H v tn SV: H QUC TRUNG Lp: HSP Ton 07B MSSV: 107121088 GVHD: Thy
THI KIM TINH TI:
CC DNG TON V H THC LNG TRONG TAM GIC V PHNG PHP GIIPHN 1: TM TT
CC NI DUNG C BN I. Kin thc c bn: 1. Cc k hiu:
II. Cc h thc lng trong tam gic vung 1) Cc h thc v cnh v ng cao
trong tam gic vung - nh l 1: b2 = a. c ; c2 = a .cc A b h B a C
- nh l 2: h = b.c - nh l 3: b.c = a.h - nh l 4:1 1 1 2 = 2 + h b
c2
2
`
H
2) Cc h thc v cnh v gc trong tam gic vung b = a.SinB = a.CosC c
= a.SinC = a.CosB b= c.TgB= c.CotgC c = b.TgC = b.CotgB1
- Nu bit 1 gc nhn th gc cn li l 900 - Tm gc - Nu bit 2 cnh th tm
1 t s lng gic ca gc - Dng h thc gia cnh v gc trong tam gic vung - T
h thc : A b = a.SinB = a . CosC b c a= a= b b = SinB CosC C C =
SinC CosBB a C
c = a. SinC = a . CosB
III. Cc h thc lng trong tam gic thng: 1. nh l hm cosin: Trong
tam gic ABC ta c:
2. nh l hm sin: Trong tam gic ABC ta c:
2
Ghi nh: Trong mt tam gic, t s gia mt cnh ca tam gic v sin ca gc
di din vi cnh bng ng knh ng trn ngoi tip tam gic. 3. nh l v ng
trung tuyn: Trong tam gic ABC ta c:
4. nh l v din tch tam gic: Din tch tam gic ABC c tnh theo cng
thc sau:
3
5. nh l v ng phn gic trong ca tam gic:
6. Hnh chiu:
A
a = b.cos C + c.cos B
c
b
B
a
C
7. Cc cng thc bin i lng gic a) Cng thc cng:
b) Cng thc nhn i:
c) Cng thc nhn ba:
d) Cng thc h bc:
4
e) Cng thc bin i tng thnh tch:
a+ b a b cos a + cos b = 2 cos cos 2 2 a+ b a b cos a cos b = 2
sin sin 2 2 a+ b a b sin a + sin b = 2 sin cos 2 2 a+ b a b sin a
sin b = 2 cos sin 2 2 sin(a b ) tan a tan b = cos a .cos bf) Cng
thc bin i tch thnh tng:
PHN HAI: CC DNG TON V H THC LNG V PHNG PHP GII A. DNG 1: CHNG
MINH CC NG THC LNG GIC TRONG TAM GIC ng thc lng gic c dng A=B Phng
php 1: dng cc php bin i tng ng bin i v ny thnh v kia Phng php 2:
xut pht t h thc ng bit suy ra ng thc cn chng minh V d 1: Cho tam
gic ABC. Chng minh
A = 2 B 2a
2
= b
b. c +
5
Gii:
2 sin( A B) a b V d 2: Cho tam gic ABC. Chng minh: = sin C
c2
2
V d 3: Cho tam gic ABC. Chng minh A B B C C A tg tg + tg tg + tg
tg 1 = 2 2 2 2 2 2
6
V d 4: Cho tam gic ABC. Chng minh:
Gii:
7
e / D A =180 C o +B tg+ = ) ( B A tgC tgA + tgB = tgC 1 tgAtgB .
tgA tgB + + = tgC tgAtgBtgC . . tgA tgB = + + tgC tgAtgBtgC . .
8
I.
B. DNG 2: CHNG MINH BT NG THC LNG GIC TRONG TAM GIC Kin thc h
tr: a/ Bt ng thc tam gic:
b/ Cc bt ng thc c bn: 1. Bt ng thc cosi;
2. Bt ng thc bunhiacopxki:
9
t cng bng 0 3. Bt ng thc c bn:
4. Bt ng thc Jensen:
chng minh bt ng thc A; ; ) ta c th thc hin cc phng php sau?
Phng php 1: Phng php 2: S dng cc bt ng thc c bn bit suy ra bt ng
thc cn chng minh. Phng php 3: Dng tam thc bc 2 Phng php 4: Dng phng
php hm s V d 1: Cho tam gic ABC nhn. Chng minh rng:10a / tan A +
tan B + tan C cot b/A B C + cot + cot 2 2 2 A B C tan A + tan B +
tan C cot + cot + cot 2 2 2Giia/ Ta c:sin( A + B ) sin( A + B ) =
cos A.cos B 1 cos( A + B ) + cos( A B ) [ ] 2 2sin( A + B ) A+ B C
2 tan = 2 cot 2 2 [ cos( A + B ) + 1] Tng t ta cng c: A tan B + tan
C 2 cot 2 B tan C + tan A 2 cot 2 Suy ra : tan A + tan B =A B C +
cot + cot 2 2 2 Du = xy ra khi va ch khi A=B=C hay tam gic ABC u 2
C b/ S dng bt ng thc cosi v tanA.tanB tan 2 Ta c: C tan A + tan B 2
tan A.tan B 2cot 2 2 C tan A+ tan B 4cot 2 C tan A+ tan B cot 2 Tng
t ta cng c: tan A + tan B + tan C cot()tan B +tan C cot tan C +tan
A cot A 2 B 2Suy ra:tan A + tan B + tan C cotV d 2 Cho tam gic ABC.
Chng minh:A B C + cot + cot 2 2 2SinA+SinB+SinC 113 3 2Xt hm s
f(x)= sinx vi x (0; ) o hm:Giif '( x) = cos x; f ''( x) = sin x 0 x
0; ) ( f ( A) + f( B) + f( C) A+ B +C f( ) 3 3 sin A + sin B + sin
C sin( ) 3 3 3 3 sin A +sin B + C sin 2 Du = xy ra khi v ch khi
A=B=C hay tam gic ABC u V d 3:Vy hm s f(x)= sinx l li trn (0; ) (0;
Do : A; B; C )GiiV d 4: Hy tm trong cc tam gic ngoi tip mt ng trn
cho trc mt tam gic c din tch nh nht. Gii12V d 5: Chng minh rng vi
mi tam gic ABC bt k ta lun c: Gii13C. DNG 3: NHN DNG TAM GICPhng
php: s dng cc php bin i tng dng hoc h qu bin i iu kin cho trc n mt
ng thc m t c th d dng kt lun c tnh cht ca tam gic V du 1:Gii Ta c:
p dng nh l hm sin14V d 2: Gii p dng nh l v ng trung tuyn:V d 3:
GiiVy tam gic ABC vung hoc cn ti C V d 4: Tnh cc gc tam gic
nu:15GiiHay du = xy ra khi tam gic ABC vung cn ti A. V d 5:Gii16V d
6: Gii Ta c:V d 7:Gii Ta c: (*) D. CC DNG BI TP V H THC LNG TRONG
TAM GIC I. NH L HM SIN V COS17Bi 1: GiiBi 2: GiiBi 3:Gii18Bi 4: Th
a2; b2; c2 l cp s cng Gii19Bi 5: Gii20Bi 6:GiiII. NH L V NG TRUNG
TUYN21Bi 1: Cho tam gic ABC c AM trung tuyn, gc AMB= ,AC=b, AB=c, S
l din tch tam gic ABC. Vi 0< < 900Gii22Bi 2:Gii23Bi 3: tuyn
BB th cotgC = 2(cotgA + cotgB) Gii:Bi 4:Gii Ta c24III. DIN TCH TAM
GICBi 1: Gii:25Bi 2:Gii:Bi 3:Gii:2627III. BN KNH NG TRNBi 1:Gii28Bi
2:Gii29Bi 3:30Gii31PHN BN: BI TP CNG C Bi 1:Bi 2:Bi 3: Bi 4:Bi 5:
Bi 6: tam gic l tam gic vung Bi 7: cnBi 8: uBi9:32Bi 10:Bi 11: Bi
12:Bi 13:Bi 14:Bi 15:Bi 16:Bi 17:Bi 18:33Bi 19:Bi 20:Bi 21:Bi 22:TI
LIU THAM KHo341. L Hng c. Phng php gii ton lng gic. NXB i hc s phm
2004 2. Phan Huy Khi. Tuyn tp cc bi ton lng gic. NXB gio dc 1996 3.
Tp ch ton hc v tui tr Cc wesite www.hocmai.com www.vnmath.com
www.tuoitreonline.com www.thaydo.net www.giaovien.net35MC LC Trang
PHN 1: TM TT CC NI DUNG C BN 1 PHN HAI: CC DNG TON V H THC LNG V
PHNG PHP GII 5 A. DNG 1: CHNG MINH CC NG THC LNG GIC TRONG TAM GIC
5 B. DNG 2: CHNG MINH BT NG THC LNG GIC TRONG TAM GIC 9 C. DNG 3:
NHN DNG TAM GIC 14 PHN BA: CC DNG BI TP V H THC LNG TRONG TAM GIC
18 I. NH L HM SIN V COS 18 II. NH L V NG TRUNG TUYN 22 III. DIN TCH
TAM GIC 25 III. BN KNH NG TRN 28 PHN BN: BI TP CNG C 32 TI LIU THAM
KHO 3536
