Nội dung

Chi/ONq II TICH VO HUCfNG CUA HAI VECTO VA LfNG DUNG §1. GIA TRI LUONG GIAC CUA MOT GOC BATKITtro^D^NlSO® 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) sao 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 ^ 0 ggi la tang cua gdc a vk dugc kf hieu la tan a= >' - o^ • Ti sd - ^ vdi Jo 5t 0 ggi 1^ cdtang cua gdc avk duoc kf hieu la >'o cot « = - ^ . 66 S - BTHHIO - B 2. Cdc he thiic luong giac a) Gia tri lugng giac cua hai gdc bii nhau 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 ra cac he thiic 2 2 sin a + cos a = I ; sma = tana (a 7^90°); cos a 1 cota = tana 2 1 +tan a = cos a = cota(a^0°;180°); sma 1 tana = cot a 1 1 + cot a = 2 cos a . 2 sm a 3. Gid tri luong giac cua cdc gdc dac Met Giatif\^^ lugng giac ^"^\^^ 0° 30° 45° 60° 90° 180° sin a 0 1 2 :/2 2 :/3 2 1 0 cos a 1 2 1 2 0 -1 tana 0 . 1 S II 0 cot a II 0 II 2 1 s 1 1 67 4. Gdc giUa hai vecta Cho hai vecto a vk b dtu khac vecto 0. Tut mdt diim O bit ki ta ve OA = a va OB = fe. Khi dd gdc AOB vdi sd do 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^ 0 ; tana = - ^ ; cota = "0 ^0 • Dua vao tfnh chit : Hai gdc bu nhau cd sin bing nhau va cd cdsin, tang, cdtang dd'i nhau. 2. Cdc vi du Vi du 1. 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 =-1. cotl35° = - l . Vl du 2. 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° = - — ; 2 , sin 150° V3 tanA= = cos150° 3 Dodd cot A = - v 3 . 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 •2 •, sin a 1 • Su dung cac he thuc cos a + sm a = 1 ; tana = cos a ; tana = cota 2. Cdc vidu Vi du 1. Cho gdc a bat ki. ChCmg minh rang sin'^a - cos'*a = 2sin^a - 1 . 69 GlAl Cdch / . Ta cd cos'^a = (cos^a)^ = (1 - sin^a)^ = 1 - 2sin^a + sin"*a. Dodd sin a - c o s a = 2 s i n a - 1 . Cdch 2. Ta bilt ring sin a - cos a = (sin a + cos a)(sin a - cos a) = 1. [sin^a- ( 1 - sin^a)] = 2sin a - 1 . Cdch J. Ta cd thi sir dung phep biln ddi tuong duong nhu sau : sin'^a - cos'* a = 2sin^a - 1 (*) <=> sin'^a - 2sin^a + 1 - cos'^a = 0 <=> (1 - sin^a)^ - cos'^a = 0 <=> cos'^a - cos a = 0. Vi he thiic cudi ciing ludn ludn diing nen he thiic (*) diing. Vi du 2. Chiimg minh rang : a) 1 + tan^a= — ^ (vdi a ^ 90°); cos a b) 1 +C0t^a: (vdia^0°;180°). sin^a GIAI . 2 2 , sm a — a) 1 + tan a = 1 + cos a ... 2 , 2 COS a b) 1 +cot a = 1 + — - — sin a 2 2 cos a +sin a _ 2 1 2 "" cos a cos a 2 2 sin a +COS a _ 1 '• T^ . 2 sm a sm a Vi du 3. Cho tam giac ABC. Chufng minh rang a) sin A = sin(fi + C); ., A . e+c b)cos— =sin ; / 2 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 2 2 c) tan A = -tan (180° -A) = -tan (B + C). 2£ VAN dE 7 Cho biet mot gia tri luong giac cua goc a, tim cac gia tri luong giac con lai cua a 1. 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 2 , sina cosa sin a + cos a = 1; tana = ; cota = ; cosa sina 2 1 , 2 1 2 ' • -^ .2 1 + tan a = — ; 1 + cot a = COS a sm a 2. Cdc vidu 2. Vi du 1. Cho biet cosa= — , hay tinh sina va tana. 3 ' Vi GlAi cosa < 0 nen 90° < a < 180°. Suy ra sina > 0 va tana < 0. Vi sm a + cos a = 1 nen thay gia tri cosa = — vao ta cd : . .2 2 2 2 4 • 9 5 sm a + — = 1 => sm a = —. 71 Vay sina= — • sina tana= 'x = ^^ = cosa _£ v5 • 2 3 Vi du 2. Cho gdc a, biet 0° < a < 90° va tana = 2. Tfnh sinava cos a. GiAi sin CC Theo gia thilt ta cd : = 2. Do dd sina = 2cosa. (1) cosa Mat khac ta lai cd : sin a + cos a = 1. (2) Thay (1) vao (2) ta cd : 4eos^a + eos^a = 1 <=> 5cos^a= 1 => cos^a= — 5 Vi 0° < a < 90° nen cosa > 0, do dd cosa = — , ma sina = 2cosa nen ta 5 . . CO sin a= 2V5 • 3 Vi du 3. Cho gdc a, biet cosa= — Hay tinh sina, tana, cota. 5 GiAi 16 4 = — ^ sina = — (vi sina > 0) 25 25 5 4 3 4 ^ . , , 3 sina = —: — = — Do do cota = — tana = cosa 5 5 3 4 7 9 Ta cd sin a = 1 - cos a = 1 9 Vl du 4. Cho gdc a biet tana = - 2 . Tfnh cosa v^ sina. GIAI Vi tana = - 2 < 0 nen 90° < a < 180°, suy ra cosa < 0. 72 Vi 1 +tan^a = 2 nen cos a = COS a Vay cosa = 1 1 + tan^a 1 1+4 1 5 'S Mat khac sin a = cosa. tan a = (-— . V5, V5 5 2 Nhan xet. Cd thi diing he thiic sin a + cos a = 1 dl tfnh sin a nhu sau sin^a = 1 - cos^a = 1 5 Dodd 2£ VAN = —• 5 2 2>/5 sina=—r= = (visina>0). >/5 5 ' dE 4 Cho biet mot gia tri luong giac cua goc a, hay xac dinh goc a do /. Phuang phdp Sir dung dinh nghia gia tri lugng giac cua gdc a di dung gdc a vk trong 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. 2. Cdc vidu Vi du 1. X^c djnh gdc nhgn a biet sin a= —• 5 GIAI Cdch I. Trtn true Oy ciia nira dudng trdn don vi ta liy diem / = | 0 ; — va qua dd ve dudng thing d song song vdi true Ox (h.2,3). 73 Dudng thing nay cit nira dudng trdn don vi tai hai diim M vk N trong dd xOM la gdc til va xON la gdc nhgn. Ta xac dinh dugc gdc a = xON cd 3 sma= — • 5 Cdch 2. Ta dung tam giac ABC vudng tai A, cd AB = 3, BC = 5 (h.2.4). Ta cd a= ACB vi sin ACB = AB BC 3 5 Cdch 3. Dung may tfnh bd tui (Casio fx-500MS). • Chgn don vi do : Sau khi md may Sin phfm len ddng chu iing vdi cac sd sau day : nhilu lin dl man hinh hien Sau dd in phfm 9 0 1 de xac dinh don vi do gdc la dd. 3 • Ta tfnh sina = — = 0,6 : 5 An lien tilp cac phfm sau day : SHIFT tin' Ta dugc kit qua la : a « 36°52'11". . 1 Vl du 2. Xac djnh gdc a bi§t rang cosa= — • o GlAl Cdch 1. 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 nay cit nira dudng trdn don vi tai M. Ta cd gdc a= xOM. lA Cdch 2. 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 = 3 (h.2.6). Ta cd cos ABC = - ntn cos (180° - ABC) = --• 3 3 vay a = 180° - ABC = ABC' (tia BC ngugc hudng vdi tia BC). Cdch 3. Dung may tfnh bd tiii (Casio fx-500MS) Tuong tu nhu tfnh sina. Vi cos a < 0 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