EBOOK bài tập HÌNH học 10 PHẦN 2 NGUYỄN MỘNG HY (CHỦ BIÊN)

Chi/ONq II TICH VO HUCfNG CUA HAI VECTO VA LfNG DUNG Đ1 GIA TRI LUONG GIAC CUA MOT GOC BATKITtro^D^NlSOđ A CAC KIEN THLTC C A N NHO / Dinh nghia : Vdi mdi gdc a (0° < a < 180°) ta xac dinh dugc mdt diim M tren nira dudng trdn don vi (h 2.1) cho xOM = a Gia sit diim M cd toa dd la M(XQ ; y^) Khi dd : • Tung dd y^ cua diim M ggi la sin cua gdc a vk dugc ki hieu la sin a = y^ • Hoanh dd JCQ cua diim M ggi la cosin cua gdc a vk dugc ki hieu \k cos a = JCQ y • Ti sd — vdi JCQ ^ ggi la tang cua gdc a vk dugc kf hieu la tan a= >' - o^ • Ti sd - ^ vdi Jo 5t ggi 1^ cdtang cua gdc avk duoc kf hieu la >'o cot « = - ^ 66 S - BTHHIO - B Cdc he thiic luong giac a) Gia tri lugng giac cua hai gdc bii sina=sin(180°-a) cosa=-cos(180°-a) tana=-tan(180°-a) cota = -cot(180°-a) b) Cac he thiic lugng giac co ban Ttt dinh nghia gia tri lugng giac ciia gdc a ta suy cac he thiic 2 sin a + cos a = I ; sma = tana (a 7^90°); cos a cota = tana +tan a = cos a = cota(a^0°;180°); sma tana = cot a 1 + cot a = cos a sm a Gid tri luong giac cua cdc gdc dac Met Giatif\^^ lugng giac ^"^\^^ 0° 30° 45° 60° 90° 180° sin a :/2 :/3 cos a 2 -1 tana S II cot a II II s 1 67 Gdc giUa hai vecta Cho hai vecto a vk b dtu khac vecto Tut mdt diim O bit ki ta ve OA = a va OB = fe Khi dd gdc AOB vdi sd tii 0° din 180° dugc ggi li gdc giUa hai vecta a vd b (h.2.2) va kf hieu la (a,fe) Hinh 2.2 B DANG TOAN CO BAN Tinh gia tri luong giac cua mot so goc dac biet I Phuang phdp • Dua vao dinh nghia, tim tung dd y^ vk hoanh dd x^ cua diim M tren nira dudng trdn don vi vdi gdc xOM = a va til dd ta cd cac gi^ tri lugng giac : _>'o - ^ sin a = j„ ; cos a = x^ ; tana = - ^ ; cota = "0 ^0 • Dua vao tfnh chit : Hai gdc bu cd sin bing va cd cdsin, tang, cdtang dd'i Cdc vi du Vi du Cho gdc a= 135° Hay tinh sina, cosor, tana va cota GlAl Ta cd sinl35° = sin(180°- 135°) = sin45° = — ; /? cos 135° = -cos(180°- 135°) = -cos45° = - ^ ^ ; 68 ,.,^0 tanl35°= Dodd sinl35° COS 135o =-1 cotl35° = - l Vl du Cho tam giac can ABC cd B = C = ^5° Hay tfnh cac gia tri li/dng giac cOa gdc A GlAl Tacd A = 180°-(B + C) = 180°-30° =150° vay sinA = sin(180° - 150°) = sin30° = - ; cosA = -cos(180° - 150°) = - cos30° = - — ; , sin 150° V3 tanA= = cos150° Dodd cot A = - v 2£ VANdE2 Chiing minh cac he thiic ve gia tri luong giac / Phuang phdp • Dua vao dinh nghia gia tri lugng giac cua mdt gdc a (0° < a < 180°) • Dua vao tfnh chit ciia ting ba gdc cua mdt tam giac bing 180° o' J ' UA 1.' •2 •, sin a • Su dung cac he thuc cos a + sm a = ; tana = cos a ; tana = cota Cdc vidu Vi du Cho gdc a bat ki ChCmg minh rang sin'^a - cos'*a = 2sin^a - 69 GlAl Cdch / Ta cd cos'^a = (cos^a)^ = (1 - sin^a)^ = - 2sin^a + sin"*a Dodd sin a - c o s a = s i n a - Cdch Ta bilt ring sin a - cos a = (sin a + cos a)(sin a - cos a) = [sin^a- ( - sin^a)] = 2sin a - Cdch J Ta cd thi sir dung phep biln ddi tuong duong nhu sau : sin'^a - cos'* a = 2sin^a - (*) sin'^a - 2sin^a + - cos'^a = (1 - sin^a)^ - cos'^a = cos'^a - cos a = Vi he thiic cudi ciing ludn ludn diing nen he thiic (*) diing Vi du Chiimg minh rang : a) + tan^a= — ^ (vdi a ^ 90°); cos a b) +C0t^a: (vdia^0°;180°) sin^a GIAI 2 , sm a — a) + tan a = + cos a , COS a b) +cot a = + — - — sin a 2 cos a +sin a _ 2 "" cos a cos a 2 sin a +COS a _ '• T^ sm a sm a Vi du Cho tam giac ABC Chufng minh rang a) sin A = sin(fi + C); , A e+c b)cos— =sin ; / 2 c) tan A = -tan {B + C) 70 GlAl Vi 180°-A = B + C nen tacd: a) sin A = sin (180° -A) = sin {B + C); b) cos— = sin vi — + = 90° (hai gdc phu nhau); 2 2 c) tan A = -tan (180° -A) = -tan (B + C) 2£ VAN dE Cho biet mot gia tri luong giac cua goc a, tim cac gia tri luong giac lai cua a Phuang phdp S& dung dinh nghia gia tri lugng giac cua gdc a vk cac he thiic co ban lien he giiia cac gid tri dd nhu : 2 , sina cosa sin a + cos a = 1; tana = ; cota = ; cosa sina , 2 ' • -^ + tan a = — ; + cot a = COS a sm a Cdc vidu Vi du Cho biet cosa= — , hay tinh sina va tana ' Vi GlAi cosa < nen 90° < a < 180° Suy sina > va tana < Vi sm a + cos a = nen thay gia tri cosa = — vao ta cd : 2 2 • sm a + — = => sm a = — 71 Vay sina= — • sina tana= 'x = ^^ = cosa _£ v5 • Vi du Cho gdc a, biet 0° < a < 90° va tana = Tfnh sinava cos a GiAi sin CC Theo gia thilt ta cd : = Do dd sina = 2cosa (1) cosa Mat khac ta lai cd : sin a + cos a = (2) Thay (1) vao (2) ta cd : 4eos^a + eos^a = 5cos^a= => cos^a= — Vi 0° < a < 90° nen cosa > 0, dd cosa = — , ma sina = 2cosa nen ta CO sin a= 2V5 • Vi du Cho gdc a, biet cosa= — Hay tinh sina, tana, cota GiAi 16 = — ^ sina = — (vi sina > 0) 25 25 4 ^ , , sina = —: — = — Do cota = — tana = cosa 5 Ta cd sin a = - cos a = Vl du Cho gdc a biet tana = - Tfnh cosa v^ sina GIAI Vi tana = - < nen 90° < a < 180°, suy cosa < 72 Vi +tan^a = nen cos a = COS a Vay cosa = 1 + tan^a 1+4 'S Mat khac sin a = cosa tan a = (-— V5, V5 Nhan xet Cd thi diing he thiic sin a + cos a = dl tfnh sin a nhu sau sin^a = - cos^a = Dodd 2£ VAN = —• 2>/5 sina=—r= = (visina>0) >/5 ' dE Cho biet mot gia tri luong giac cua goc a, hay xac dinh goc a / Phuang phdp Sir dung dinh nghia gia tri lugng giac cua gdc a di dung gdc a vk mdt sd trudng hgp cd thi sir dung ti sd lugng giac cua gdc nhgn dl dung gdc a Tap sir dung may tfnh bd tui dl xac dinh gdc a Cdc vidu Vi du X^c djnh gdc nhgn a biet sin a= —• GIAI Cdch I Trtn true Oy ciia nira dudng trdn don vi ta liy diem / = | ; — va qua dd ve dudng thing d song song vdi true Ox (h.2,3) 73 Dudng thing cit nira dudng trdn don vi tai hai diim M vk N dd xOM la gdc til va xON la gdc nhgn Ta xac dinh dugc gdc a = xON cd sma= — • Cdch Ta dung tam giac ABC vudng tai A, cd AB = 3, BC = (h.2.4) Ta cd a= ACB vi sin ACB = AB BC Cdch Dung may tfnh bd tui (Casio fx-500MS) • Chgn don vi : Sau md may Sin phfm len ddng chu iing vdi cac sd sau day : nhilu lin dl man hinh hien Sau dd in phfm de xac dinh don vi gdc la dd • Ta tfnh sina = — = 0,6 : An lien tilp cac phfm sau day : SHIFT tin' Ta dugc kit qua la : a « 36°52'11" Vl du Xac djnh gdc a bi§t rang cosa= — • o GlAl Cdch Tren true Ox ciia nira dudng trdn don vi ta liy diim H = vk qua dd ve dudng thing m song song vdi true Oy (h.2.5) Dudng thing cit nira dudng trdn don vi tai M Ta cd gdc a= xOM lA Cdch Ta bilt ring cos a = -cos (180° - a) Theo gia thilt cos a = — , vay cos (180° - a)= -• Ta dung tam giac ABC vudng tai A cd AB = 1, BC = (h.2.6) Ta cd cos ABC = - ntn cos (180° - ABC) = • 3 vay a = 180° - ABC = ABC' (tia BC ngugc hudng vdi tia BC) Cdch Dung may tfnh bd tiii (Casio fx-500MS) Tuong tu nhu tfnh sina Vi cos a < nen a la gdc tu An lien tilp cac phfm sau day : SHin cor' lOoo'l £ " Ta dugc kit qua la : a « 109°28'16 C 2.1 CAU HOI VA BAI TAP Vdi nhiing gia tri nao ciia gdc a (0° < a < 180°) thi: a) sin a vk cos a ciing diu ? c) sin a va tan a cung diu ? b) sin a va cos a khac dau ? d) sin a va tan a khac diu ? 2.2., Tfnh gia tri lugng giac ciia cae gdc sau day : a) 120°; b) 150°; c) 135° 2.3 Tfnh gia tri ciia bilu thiifc : a) 2sin 30° + 3cos 45° - sin 60° ; b) 2cos30° + 3sin 45° - cos 60° 75 a^ = 9b^ => a^ = 9(a^ - e^) => 9e^ = 8a^ =» 3e = 2V2a v a• ^v a' ^ 3^ b) FjBjF2=90° ^ ^ l ^ O OB, = - ! - ^ => fe = e ^ f e = e2 ^a^-c^ = c^ =» a^ = 2c^ => a = cV2 a V2 c) AjBj = 2e =i> AjBj^ = 4e^ => ^2 + ^2 ^ 4^2 «2 + ^ _ ^ ^ ^ => => 2a^ = 5c^ => V2a = VSe ^*>'f=|167 3.36 (F): 4x^ + 9y^ = 36 (1) Xet dudng thing (d) di qua diim M(l ; 1) va cd he sd gdc it Ta cd phuong trinh eua (d)-.y - I = k(x- I) hay y = k(x-l) + I (2) Thay (2) vao (1) tadugc A^ + 9[k(x-l)+lf ^ = 36 (9k^ + A)x^ + lSk(l - k)x +9(1-k)^-36 = (3) Ta cd : (d) cit (F) tai hai diem A, thoa man MA = MB va ehi phuong trinh (3) cd hai nghiem x^, Xg cho : "'A+^'B -lSk(l-k) X/^ 2 3.43 a ) ( F ) : — + ^ = 1; 2 b)(F): — + ^ = 25 16 2 3.44 (F): — + ^ = 25 Ta cd : a^ = 25,fe2= ^ e^ = a^ -fe^= 16 : ^ e = vay (F) cd hai tieu diim la Fj(-4 ; 0) va F2(4 ; 0) Ta ed : cfj = cf(F, , A) = J, = d(F^ , A) = I-4A + CI |4A + C| a, = C -16A^\ buyra a, r— ' ^ A^+B^ Thay C^ = 25A^ + 9B^ vao (1) ta dugc : |25A^+9B^-16Al 9(A^ + B^) ^1-^2 = Vay 170 (1) ^1.^2 = 9- A' + B' A^+ B^ 2 3.45 (F): x^ + 4y^ = 16 « — + ^ = 16 Ta ed a^ = 16 ,fe^= ^ c^ = a^ -fe^= 12 =*e = 2V3 vay (F) cd hai tieu diim : Fj(-2 Vs ; 0) va F2(2 Vs ; 0) va cae dinh Aj(-4 ; 0), A2(4 ; 0) Bj(0;-2),B2(0;2) b) Phuong trinh A cd dang = hay x + y - = l(x-l)+2 y c) Toa dd ciia giao diim ciia A va (F) la nghiem cua he : I x^+Ay'^ =16 [x = 2-2y (1) (2) Thay (2) vao (1) ta duge : (2 -2y)^ + 4y^ = 16 « ( l - y ) ' + y' = 2y^ - 2y - = (3) Phuong trinh (3) cd hai nghiem y^, y^ thoa man IA1IB_-1-1.2 ~4~2"^^vay MA = MB 1-V7 + V7 Tacd y^ = = + V7, x „ = l - V7 vay A cd toa la l + Jl ; 1-V71 _ .r, f, i+j^' , B cd toa la 1-V7; 171 CAU HOI T R A C NGHlfiM 3.46 BA = (-2 ; 2) 'BC =.(2; 2) 'BA.'BC =0=> ABC = 90° Dudng trdn ngoai tilp cd tam la trung diim / cua AC ntn cd toa dd (3 ; 4) Chgn(D) 3.47 Chgn (A) a, • 3.48 Dudng thing A : 6x - 4y - 12 = cit Ox vk Oy lin lugt tai A(2 ; 0) va B(0;-3) Ta CO AB = Jl3: Chgn (Q 3.49 Chgn (A) 3.50 Dudng thing A:2x + y - = song song vdi du^g thing d:Ax + 2y+ 1=0 vk di qua diim M(l; 2) Chgn (C) 3.51 Dudng thing A : 3x + 5y + 2006 = cd he sd gdc lkk= — Phat bilu (C) sai Chgn (C) 3.52 Diim C(2; 2) cd toa dd thoa man phuong trinh dudng thing A: x - 2y + = Ta lai ed TlC = (1 ; -2), n^ = (1 ; -2) suy MC vudng gdc vdi A Vay C(2 ; 2) la hinh chilu vudng gdc cua M xudng A Chgn (C) 3.53 Dudng thing A di qua A(l ; 1), B(2 ; 2) cd vecto chi phuong AB = (1 ; 1) ^ {x = l + t vay A ed phuong trinh tham sd -^ b=i+^ Diim 0(0 ; 0) thoa man phuong trinh cua A (dng vdi t = -1) Vay phuong , , , , \x = t tnnh tham sd cua A cd the viet la < {y = t Chgn(D) 3.54 K = d(0;A)= Chgn (D) 172 , ^^ = 10 V64 + 36 1-6 3.55 cos(A A-)= ^L— ^—= = -f= ^ ^ JI + AJI + J2 Chgn(C) 3.56 (Ox, AJ ) = 45°, (Ox, A^) = 60° Suy (Aj, A2) = 15° Chgn (B) 3.57 Phuong tiinh x^+y^+x + y + = khdng la phuong trinh cua dudng ti-dn vi khdng thoa man dilu kien a + fe - c > Chgn (B) 3.58 Toa dd ba diim A(-2 ; 0), B(>^; V2), C(2 ; 0) diu thoa man phuong trinh dudngti-dnx^ +y^ = Chgn (A) 3.59 Dudngti-dnndi tilp tam giac OAB cd tam I(a ; a) Ta ed d(I, AB) = d(I, Ox) suy 7(1 ; 1) Ta cd B = d(I, Ox) = l vay phuong trinh cua dudng trdn ndi tilp tam giac OAB la.: • x^+y^-2x-2y +1=0 Chgn (C) 3.60 (Cj) cd tam /i(-l ; 3) va bdn kfnh Bj = (C2) cd tam /2(2 ; -1) vk ban kfnh B2 = Tacd/j/2=Bj+B2 vay (Cl) tilp xiic ngoai vdi (Cj) Chgn(D) 3.61 Tilp tuyln A cd vecto phap tuyln OM^ = (1 ; 1) Phuong trinh A cd dang l.(x-l) + l.(y-l) = hay X + y - = Chgn (A) 3.62 IM>R suy diim M nim ngoai dudng trdn Chgn (C) 3.63 Dudngti-dn(C) di qua gd'c 0(0 ; 0) Cljgn (B) 3.64 Chgn(B) 3.65 Chgn (C) 173 3.66, (F) di qua cac diim Mj, Mj, M3 Chgn (D) 3.67, Chgn(D) 3.68, Chgn(C) 3.69, (C) tilp xuc vdi (F) tai Ai(-5 ; 0) va A2(5 ; 0) Chgn (C) 3.70, J(Fi, A) X d(F2, A) = b^ = Chgn (B) 3.71, 0A = OB = OC = Dudng trdn ngoai tilp tam giae ABC cd phuong trinh x^ + y^-9 = Chgn (D) 3.72, Atie'pxucvdiC(0; l)rf(C; A)=l 9 a) Chiing minh rang tap hgp cae diem M(x ; y) thoa man MA + MB = MC la mdt dudng trdn b) Tim toa dd tam va ban kfnh cua dudng trdn ndi tren 174 Cho hai diim A(3 ; -1), B(-l ; -2) va dudng thing d ed phuong trinh X + 2y + = a) Tim toa dd diim C tren dudng thing d cho tam giac ABC la tam gide can tai C b) Tim toa dd eiia diim Mti-endudng thing d cho tam giac AMB vudng taiM Trong mat phing Oxy cho dudng trdn (J) cd phuong trinh x2 + y2 - 4x - 2y + = a) Tim toa tam va tfnh ban kfnh ciia dudng trdn (F) b) Tim m dl dudng thing y = x + mc6 diem chung vdi dudng trdn (T) e) Vilt phuong trinh tilp tuyln A vdi dudng trdn (T) bilt ring A vudng gdc vdi dudng thing dcd phuong trinhx-y + 2006 = Trong mat phing Oxy cho elip (F) cd tieu diim thii nhat la (-V3 ; 0) va di qua diem M 1; V^^ a) Hay xac dinh toa eac dinh cua (F) b) Vie't phuong trinh chfnh tic ciia (F) c) Dudng thing A di qua tieu diim thur hai cua elip (F) va vudng gdc vdi true Ox va cit (F) tai hai diim C va D Tfnh dai doan thing CD HUCnSTG DAU GIAI VA DAP SO' a) AM = -AG o ^A^-l=-(l-l) M Xu = M yM-2 vay M cd toa dd la 1; (h.3.11) Diim A^(x ; y) thoa man he phuong trinh rx + y = ^\x = l-x + y = l y = vay N cd toa dd la (0 ; 2) x ^ - = 2(1-0) b) AB = 2iVM ã y,-l-2 |jô=2 175 Eằudng thing chiia canh AB di qua hai diim A(l ; 1) va B(3 ; 2) nen cd phuong trinh : x - 2y + = Dudng thing chiia canh BC di qua hai diim B(3 ; 2) va Ml 1; — j nen ed phuong tiinh : x + 4y - 11 = (Xem hinh 3.12) MA = 3MG