Hướng dẫn giải bài tập Hình học 10: Phần 2

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 ... cd a2=fe2+c2_2feccosA = 7^+5^ -2. 7.5.- = 32 =>a = 4V2 (cm) 2 16 = — =>sinA = —(vi sinA >0) sm A = l-cos A = l 25 25 5 = -fecsinA = - - = 14 (cm^) 2 - , b) h = — = —7= = '' 2. 5 28 a 4V2 7V2 ,... = 2R (R la ban kfnh dudng trdn ngoai tilp tam giac ABC) Dp ddi dudng trung tuyen cua tam giac b +c m = ' ' 2 , 2 a +c m = ' ' 22 , , m2 a + f e a 2( fe +c )-a = 4 ,2 ^ , b 2( a +c )-b c = 2^ ,2. .. AM = m , BN = m^^, CP = m^ (h .2. 14) Dinh li cosin a =b +c -2feccosA fe =a +e 2 -2accos8 c =a +b -2abcosC He qua: 1 ,2 , 2 cos A = fe +c -a 2bc + c'- b' 2ac + b'- -c 2ab cosB = a cos C = a Dinh li