiern cua (l),nghia Iasin zz = a [ x =a+ k2Jr (1) � sinx = sin a ke Z x = Jr - a + k2Jr •cosx =a (2) neu a la l nghiern cu a (2),nghia lacosa = a thl (2) � cosx = cos a x =±a+ k2Jr,k E Z • tanx = a (3) neu a la I nghiern cua (3),nghia la tana = a thl (3) � tan x = tan a x = a + kst k E Z •cotx = a (4) neu rz la l nghiern cu a (4),nghia la cot a= a thi (4) � cotx = cot a x =a+ kst , kEZ Chu y: sin x = a, cos x = a c6 nghie m I al s; tanx = a, cotx = a c6 nghie m vdi Va 14.Phtidng trinh b�c nhat do'i vdi sinx va cosx * 2 a sin x ± b cos x = c -J a + b sin( x ± a) = c acosx±bsinx =c -Ja2 +b2 cos(x+a) =c (cos nhtr t/Ji dilu) a b ) 01 cos a= ,sma = (v, 1 -va2+b2 -va2+b2 * 22.Cong tlurc nh] tlnrc Niu-To'n ( a+ b)n = cona n + clna n-1b + + ckna n-kbk + + n en: ca + b r = L c� an-k bk n k=O 23.Bang cong thirc d�o ham Ca hai PT tren muon tim a barn shif cos -J a' + b' (c)' , PT tren " co, ng hiiern " Ch , , C ac !dll!.J! : (x)'=l 15 PT thuan nha'tb�c hai do'i vdi sinx va cosx Dang: asin2x+bsinxcosx+c cosix = d (6) each giai: Bl:thu' vdi cosx=O co thoa (6) khong? B2:Chia v€ cua (6) cho cosx :;t: ta duce pt: d atan x +btanx +c = )' cos- x ¢=> atan 'x +btanx +c =d(l +tanx ) ¢:> (a-djtanx +btanx +c -t cong viec duoc hoan boi hanh d(>ng.NSu HDl c6 m each thuc hien, HD2 c6 n each thirc hien khong trung voi bky each nao cua HD thi cong viec d6 c6m+n each thuc hien 18.0ui tic nhan: Mot cong viec duce hoan boi hanh dt tlur tu nao d6 duoc goi la mdt chinh hop chap k cua n phan tu S6 cac chinh hg ch� k cua n p.tir ki hi�u:A\ ta c6 : n! Akn = (n-k)! ma 21.T6 hQ'P:M(>t t�p g6m k p.nr cua A (1 s ks n) duce goi la mot t6 hop chap k cua n p.tir S6 cac t6 hgp ch�p k cua n phan tu ki hieu.C" n ta c6 : n! c = k!(n -k)! Tinh chftt: V6'i u ( C.x)' = C , , (xii) = n.x'"" ( u II) _-n.u 11-J u , ( t) � - :, ( (Fx)' = , ta m9t ham s6 \- f x < 0) [x c- O] 2'\/X (sin x) = cosx (Fu)'= u� 2'\/U , (sin u) = u' cosu (cosx)' = -sin x (cos u )' = -u' sin u , , u' (tanx) (tanu) cos- x cos- u ' ' u' ( ( cotx) = cotu) = 2 sin x sin u D�o ham t6ng ,Hi�u,Tich va ThuO'ng =-)' =-') a + b2>_ c = (C: hang so') * ( u ± v) , * (l_! ) = u' ± v' = u'.v�u.v' v * , , v- * ( u.v) = u'.v + u.v' * (k.u )' = k.u' (k la hang so) PTTT cua d6 thi hs :y=f(x) te;1i diSm M(xo;Yo): y = y(x0).(x-x0)+ Yo 24.Bi�u thrrc t9a dQ cu.a phep tjnh ti�n: Trong mp oxy cho diSm M(x;y),M'(x';y') va � (a;b) {x'= x+a , T-(M) = M ¢:> y'=y+b 1' 25 Bi�u thrrc t9a dQ cu.a phep D6i xrrng tr\)C: • Trong mp oxy cho diSm M(x;y) goi M'(x' ;y')= Dd(M) { * NSu ch9n d la t[\lc ox,thi ¢=> �· = x y =-y * NSu ch9n d la t[\lc oy,thi¢=> {x'� -x y=y 26 Bi�u thrrc t9a dQ cu.a phep D6i tam: • Trong mp oxy cho diSm M(x;y),I(a;b) goi {x'=2a-x M'=D1(M)=(x';y'),khi d6 y'= 2b- y * NSu chon I la g6c toa 0(0;0) thi: {x:= -x M'=D0(M)=(x';y'),khi d6 y =-y ... hieu.C" n ta c6 : n! c = k!(n -k)! Tinh chftt: V6'i u ( C.x)' = C , , (xii) = n.x'"" ( u II) _-n.u 11- J u , ( t) � - :, ( (Fx)' = , ta m9t ham s6 - f x < 0) [x c- O] 2'/X (sin x) = cosx (Fu)'=