-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
1/341
8.HHHPHThc s. nh gio u t
MI
Bin son theo ni dung v hng ra thi mi ca B GD&T.Dnh cho HS
11, 12 chng trnh c bn v nng cao.
o r a _mkXHT BN I HCquc GIAH \l
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
2/341
1
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
3/341
NII XT EN DI HC QUC GIA SN16 Hng Chui - Hai B Trng - H Ni
T (04) 9715013; (04) 7685236. Fax: (04) 9714899
Ch u t r c h n h i m x u t bs%:
Gim c PH N G QUC BO
Tng bin tp NGUY N B TIINII
Bin tp ni dungHU NGUYN
Sa bi
L I-IO, ANH TH
Ch bn
CNG TI ANPXA
Trnh by biaSN K
i tc lin kt xut bnCNG TI ANPI-A
1234 B TP T LUSVS M h n h hMH HC - LNG CC
M s": 1L-01H2009In 2.000 cun, ki 16 X24 C i lai Cong ti TNH n
liau ) s!j'tii r'itijS'xut bn: 920-2008/CXB/0:-162/DHQCHN, ngy
02/10/200Quyt ti I1I xu L>in s": 01LK-TN/XB
n xong v up lu chiu qu nm 2009.
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
4/341
e^ u^ ^ eK C cc-c thi quc gia (tt nghip / tuyh sinh...) do B
t234icMtv-. bi tp trc nghim in
bi- y b.-.tp trc nghim in hnh V t ?',^ cun sch: "2234 bi t p t
lun n hnh.
^^-^Gt0jH.-^--?234 bi tp tun ih hnh Mink hc - Lng gic" o xi g i
a p t hc s L H onh Ph t chc bin son.' :yNm:l^sch^bm ^st cc k nng,,
kin thc chun ca cKng trnh:
%^;SC^^br|fcai rr (chng trnh chun v chng trnh nng cao).
Tcgif:^ec^on v gi iiu mt khi lng ln cc dng tOkr in hnh
vlt^ r^ gip-trng cc thi d B GD&T ra trong cc nm qua.
Cc^:'to^| ^ c chn lc t cc dng ton cng c l thuyt v rn
n t^uy..v bn.c cn tm thy nhiu dng bi tp nng cao t d h
:v: cho vic h thng ton b kin thc trng tm mn
sirih rn luyn v nng cao k nng t duy cht ch iiieo
lpluntOc^hc.
nghim c mt thy gio c thi gian di trc ip ng lp i^^^ y bi dfng hc
sinh gii chuyn ton, chng ti tin. rang cuh
schM s ern li nKng iu b ch cho bn c v t c nhng, kt!: ;^;trpng
rihtmg k.thi sp fr.. i- o thi gian b in son c hn, mc d rt c'gng
nhng cun sch ny
G0 the_c0 ring khim khuyt, Ttmong nhn c nhng ng gp chn H rhicu c
c n g n g h i p v c c b n h c, s i n h g r v x a b s c h c h o
n
thin/ hn trng ln ti bn sau.
}--::'T&;t gp- xin g v:- Trung tm Sch Gio dc Anpha
;v. 225C Nguyn Tr Phng, P.9, Q.5, TpJHCMT: 08.8547464 ;
- Cng ti Sch - Thit b Gio dc npha
50 Ng uy n Vn Sng, qun Tn PhyTp.HCM.T:08.2676463, 08.38107718
;
Email: atphabokeenteryah oo.com
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
5/341
j
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
6/341
D A N G 1: H T H d & B N - C N G T HC-Grs G S
1. L THUYT V PHNG PHP GII TON
- Hm s lng gic:' {ng trn lng 'gic (O; 1); im cui Mc, y ) e (;
1)' ca gc lng gic a = (Ox, OM). '. - ' /...5 ; , ^
- y ^ . sina =y; co sa = x; tana . , ' 1 ( 1 ' - X . . -V-" .' '
. 1 "1 - " ^ - - * \ o-
a 5 + k?t; cota =r , a ^ k X . X- . . ' , 2^ . X . '- C un ggoc
c bit: \ ' -
\ ::: w ' \ _ ' ... V' v. -
irtrxV-- y. :
0 ' n/6 riz/4 '1Z2 "2n/3 3tc/4' *- >
- 0 30' 45 60 90* 120 , 135 .150. 180:sin .> .0 . - 1/2. V2/2
&/2 *- 1 ' \ V3 / 2 t V2 /2 -- '-
cos i \ 1 -V/2 42. /2 - 0 - -1/2 -> /2 /2 ,.. s i A ( 2 ~.
-
* ta*}-' ' 0 / S \ V3 . . |v.-V5 1 -4-: -'. .
cot *' s : ' 1 ~i a /3~ ' ' l ' 1 1 ~
Chil : Du hm s"phn t I, II, HI, IV
-r: H thc co: bn: v isin2x + co s^ :^s i nx tanx = cosx
^ cosx cox =
sinx^ Cun g lin qu an c
Hai gc i nhausin (x) = -sin x tan(-x) = -tan:
Hai g c b nhau:; sin(7T - ) = sir
tan(jt a) ~ - t
lang gic tu thuc to im - cui: M thc gc'
cc iu kin xc nh-th ' Y- ~1 tanx. cotx 1
1 + tan2x = , X - - 'COS X
l + c o t 2x = --- \ ' - r . -rsin X _ " ;
b it:' ^
, cos(x) = cosx, . ' . -..7 . .. ,
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
7/341
Hai g c ph nhau:
sin ( - a) = cosa, os( a) = sina ,2 2
tan(-^ a) = co ta , c o t a) = tana2 . - 2 .
Hai gc hn km 7t:s in(7T + x) = si n x , c o s ( tt + x ) = c o
sx tan( + x) = tanx, cot(jr + x) = cotx
Hai gc hn km :2
sin(-^ + x) cosx, co s( + x) = sinx2 2
ta n (^ x) = cotx, cot( + x) = -tan x Z 2
Ch : "cos i sin b ph cho tan, cot 7t'
Cng thc cng:. . sin(a b) = sinaxo sb sinb.cosacos(a b) =
cosa.cosb + sina.sinb
. > . TLS' t a n a i t a n btan(a b) = 1 + tan a. tan b
CS : sin x cosx = y2s in (x );
cosx sinx = V2cos(x T )4
2 . CC B TON IM HNH
Bi 1: Rt gn biu thc A = 1 + cosx
Ta c: A =1 + cosx
s i n x
Gii:V2 >
(1 - c o s x ) s
s i n 2 X
-s m x
_ 1 4- c o s x 2 c o s x
( l - c o sx )
1 - COS2 X
1 + c o s X
s i n X1 -
1 c o s x
1 + c o s x
= 2cotx.s i n X 1 -H- c o s XBi 2: Chng mnh biu thc sau khng ph
hc X
B = 3(sinsx - cossx) + 4(cos6x 2 sinx) + sin4xGii:
Ta c: B 3(sin4x cos4x)(sin4x + cos4x) + 4cos6x 8sin6x5sin 6x +
cos6x + 6sin4x 3sin4x cos2x + 3sin2x
+ s i n X
cos4x
6
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
8/341
= sin 6x + cosGx + 6sin 4x6sin6x + 3sin2x cos4x - sin*x cos2x= 1
3 s i n 2x c o s 2x 3 s i n 4x c o s 2x + 3 s i n 2x c o s 4x = 1
.
Ch : Loi ton ny, kt qu bit trc bng cch chn 1 gi tr ca X,chng hn
X = 0.
Bi 3: Cho tana = . Tnh c = : 2---------:----- ----------------
24 sin a - sm cc COS a + COS a
Gi:Vi gi thit th cosa 50, nn cha c t ln mu cho co s2oc:
1 _ _______________ COS2 _______________
s i n 2 a s in a COS a COS2 a
c o s2 a c o s2 a COS2
__ ' 1 + tan 2a
t a n 2 a t a n a + 1 f l Y 1U J ~ 4 + 1
Bi 4: Cho sinx + cosx = a. Tnh D = sin5x + cos5x.Gii:
Ta Ca2(sinx + cosx)2 = 1 + 2sinx.cosx, nn
sinx.cosx = (a21) v sin2x.cos2x - (a2- I) 2 .y 4 V
D o : D = s i n 5x + c o s 5x
= ( s i n x + c o s x ) ( s i n 4x s i n 3x c o s x + s i n 2x c
o s 2x s i n x c o s 3x + c o s 4x )
= (s in x + co sx) ( [ s in 4x + co s4x] sinx.cosx . [s in /Sc +
co s2x] + s in 2jc co s2x)
= (sinx + cosx)([sin2x + cos 2- 2sin^.cos^K - sinx.cosx +
sin2x.cos2x)= (sinx + cosx)(l - sinx.cosx - sin2x.cos2x).
= a ( l - [a2- 1 ] - ~ [a2- 1 ] *) = i ( 5 a - a5).2 4 4
Bi 5: n gin biu thcT 3 T
E = co s ( t t a) + sin(----- + a ) - t a n ( + ) , cot ( a )
.
Gii:Ta c: cos(7E ) = cosa
sin( + a ) = sin[ 2n + ( + a)] = sin( + a) = cosa
tan( + a ) = cota v
co ( a ) = cot(271 [ 4- a]) = cot( + a ) = ~ta.no..2 2 2
V y E = c o s a + c o s a + c o t a (t a n a ) = 1.
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
9/341
Bi 6: om gin biu thc G = 2co s( + a) .co s( - a).4 4
Gii:
G = 2 (cos cosa - si n sina)(cos cosa + s in sina)4 4 ' 4 4
' 2 . w n/2 V2 . .= 2 ( c o sa sin a)( co sa + sina)
2 2 2 2 ;V2' V2= 2 . . ---- (cosa sin a)(c osa + sinct) = cos2ot
sn 2a .
2 2Bi 7: n gin biu thc E = s in 2(30 - a)+ sin 2(30 + a) sin
2a
Gii:E - (sin30cosa - sinacos30)2 + (sin30cosa + sinacos30)2 -
sin2a
_ . L___ V3 ,1 ___ ys . -2 _-_2= (--cosa -sina) + ( co sa
----sina) sin^a
2 2 22 .
= -- (cos2a + 3sin2a 2 /3 cosa.sina + cos2a + 3sin2a + 2
\3cosa.sia) sin2a4
-- (2cos2a + 6sin 2a) sin 2a = cos2a + sin2a sin2a = .4 2 2
2
Bi 8: Chng minh:cosa sin(b c) + cosb sin(c a) + COSC sin(a b) =
0 .
Gii:VT = cosa(sinbcosc sinccosb) + cosb(sinccosa sinacosc)
+ cosc(sinacosb sinbcosa)cosa sinbcosc cosasinc cosb
cosbsinccosa cosbsinacosc
+ coscsinacosb coscsinbcosa = 0 = VP.Bi 9: Chng minh: nu 3sin3 =
sin (2a + (3) th tan(a + P) = 2in a
cos aGii:
Ta c 3 sinp sin (2a + 3) nn 3sin((a + |3) - a) = sin ((a + 3) +
a)3sm(a + 3)cosa 3sinacos(a + p)= sin(a + P)cosa + sina.cos(a + p)
hay
2 s i n ( a + p) c osc t = 4 s i n a c o s ( a + P) d o s m ( a
+ . P) = 2 ? ln a ( p c m ) .co s(a + P) cos a
Bi 10: Cho sin(40 + ct) = b v 0 < a < 45. Tnh cos(70 + a
)
Gii: T gi thit sin(40 + a ) = b v 0 < 40 + oc < 90
nn cos(40 + a ) = y1 - b2 .
Do cos(70 + a) = cos[30 + (40 + a)]
= cos30. sin(40 + a) + cos(40 + a) sin 30 = +yl b2 .
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
10/341
Bi 11: Tnh tan( - a), bit sina = v TC< a < .4 13 2
Gii:_ , X _ _ '5 _ 12T gi thit suy ra cosa = => tana = .
1 3 5
1 -
Do t a n ( - a) - = J L .4 1 + ta n a 1 , 1217 1 -- ;
B i 12: Chng minh: tan o0- tan40 = 2tanl0.Gii:
tan 50 tan 40Ta c: ta n io 0= tan(5p - 40) -
1 + tan 50! tan 40
tan 50 - tan 401 + cot 40. tan 40
tan 5 0 tan 40 ^ _ ---------------- ---- nn suy ra pcm.
Bi 13: Chng minh: tan30 + tan40 + tanO0+ tan60 = cos203Gii:
Ta c:
V T = ------- - ir?-Z ---------'+ --------= 2 c o s 2 0 (
--------- - + ----- 0COS30 cos40 COSO cos60 V 3cos40 COS0
o crnO , ,0 o oa0 cos50 + -^^COs400_ 2 -c o s 50 -h V3cos4Q _
8cos20 2_________ 2s/3 COS50 -COS40 y/s 2 s in 40 .COS40
8 c o s 20 s i n 30 COS50 + COS30 s i n 50y/s sin 40. COS40 +
sin 40 .COS 40
= VP.
3. B1LUYNTP
Bi 14: Chng minh ng thcsin2a ta na + cos2a cota + 2snacosa =
tana + cota
HD: Dng h thc c bn.Bi 15: Tnh tng F COS20 + cos40 + . . . +
COS180.
S: 1.B i 1 6 : T m k e z b i t r ng : s i n ( 2 1 - k 9 0 ) >
0 .
S: k = 4m hoc 4m + 3 vi k nguyn.Bi 17: Cho sina + c os a = m.
Tnh: sina cosa; sina cosa ; sin4a + cos4a .
HD: Bnh phng ca ng thc gi thit.
9
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
11/341
/11* _________ o ' A A 0 ^ ^ . 1On0 T ' _ A COS3 a + COS a . si
n 2 a - si n aBi 18: Cho tana = 2 va 90 < a < 180 . Tnh: A
------------- ---- ------------s in J a - c os a
_ 53
B i 19: Cho 3s in4x + 2cos4x = . Tnh A = 2sin4x + 3cos4x81HB:
Dng h thc ca bn.
Bi 20: Tnh: sina; sinb, sinc m a, b, c tho mn:cosa = tan b, cosb
= tan c, cosc = tan a
HD : Dng h thc ca bn v bnh phang.T O i - / - t u - ______7 s in
x + c o s x - 1 2 c s x -Bi 21: Chng minh
Bi 23: Chng minh cc biu thc sau c lp vi bin:B = 2(sin4x + cos4x
+ sin2xcos2x)2- (sinsx + cossx)
BS: B = Bi 24: Chng minh cc biu thc sau c lp vi bin:
c = 3(sn8x cos8x) + 4(cosGX - 2s in6x) + 6sin4xS: c = -2
Bi 25: Chng minh: Nu ^ - 1 - th l + = . J .
a b a + b a3 b3 ( a + b )3HD: Dng h thc c bn.Bi 26: Tnh gn
D = cos(270x) 2sin(x 450) + cos(x + 000) + 2sn(720~x)S:D = cosx
Ssinx
Bi 27: Tnh con E ~ sin (-4,8)i).sin(-5,7 ir) t
cos(-6,7tt).cos(-5,8iQcot(-5,2it) tan(-6,27)
S : E = 1 .Bi 28:Tnh gi tr lng gic ca gc 105
HD: 105= 45+60Bi 29:Tnh gn G = co s(- 53 )sin(- 337) +
sin(307)sin(l 13)
1BS: G =
2
, T ' u T T _ ta n 225 cot8 1.cot69Bi 3:Tmh gn H =C0t261 + ta n
201
BS: H =y/.
10
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
12/341
Tnh sin(a - b); cos(a + ib); tan(a - b)15 , _ 2
H D: c os a = ~ ; c o s b =
Bi 32:Cho 0 < a , b < - , a + b = - v tana. tanb = 3 -
2V22 4
Bi 31:Cho a, b l cc gc nhn vi sina ; tanb = .
Tnh tana v tanb.S: tana ~ tanb = V2 .
Bi 33:Chng minh: tan75 + cot75 = 4HD: Dng h thc c bn v cng thc
cng.
_ , . , tan(a b) 4- tan b cos(a + b)Bi 34: Chng minh l= - 7
ta n(a + b ) ~ ta n D co s(a - b)HD: Dng cng thc cng.
Bi 35: Chng mnh biu thc sau khng ph thuc vo x:I = cos^
2cosa.cosx.cos(a + x) + cos2(a + x)
BS: sin2a.B i 36: Chng minh biu thc sau khng ph thuc vo x:
J = tanx.tan(x + ) + tan(x + )tan(x + ) + tan(x + }tanx3 3 3
3
BS: J = -3 .Bi 37: Tnh gn:
K = co s(x ). cos(x + ) + cos(x + ) co s(x + )
/B S K = ( l- > / 3 ) .
4Bi 38:Bin i thnh tch: L = a.snx + b .cosx
HD: Chia 2 v cho \/a 2 + b2Bi 39:B in i thnh ch: M = sinx + sin2
+ sn3x + sin4x
HB: Ghp 2 nhmBi 40: Chng minh: N u sina + sinp > th cosa +
CQSPI < 1
HD: S ng tng bnh phng ca 2 v tri.
B SM o f
. L THUYT V PHOMG PHP G TON __________:~Ci) |^^ :'.v ' '^;Vf.uc
s2 a ~ c o s?a 4 sin2 2cosz - 1':?=: 2siri2a .
sin^^s^ar cosa; tn2a 4 :7 - ; - .. ; 'V 0;/ 1ta il a
11
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
13/341
- Cng thc h bc hai:' 2 l + cos 2a - 2 _ 1cos2a ^ ?- - c s2aOS a
----- -; sin a ; tan a - -
, . 2 . 2 x + cos2a \ -Ket qu:1 + cos2a = 2cos2a? 1- cos2a =
2si2a
Nhn ba (m rng): , . I V. - -sin3a = 3sina 4sin 3a; cos3a =
4cos3a r- 3cosa '
, . _ 3 tan a tan3a '!, , -- ---tan3a = ---------------r-------
;~-r-1 3 ta n a > , ... , . V -V '
rx. -__ -' 3____ 3 sin a sin 3a- . - .V 3 cos a"+ COS 3aH bc ba
(m rng): sin 3a = ------------ --------- ; cos3 - - ---------
---------- ' ' : 4v-:r.--/ *4 v~
GC ph t = tan(mxng): - ; . r, v ~ ' v :
_ 1 - t 2 "' ~ 2 t . ' , 2 t ' .cos a --------- - ; sin a = ----
--- tan a =
\ + t 2, y : -1 + 12 - - 1 - 12 ; , - -; i ' V ; ; Cng thc bin i
t ng th nh'tch: ' ' 5- * 0 ':
,__
-+b? _ _ a - b . - -{ ' - c o sa + c o sb = 2 c o s ---------
COS - ---
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
14/341
Gii:r * _ 1 /A_'_ T 7Z V 7C 1 1 . TU____ 7 N__ 1 , TC 1
Ta c A = (2sin -~cos-r )cos- j = ~ (2sinCOS ) = sin = .2 8 8 4 2
2 4 4 4 2 4
~ _ ta n 15B i 4 2 : T f a h B = ^ i _ .
Gii:B = - 2 t a - r ^ ~ = - ta n 2 (1 5 ) = - ta n 3 0 = .
2 1 - tan215 2 2 4Bi 43: Chng minh cng thc
sin 3a = 3 sina 4s in3a = 4sna sin (60 + a) sin (60 a).
.Gii:
Ta c sin3a = sin(a + 2a) = sinacos2a + cosasin2a = sina(l 2sin
2a) + 2sinacos2a= sina (l 2sin2a) + 2sina(l sin2a) = 3sina -
4sin3a.v 4sin a sin(60 + a) sin(60 a)= 4sina (sin60cosa + cos6sina)
(sin60cosa cos60sina)
= sin a( V3 cosa + sina) (-s/3 cosa sina) = s ina (3cos2a sin
2a)= sina[3(l sin2a) sin2a] = 3sin a 4sin 3a.
B i 44: Chng minh cng thccos3a = 4cos3a 3cosa = 4cosa.cos(60 +
a) cos(60 a)
Gii:Ta c cos3a = cos(a + 2a) = cosacos2a + sina sin2a
= cosa(2cos2a 1 ) + 2cosasinza= cosa(2cos2a ) + 2co sa(l cos2a)
= 4cos3a 3cosa
v 4cosa.cos(60 + a) cos(60 a)= 4cosa (cos60 cosa - sin60 sina)
(cos60 cosa + sin60 sina)= cosa(cosa V3 sina) (cosa + V3 sina)
= cosa (cos2a 3s in2a) = 4cossa 3cosa.Bi 45: Chng minh C0t2x =
-c0- - - = (cot X tan x ) .
2 cot X 2 , 1 , a , l A_ _ a , 1 ,__a , 1 , aSuy ra tng c tana +
ta n + ta n + ta n + tan
* 2 2 4 4 8 8 16 16Gii:
, c o t 2 x - 1 _ 1 , c o t 2 X 1 X _ 1 ,Ta c = (
------------------ ) = (cotx - tanx)2 co t X 2 co t X co t X 2
_ 1 c o s X _ s i n X ^ _ c o s 2 X s i n 2 X _ c o s 2 x =
2 sin x cos x 2 sin X cos X sin 2 x
do (cotx - tanx) = cot2X tan x = cotx 2cot2x.
p dng kt qu ny vi x = , v cng v theo v th:
c = cot - 2cot2a.16 16
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
15/341
Bi 46: Chng minh rng 1 +cos2x t a nx
I V . . 1 V- 1 Y- 1 "Suy ra tch D 1 + 1 + - 1 -5------ - 1 +
cosayl^ cos2a y \ c o s 4 a /1 cos8a)Gii:
sin 2x.p tan 2x _ cos2x _ sin2x GOSX
ta n x sin X COS 2x s in xGOSX
_ 2 s in XcosX co sX __ 2 eos2 X __ cos2x + 1 __cos2x s inx
cos2x cos2x
p dng kt qu ny v i X = , a, 2a, 4a th:
_ ta n a ta n 2a tan4a tan 8a _ tan 8a
t a n tan a ta n2 a ta n4 a tan 2 2
Bi 47: Rt gn: E = yj2 + y2 + 2COSCX ( 0 < a < tc).
Gii:
T ac E = -y/2 + /2 + 2cos a = ^ 2 + 2 ^cos2 j
+cos2x
= 2 + 2 cos^ = ^ 2 + 2cos|^j = 2 1C O S J = 2cos^
Bi 48: Chng minh cng thc theo t = tan2
: _ 2 t 2tsina = ----- , cosa = -----5- , tana = 1 + t 2 1 + t 2
1 - t 2
Gii:
0 , 2t a n 2 tan rr r 2 t 2 a _ aTa c ---- = -------------- =
----- = 2 s in .cos = s in a1 + t l + _ 1 _ 2 2
2 COS2 *2
1 +. 2 2 ^*_2 &1 4.2 1 - ta n cos ------- sin X 1 t 0 0 o
cos av --------------. = -------2 = cos a
1 "~t . X. 2 __23 - 2^ ^1 + tan. COS---(- sin 2 2 2
Suy ra tana = 2t1 - V
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
16/341
Bi 49' Tnh gi tr ln nht - b nht ca hm s y =cosx 4- 2sin ?: 32c
os x sin + 4
Gii:
t t = ta n-' th y = 1 " cf" (y -* 1)2- (y + 2H + 3y - 2 = 0.
Nu y = 1: phng trnh tr thnh 3 t + l = 0 = s > t = o
Nu y &1: phng trnh c nghim khi:
A 0 (y + 2)z 4(y l) (3 y - 2 ) : 0 o 115^-2^^ + 4< 0 < =
> ^ - < y < 2 .11
2 2Do min gi tr , 2] nn maxy = 2 v miny = YY .
B 50: Chng minh vi a, b, c tu th:
cos(a + b + c) + cos(b + c - a) + cos(c + a b) + cos(a + b c) =
4 cosacosbcoscTa c VT = 2cos(b + c)cosa + 2cosacos(c b)
= 2cosa(cos(b + c) + cos(c b)) = 4cosa. cosc. cosb = VP. Bi
51:Bin i thnh tch so: F = sina + sin2 a + sin 3a + sin 4a
G :F = sina + sin3 a + sin 2a + sin4a = 2sin 2a cosa + sin3aco
sa
. = 2c osa(sin2 a + sin3a) = 4cosa. sin . COS .
Bi 52: Bin i thnh tch s:
G = sina + sin 2 a + sin3a + cosa + cos2 a + cos3a
Ta c G = 2sin 2a cosa + sin 2a -- 2cos2a cosa + cos2a
= 2 (cosa + ).( sin2a + cos2a) = 2 - ^ 2 (cosa + COS) sin(2 a +
)
= 4 V2 co s( + ) cos( - ) sin(2a + ).2 6 2 6 4
Bi 53: Chng minh tan9 tan27 - tan63 + fcan81 = 4 .Gii:
VT = (tan0+ tan81) - (tan27 + tan63)
_ sin 90________sin 90 2_______________ 2________cos9 .sin9
cos27 .s in 27 2co s9.sin9 2 COS27 .sin 27
2 2 __ s i n 54 - s in l S 0 _ . COS36 .sin 18 _ _XTT>=
------------- ----------- = 2 -------------------- = 4 ------ - 4 =
VP.sin 18 sin 54 sin 18 . sin 54 sin 18. COS38
Bi 54: Tnh gn: H = 3sin l5 COS 15 + sm 60sin4 15 - COS415
15
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
17/341
Gii:Ta c:
H = - s i n 3 0 + -------------------- = - _ ^ __ = 1 _ 1 = _ I2
2 (s in 215 - COS215) 4 2.co s30 4 4
Bi 55: Tnh gn I = COS 10 cos30 cos50 cos70.Gii:
1 . /I =: cos30 (coslO0. COS 50). COS 70 = -(cos(- 40) +
cos60).cos702, 8
Cos40-Cos70 + cos70 = -^-c os(30) +COS 110 + ~ COS 7 08 16 16 16
' 16
- sin 20 + ^ s i n 20 = 32 16 16 32
Ta c sin . nsin . COS7 7
1 . 71 . 1 . 3nsin + sin2 7 2 7
_ 1 . 71 Suy ra J :2
sin.7
Bi 5'S: Khng dng bng hy tnh tng: J = COS + COS + COS
Gii:
2 n , 7Z___
4tc , . 7U 67U+sin. COS-" + sin-r . COS-3-7 7 7 7 7
1 . 1 . 5n 1 5jt , 1 . 7ir-----sin- + sinsinr- + 'Sin-3-
2 7 2 7 2 7 2 7
= _ 2
Bi 57: Tnh tng vi a k2 7TK = sinh + sin(h + a) + sin(h + 2a)
+sin(h + 3a) + sin(h + 4a)
Gii:
7'a c2 sin K = 2sinh. sin + 2sin(h + a),s in +2sin(h4- 2a).
sin
2 2 2 2
-f*2sin(h + 3a). s in + 2s n(h + 4a) sin .2 2
== c o s (h ~ ) - c o s (h + ) 4- c o s (h + ) c o s( h + ) + c
o s (h + )2 2 2 2 2
- cos(h + ) + cos(h + ) - cos(h + ) + cos(h + ) - cos(h + )2 2 2
2 2
9a
cos h - cosh +
2 s i n
16
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
18/341
Bi 58: Tnh tng vi a ^ k nL = cos2a + cos22a + cos23a + cos24a +
cos25a + cos26a.
Gii:
Ta c: L = (1 + cos2a) + (1+ cos4a) + (1 + cosSa)2 2 2
+ (1 + cosSa) + -(1+ coslOa) + ( 1 4 - cosl2a)2 2 2
nn: 2L 6= cos2a + cos4a + cos6a + cos8a + COS1 Oa + co sl2 a .Do
: 2sna. (2L - 6) = 2sina cos2a + 2sina cos4a + 2sin a cos6a
+ 2sina cos8a + 2sina coslOa + 2sin a Csl2a= - sina + sin3a -
sin3a + sina - sina + sin7a sin7a
+ sin9a sin9a + sin lla s in lla + sinl3a- T _ s in l3 a + l l s
i n a
= sin l3 a - sina . Vy L = ---------- --------------- .4s
ina
3. BI LUYN TP
Bi 59: Chng minh, nu sina ^ 0, th cosa cos2a cos4a cos8a COS la
= .32 sin a
HD: Dng cng thc nhn i.Bi 60:Chng minh vi , b , c tu th:
. a + b a + b . c + a .sina + sinb + sinc = 4 sin ---- -----sin
---------- sin ----- + sn(a + b + c)
2HD: Bin i tng thnh tch.
Bi 61: Tnh cc gi .t lng gic ca gc 8
HD: Dng cng thc h bc
Bi 62: Cho tan a = , tan b = v i 0 < a
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
19/341
B 68 : Tnh: D = sinl0sin50sin70
TT_ n 271 4 8 _ 167__32tc _ B! 69:Tmh: E = COS--COS-.COS-
.COS.COSr^-.cos S: 2.65 65 65 65 65 65
1* A , * f ,__- 4 7U . 4 3tc _ 4 4 T7Bi 70:Tnh F = sin r + sin +
sm + sin -f
16 16 16 16HD: H bc lng gic.
Bi 7:Tnh G = sinc~ + COS6 B$ : 24 24 16
Bi 72: Tnh H = tan2 + tan2 + tan212 12 12
HD: + ^ 1 12 12 2
Bi 73: Chng minh cng thc:
t a n 3 a = ^ t a n a a = t a n a .t a n ( 6 0 + a ). t a n (6 0
- a )_ 1 - 3 tan2aHD: 3a = 2a + a
B 74: Chng minh cng thc: COS 4a = 8cos4 a - 8cos2a + 1HD: 4a = 2
.2a
Bi 75:Chng minh cng thc: sin. 5a =16 sin 5 a - 20s in 3a + 5sin
aHD: 5a = 2a + 3a
Bi 76: Chng minh: COS3 X. sin X - sin 3 X. COS X = 5" 4x4
HD: t tha cho chung v tri.
Bi 77: Chng minh: 128(sin10x + cos10x) - 5es8x +60cos4x + 63H:
Dng cng thc h bcBi 78: Chng minh biu thc sau khng ph thuc vo x:
I = sin4x + sin^x + ) + sin 4(x + ) + sin^(x + ) S: -4 2 4 ,
2
1 1 2 3Bi 79: Chng minh: cosx - COS 3x - cos5x = 8sin x.cos X2
2
KD: Ghp (cos 3x + cosx)2
Bi 80: Chng minh: COS12 + COS18 4cosl5 0.cos21Q.cos24 = -2
HD: Dng cng thc bin i tch thnh tng
Bi 81: Tnh tch: J = cos75.cosl5 S: .4
Bi 82: Tnh tch K = sin .COS07112 , 12
H: Dng cng thc bin i tch thnh tng
18
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
20/341
Bi 83:Tnh tch L = tan20o.tan 40o.tan60o.tan80o S: V.Bi 84: Tnh
tng: M = COS 85 + cos35 - cos25 BS: 0.
Bi 85:Tnh tng: N = ta n 75 - ta n 15 BS: 2V.
Bi 86: Tnh tng P = s in -^-s in -s in S: .
_ ^ 6n 8t ~ ,B i 8 7 : T n h to n g Q = COS - + COS + COS 1 +
COS S : 1
5 5 5 5
Bai 88:Tinh tong^ ^ ^ 2 sn 2a sn 3a sn na sn(n +
BS:sin a. i \n(n +1)a
Bi 89: Tm gi tr 11nht v nh nht ca hm s: y = sin 4X + c o s 4
x
Bi 90: Tm gi tr ln nht v nh nht ca hm s:y = a.sin2x + b.cosxsinx
+ c.cos2x
HO: Dng cng thc h bc a v gc 2x.
.. D N G' 3 ' ; v NG . D N G L N G CSC
1. L THUYT V PHNG PHP GI TONua hm s lcig; g
N: : X iac vaobai tn.i s- v. t X ==sint hay X =: cost
.. -N eux --K y?^= ]^Q ;^t:X ^ E:sint;hay X r.cost ;thi^ct-X: s
sirit va y ^ co st/
- Neu-x2:+;y? 2thTc;x==.r siiit va y = r.cost
rr N .l-x-^'d t.x 3 hay ' cos t .
: ;;; t x ^ t a r t h y x - c o t t 1; ..rr.N eu c +JB; c
gc nh cac%oc-;
w ab'c^ab'^ibc^ c a ;==1 th t cc i lng an ca cc
^atangic,.::
2. CC B TON IN HNHBi 91:Cho ab * 1 ; bc 9*- 1; ca * - 1 .
,____ a b , b c , c - a_ a - b b - c c - aChng m in h ---------
+ ------- ; + ------- = -------. ------------------- .---
1 + ab 1 + DC 1 4- ca 1 + ab 1 + be 1 - ca
Gi:t a = tanA; b = tanB; c = tanC, th ng thc bi thnh
19
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
21/341
tan A tan B fcanB tan C tan c tan A + tan A tan B 1 + tan B tan
c 1 4- tan c tan A
_ tan A tan B tan B tan c tan c ta n A1 + tan A tan B 1 + tan B
tan c 1 + tan c tan A *
tan(AB) + tan(B C) + tan(C A) = tan(A -B).tan (B - C).tan(C - Ta
chng minh: a + (3 + Y= k o> tana + tanp + tany = tana. tanp.
tany
rng (A - B) + (B - C) + (C A) = 0 => (pcm).Bi 92: Cho b <
a. _ _
Chng minh: | a + b | + a - b = l a + >/a2b2 + a -
Va2b2Gii:
Neu a = 0 th b = 0 nn ng thc ng.Neu a t 0, ng ic tng ng:
I I + - I + I 1 - - I = I 1 + J i ~ I + 1 1 - J i - K I (1)a a M
a \ .
V |b[ < aj nn t = cost, 0
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
22/341
B 95: Gii bt phcmg trnh y / + X - y[ l -~xT < X.
Gii:
iu kin: - 1 < X < 1 nn t X = cos2t, t e [0, J.2
Bt phng trinh tr thnh -v/l + COS2 t -J~^ ~c s2 t cos2 t
V2 1cost - -s/2 sint < cos2t - sin2t V 2 ( c o s t s i n t )
< ( c o s t + s in t ) ( c o s t si n t )
ce> cos(t + )(cos(t - ) 1 ) 0cos(t + ) ^ 04 4 4
. < t < < 2 t < 7 t o - X < x < 0 .4 2 2
Nghim ca bt phng trnh l: - 1 X < 0.
Bi 96: Gii phng trnh: X3 + yj(1 X2)3 = Xyj2 (1 X2)
Gii:
iu kin: IXI < 1 nn t X cosu, u e [0, jc].Phng trnh tr thnh
cos3u + sin3u = /2sinu cosu (X)t t = sinu + cosu, 1 1 < V2 .
(1) (sinu + cosu)(l - sinucosu) = V2sinu cosu
t ( l - = V2 o t3+ V2t2- 3t - -J2 - 02 2
{ t - )(t2+ 2V2 t + l ) = 0 t = V2 hay t = -V 2 X/2
Chon t = V2 thi c X = ----2
r
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
23/341
T iu kin 0 < t < , suy ra c bn nghim thch hp l:2
__ 7E __- OTZ ___. 71 , 7Tx = sm -x -;x = sxn-; X = sin- v X
18 18 14 14
Bi 98: Gii phng trnh: 4x33x = Gii:
Ta c = cos-^ = 4 c g s3 3 eos^ nn X = COS l 1 nghim ca
phng trnh.
Tng t - = cos = COS = COS nn cos5^ v c o s 7 c n g nghim.2 3 3 3
9 9 & V
V 3 nghim phn bit v phng trnh cho bc 3 nn phng trnh c1 - A TZ _
In
3 ngum X, = C0S~?3C>= cos ,x
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
24/341
s u y r a ( t - y ) ( t 2 + t y + y 2 + 3 ) = 0 n n t y d o y 3
- 3 y = 1 ( 1 )
Xet 2 y 2, t t = 2cosa
(1 ): 8cos3a 6cosa = 1 hay cos3a =
T gii v chn 3 nghim a l J
9 9VI (1) l phng trnh bc 3 nn c. ng 3 nghim y l
2 cos-T-:2 cos ; 2 COS suy ra 3 nghim (x,y) ca h.9 9 9
Bi 101: Cho X, y thay i tha mn X2+ y2= 1.
Tm gi tri ln nht, b nht caP = 2(x +1 + 2 x y + 2 y
(H khi B nm 2008)G:
V X2 + y2 =1 nn t X = cost, y = sint,_ 2(cos2 t + 6 cost .s int)
_p = ---------- ---- ( P -6) . s in t- (P + l)cos t = 1- 2P
2 cost .s in t + 2 COS tiu kin c ngkm ca phng trnh
(P- 6)2+ ( P - I )2 > (1 - 2P) p2+ 3P - 18 < 0 6 < p
< 3Vy min p = 67max p = 3.
Bi 102: Tm gi tri ln nht b nht ca biu thc p =(1 + X 2 ) + y
)
Gii:
t X tancx, y = tanp vi a, |3 e (, ). p __ ( t a n a + t a n 3)(l
t a n a t a n p)
(1 + tan2ct)(l + tan 2P)
= sin (a + 3).cos(a + p) = sn2(a + 3) nn < p < .2 2 2
Vy, max p = chng hn a +p = ; rain p = chng hn oc 3 = .
Bi 103: Tm gi tr ln nht, b nht ca hm s y =12x + 8x 2 - 3
(2
x2
+ l)2
tX\2 = tant, vi t
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
25/341
Cho t = 0 th y = 3, cho t = th y = .4 25
Vy max y = 3 v miny = .2
Bi 104: Gi X, y, z , t l nghim ca h
X + y = 4
z2 + t2 = 9 .
x t + yz 6
Tm X, y, z, t sao cho p = zx t gi tr ln nht.Gii:
t X = 2cosa, y = 2 sina; z 3cos(3, t = 3sin p,a, p e [0, 2n]
th
xt + yz > 66(cosasinp + sinacosp) > 66sin(a + P)>6a+3 =
2
Lc , p = xz = 6cosacosp = 3[coss(a 3) + cos(a + [3)3 = 3 co s(a
pt gi tr ln nht khi cos(a - 0) = 1 a - (3 = 0.
Suy ra a = B = x = y = V2 ; z = t = .4 2
Vy maxP = 3, khi X = y \2 ,z . t = 2
3. BI LUYN TP,Bii 105: Cho X2+ y2+ Zz + 2xyz = 1; X, y, z >
0. Chng minh:
1+ x y z =X^CL-Vxi -z? ) + y ^ /c i-^ x i-X 2) + Z y ( l - i X l
- y2)
HD: t X = cosa, y = cos(3, z = cosy,0 < a, p, Y< Bi 106:
Cho xy + yz + zx = 1v xyz ^ 0 .
Chng minh: (x - )(y - ) + (y - )(z - ) + (z - )(x - ) = 4.X y y
z z X
HO: t X = tana , y = tanP; z = tany.Bi 107: Cho xy + yz + zx =
1.
C h ng m inh: X + y + z - 3 x y z = x (y ^ + Z2) + y ( z2 4- X2)
+ z ( x 2 + y 2) .
HD: t X ~ tana, y = tanP; z = tany.
B 108: Cho X2+ y2= 1, u2+ V2= 1- Chng minh: |x(u - v) + y(u + v)
1 th (1 - x)p+ (1 + x)p< 2P.b) nu 0 < p < 1 th (1 - x)p+
(1 + x)p > 2P.HD: t X = COS 2a.
24
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
26/341
Bi 11: Cho cc s x>y> z tho mn 0 < X, y, 2< 1 v xy +
yz + zx = 1., , ____ - u X y z 3v3c h to g m in h _ ^ ? . + _ ^ _ ?
+ r ^ _ s
HD: t X = tana , y = tanp; z = tany vi cc gc thuc (0; )
Bi 111: Chng minh rng t bn s cho trc lun c th chn ra hai s
X,
y sao cho: 0< 1 .1 + x y
HD: t X = tana, y = tanp;Bi 112: Gii phng trnh: 8 X 3 - 4x2- 4x
+ 1 = 0
TTK _ ___5 zS: xx = cos-7x2 = COS- ,x 3=cos
Bi 113: Gii phng trnh: \ j l + = x(l + 2yl - X2 )
HD: t X = sint
Bi 114: Gii phng trnh: (y( l + x)3 - 7(1 - x )3 ) = 2 + V1 ~
X2HD: t X = cost
X2Bi 115: Gii bt phng trnh: -y/l + X + -71 - X < 2
--------
HD: t X = cos2t
Bi 116: Gii h phng trnh:W i - y2 = 74
y V l - x2 = - 4
HD: t Xsina, y sinb
Bi 117: Tm gi tr ln nht, b nht ca hm s: y =1 + Xs
(1 + X2)3
S: 1 v -4
Bi 118: Tm gi tr ln nht, b nht ca hm s: y = X + -s/lX2HD: t X =
sint
Bi 119: Cho a, b, c > 0 v abc + a + c b = 0., 2 2 3
Tm gi tri ln nht ca p = ----- -- 4-
a + 1 b + 1 c + 1
Bi 120: Cho a, b, c thay i tho mn a 2+ 9bz + 9c2= 16.Tm gi ln
nht:ca Q = 9ab + 6bc + 9ca
HD: T gi thit vit dng tng bnh phng bng 1.
S: 3
25
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
27/341
D N G i:
1. L THUYT V PHNG PHP GII TON
jnjfL ;
.~ J$ ?^^ '7
Gitg thc;' hjeu
A ( H r ng )
Chi/^Bin;oi:phi
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
28/341
2. CC B TON IN HNH
Bi 121: Chng minh rng trong tam gic ABC a un c:: B ^ - C B _
A
s i n c o s + s in COS = c o s - i r 2
2
2
2
2Gii:
Ta C:A _ ,n , B C B , c - B c . o B
c o s = c o s( -r ( + - - ) ) = s m ( + -T-) s m ~ c o s + s m -
- COS .2 2 2 2 2 2 2 2 2 2
Bi 122:Cho tam gic ABC khng vung.Chng minh rng; tanA -- tanB +
tanC = tanA tanB tanC.
Ta c tan (A -f C) = tan (n B) = tan B nn tan B1 - tan A tan
c
Ihay tanA + tanB + tanC = tanA tanB tanC.
Bi 123: Chng minh rngtrong tam gic ABC talun c, A . B B c c , A
_ .tan-r- tan + tantan-1 + tan ta n -- = 1.
2 2 2 2 2 2Gii:
Ta C:
* / A ^ B , .7t _ c c . t a n f + t a n f 1tan(- + ) = ta n (
------) = cot nn =
2 2 2 2 2 - . A B c1 - tan ta n tan 2
2
2
u . c /+ A B . t A B * _hay tan (tan11+ tan) I tan tan =>
pcm.2 2 , 2 2 2Bi 124: Chng minh rng trong tam gic ABC ta iun
c:
_ A B C sinA 4- sinB + sinC = 4COS---COS COS
2 2 2
VT = 2s i n A + - COS------ -- + sin(A + B)2 2
_ . A +B __A - B - r t f A + B= 2sin ---- ----- C O S----- -----
+ sn2
A -- B A - B . A + B A + B= 2sin ---- ----- COS----- ----- + 2sn
----- -----COS-----------
2 2 2 2A + B , A - B A -f- B
= 2 s i n ------ -------( c o s -------------- + COS------
------ )2 2 2c 0 A , B A B C
2cos . 2 COS cos(------ ) = 4eos COS COS .2 2 2 2 2 2
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
29/341
Bi 125: Chng minh rng trong tam gic ABC ta lun csin 2A + sin 2B
+ sin2C = 4sinA snB sinC
Gi:Ta c: VT = 2sin(A + B)cos(A - B) sin (2tc - 2(A + B))
= 2sin(A +B)cos(A B) sin2(A + B)= 2sin(A + B)[cos(A B) - cos(A +
B)]= 2sinC.(2)sinA.sn(B)
= 4sinA sinB sinC.Bi 126: Chng minh rng trong tam gic ABC ta lun
cCOS2A + cos2B + cos2C = - 1 - 4cosA cosB cosC
Gi:\TT = 2cos(A + B) cos(A B) + c o s2(tt(A + B))
= 2cos(A + B) cos(A B) + c o s(2tu2(A + B))= 2cos(A + B) cos(A
B) + cos(2 (A + B))= 2cos(A + B) cos(A - B) + cos(2(A + B))= 2cos(A
+ B) cos(A B) + 2cosz(A + B) 1 = 1+ 2cos(A + B)(cos(A B) + cos(A
-+- B))
14cosC. cosA cos(B)= 14cosA cosB cosC.Bi 127: Chng mnh rng trong
tam gic ABC ta lun c
cos2A + cos2B + cos2C = 12cosA cosB cosC.Gii:
1 + co s 2A , 1 + cos 2B , _ 2/ A . T>\\tT = ----------------
+ ------------- + C0S2(A + B)2 2
= 1 + (cos2A 4- cos2B) + cos2(A + B)
= 1+ cos(A + B)cos(A - B) + cos2(A + B)
= 1+ cos(A + B)[cos(A B) + cos(A + B)]= 1cosC. 2 cosA. cos(B)= 1
2cosAcosBcosC.
Bi 128: Chng minh rng trong tam gic ABC ta lun csin2A + sin2B +
sin2C = 2+ 2cosAcosBcosC.
Gii:
VT = (1 cos2A) + (1cos2B) + sin2(A + B)2 2
1 (cos2A + cos2B) + 1 cos2(A + B)
~ 2cos(A + B) cos(A B) cos2(A + B)2 - cos(A + B)(cos(A - B) +
cos(A + B))= 2+ cosC.(2 cosA cos(-B)) = 2+ 2cosAcosBcosC.
28
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
30/341
Bi 129: Cho tam gic .ABC, chng minh: = 5-^-s n c c
Gii:p dng nh l sin:
4R2(sin2A sin 2B) __ sin2A sin 2B4 R 2 . s i n 2 c ~~ s i n 2
c
1 cos2A 1 cos 2B co$2B c os2A
VP =
sin2c 2 sin2 c
_ 2 sin(B + A )sin(B A) _ sinC .sin(A B) = yrji2 sin 2c ~~
sin2c
Bi 130: Cho tam gic ABC, chng minh rng: cotA + cotB + cotC =
Gii:p dig nh l sin v cng thc din tch:
2 + b2 + c2 _ T5 4R2(sin2A + sin2B + sin 2C)
a2 + + C243
VP = abc4R
= R-8RS sin A sin B sin c
sin A + sinB*2 sin B sin c 2 sin c sin A
+ sinC 2 s inAsiB
1 sinBco sC + sinCcosB 1 sin c COSA + sin A COS c2 s inBs inC 2
s inCsinA
+ 1 sin A cos B + sin B COSA2 sin A sin B
= (co te + cotB + cotA + co te + cotB + cotA) = VT.
Cch khc: p dng nh l cosin v cng thc in tch:cos A b2+ c2- a2 b2+
c2a2
cot A =sin A 2bc.sinA
Bi 131: Cho tam gic ABC, chng minh rng:4S
m* + = (a2 + b2 + c2).
p dng nh l trung tuyn:
4Gii:
_ 1 (12 , 2b 4- c
'
1
d
11 + 2 b2'c + a - + i ra2+ b2 -
~~2 2) 2 l 2J 2 l 2 J
= ( a 2+ b 2 + c2).
29
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
31/341
'Bi 132:Cho tam gic ABC, chng minh nu cc cnh a, b, c ca tam
giclp thnh mt cp s cng th ac 6Rj.
Gii:' , __ abc _6br c _ 3 u s
Ta c ac= 6Rr = 7- s b4R 4 2 p
2 p = 3b 2b = a + c a, b, c lp thnh cp s cng.
3. BI LUYN TPBi 133: Vi A ,B, c l ba gc mt tam gic, chng
minh:
. A . B . V A B csin + sin+ sin + COS + COS + COS
4 4 4 4 4 4
= 4 -s/2 c o s ( + ). c o s ( + ). c o s ( + ).8 8 8 8 8 8
HD: sin COS = V2 sin( + )4 4 4 4Bi 134: Cho tam gic ABC, chng
minh rng:
A B cc o s COS COS-T- - 1 -Z 1 2 . +_ 2 - + - 2 . . I + 1 +
1
eA eB ec 3 b c .
2bccos,HD: a = --------- 2-
b + c
S 135: Cho , B , c l ba gc mt tam gic.. A , B . c
s in s i n s in
Tnh gn 2 + C 2 a + A 2 Bc o s COS COS COS COS COS
2 2 2 2 2 2
S: 2.B 136:Cho tam gic ABC, chng minh ng thc:
cosB cosC sinB sinC + cosA 0.D: T A + B + c 7C=> A T CB +
C).
Bi 137:Cho tam gic ABC, chng minh ng thc:
: A . B c . A B , c A . B . csin sin cos +COS sin + COS-sin sin
2 - 2 2 2 2 2 2 2 2
_ A c= cos . COS COS .2
2
2A B c
HD: ng ictng ng tan( + ) = cot : ng
30
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
32/341
Bi 138: Chng minh rng trong tam gic ABC bt k ta cA B C
sinA - sinB + sinC = 4sin .COS .s in 2 2 2
HD: Ghp sinA - sinBBi 139: Chng minh rng trong tam gic ABC bt k
ta c
cotA.cotB + cotB.cotC -- cotC.cotA 1H: Dng quan h A + B = 7 -
c
Bi 140: Chng minh rng trong tam gic ABC bt k ta c
*. A . . B * c _ 4. A ,B . Cc o t g ^ + c o t g + c o t g = c o
t g c o t g . c o t g z z z z z
- A B 71 cH: Dng quan h ^ + =
lii 141: Chng minh rng trong tam gic ABC ta lun c
A JB c , . B C A , . c A Bs in --c o s COS + sin COS C O S + sin
COS COS 2
2
2 2 2 2 2 2 2
_ . A . B . c A B B c A= sin -^-sin-sin + tan tan + tantan +
tan-tan
2 2 2 2 2 2 2 2 2
HD: 1 = sin = sin[( + ) + ]2 2 2 2
Bi 142: Chng minh rng trong tam gic ABC ta lun c
+ 4- 1 = ( t a n + t a n + t a n + c o t c o t c o t )sin A sin
B sin c 22 222 22
HD: Ghp tan + c o t t a n + cot , tan + co t v 2 2 2 2 2 2
Bi 143: Chng minh rng trong tam gic ABC bt k ta ca = b.cosC +
c.cosB
HB: Dng nh l sin hoc cosin.
Bi 144: Chng minh rng trong tam gic ABC bt k ta c a = r(cot + c
o t )
HB: Chn ng vung gc t tm ni tip n cnh BC l K chia BCthnh 2on.
Bi 145: Gi H l trc tm ABC. Chng minh rng 3 ng trn ngoi tip tam
gic HBC, HCA, HAB c cng bn knh vi ng trn ngoi tip RH: Dng nh
sin.
Bi 146:Gi I l tm ng trn ni tip A ABC v R, R2, R3 l 3 bn knhng
trn ngoi ip A IBC, IC A, IAB.Chng minh: RiRal = 2rR2HD: Dng nh l
sin.
31
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
33/341
Bi 47: Chng minh rng ixong tam gic ABC bt k ta c
HD: Dng cng thc din tch.Bi 14: Cho A ABC c 2 trung tuyn AM
_!_BN.
Chng minh: cotc = 2(cotA 4- cotB)
HI): Chng minh a2+ b2 5c?Bi 149: Cho A ABC vung ti A, tm ng trn
ni tip I v c tch 2____ /2
ng phn gic BK.CJ /2. Chng minh; IB .IC= 2
Hl: Chng minh s in s in = -r^ 2 2 4a
Bi 150: Chng minh rng trong tam gic A BC bt k ta c:
Bi 151:Chng minh rng trong tam gic ABC bt k ta c
HD: Dng nh i sin.
Bi 153: Cho tam gic ABC. Chng minh nu 4A = 2B c th = + -
HO:: Tnh trc cc gc t gi thit cho.Bi 154: Cho tam gic ABC.
Chng minh nu b = c, A = 20 th a 3+ b3- 3ab2= 0H 0: Dng ng cao AH
chia i tam gic ABC.
Bi 155: Chng minh rng trong tam gic ABC bt k ta c
1 + -r- = cosA + cosB + cosCR
H: Dng cng thc din tchBi 156: Chng minh rng ong tam gic ABC bt k
ta c
tan.tan =2 2
s = (a2sin2B + b2sin2A)
a b c
b e c o t ' + c a c o t + ab c o t =2 2 2
1 _ .a b c p
+
HB : Dng cng thc din tch.
32
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
34/341
Bi 157: Chng minh rng trorig tam gic ABC bt k ta c a cotA +
bcotB + c cotC = 2(R + r)
HD: a v cosA + cosB + cosC.Bi 158: Chng minh rng trong tam gic
ABC bt k ta c
be cos2 + ca cos2 + abcos2 = p2.2 2 2
HD: Dng cng thc h bc v nh l cosin.
Bi 159: Chng minh rng trong tam gic ABC bt k ta c(b c)(p - a)
cosA + (c - a)(p - b) cosB + (a b)(p c) cosC = 0.
HD: Dng nh, cosin.Bi 160: Cho A AB C c ng trn ni tip tip xc 3
cnh ti A \ B%
Cca tam gic A BC cnh a%b%c v din tch s \
Chng minh: =2 (sin s in s in )s 2 2 2
HD: Chng minh + = 2sin (sin + sin )& a b 2 2 2
D N G 2 : c p f e T R r y f e T A M g e
1. L THUYT V PHNG PHP GII TON
Bt xng thc v b'tc :cc'':tq;c-^u;phngph^^nb^gMV:chag minh
o Phng php so snh ' ' J .
Phcmg php bin i tng ng . 'Bin i bt ng thic v dng tng cc binh
phng khng m hoc
tch cc tha s khng m.
Phng php'dng bt ng tic ccr bn:. - :Bt ng thc gi tr tuvt di ;
. . Vi a, b tu. thc: |a'Vb: 0,; - ' ,
a + b > c, b + c > a, c .+ \a > b , ' - . ;a b < c,
b G< a, c a < b , .. J V| a - b | < c , I b C < 2L,.. C
a . - V-v (a - b) (A - B);io>(b - c) (B - C) i o, (c - a) (C -A
) 2: 0, . '. : . ;(A B) (cosA cosB) :< 0?~(B - C).(cosB - cosC)
< Q. r: ':. ri
33
if
.
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
35/341
Bt ng thc Gsi (bt ng ic gia trmg bmrcQg v traing brih nhn)
Nu a, b > 0 th a + k > Vb , h ocvitcch khc: :
;i;^ng' ch xy m -l^-;-'chi;-iiva b;-
Nu a, b, c > 0 th -a + -k + c- > >/abc hbc vit cch
khc:,
a + b + c > a +,b + c j
-Du bang chi x ra khi va
Phng php tam thc bc; hai f(x) x? + bx + c, a 560
Nu f(x) = 0 c nghim: th A >: 0Nu A < 0 i a.f(x) ^ 0,
V-x.
Nu A < 0 th .f(x) > 0, Vx' . .-'
Phng php min gi tr, i.kiri c nghim, . .
Phng php to , yct,
Phng php o hm
Nu y f(x) - > 0 trn K th f(x) ng bin trn K:
. . . rx > a => f(x).> f(a); X < b => f(x) <
f(b)
- i vi yf < 0 tn.K. th ta c bt ng thc ngQ i:' :v: X > a
=> f(x) < f(a): X < b => f(x) >.f(b) ^
;V;y-^Neirf t GTNN lin 0
Ni f t GTN trn P j ix t . ^Da vo bng bin thin ct 1 hm s trn D ta
cng c hh gi v
hm s tng ng.Nguyntc & n'gU 'tq ' !^n h^ jn | iot iMirm
hm/s:
^ tin GTLN x hani so f(^ k m ien D t xc lp 'bt ng thc dng
f(x) < A v i l hng s.cj: Bc '-2 :i xem xt d ng ic ixay r Mil
nao.' au th nu tnij-hc nu chng hn gi tr X no thuc D.
Cui-cngl kt lun-GTLN'f(x)i=>A:.:'-:K h i u m a xf(x) ; nu
khhg nhm lnio ghi max f(x).
' . ; . xcD .-vi ; '
. :T ong t cho vic tm G T N N . - -
34
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
36/341
Nguyn: t c . tm gi r | Sm m2f, nh sali ca mt biaa thc:
E)e tm GTLN ca mi biu the T .= g(a; b; c;.) trn min D CSCiu
kin cabiri , b7 C,... ta xc lp bt ng thc dng: g(a; b; c;..) <
Avi Al hng s. ,'T
Bc 2 l xem xt u ng thc xy ra khi no. Du c th rCu
tntai--hpo:hcigincc gi-tr.a,-b,-c.,... no -thuc D.
' Ciii ciig kt ln GTLN g(a; b; c ;...) = A.
Tiic^g tir chd vic tm GTNN.
Ch : . . .. . :1) Gc bin i ig gic, phi hp cc cng ic ng gic.
2) Gc h thc lng gic c bn troig .tam gic phn trc.
3) V; B, G l ba gc mt tam giac th
A B c> sinA; sinB; sinC > 0, sin ; s i n ; sin > 0, - ,
; - : 2 2 2
: .. A , B cCOS ; c o s ; COS > 0 .
2 2 2
4) Bt ng thc JENSEN-y hm li khng c s dng gii ton nhng li c nhng
nh gi rt tt, tin li nh hng:
Cho y = f(x) c y . 0 1 du bt ng c ngc li:
- f(a ) + f(b) + f(c) ^ a.'+? + c V
;|v I 3 J_________
2.CC B1T0N IN HNH
'3 \ /3Bi 161: Chng minh vi ABC l tam gic b k: sinA + sinB +
sinC <
2
G:e rng vi 0 < X, y < nth
n ^ X - y ^ n _ ^ X - y ^ - < ----- < -Z- => 0< co
s------< 1nn2 2 2 2
s in x+ s i n y ____ x + y _ _X - y ^ X + y-------- ------ =
sinCOS < s in
2 2 2 2
3 5
irnM!.Mwrm!Baii!aaas3asLMBHBM
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
37/341
Ap dng kt qu trn:
: A Ty sinC-t-sin^ , T c + ^snA + snB ^ . A +Jd . Q + _ sin +
sin -2 2 ^ 2 2
A + B + C + - _ n;. 3 ___- n V3 ,A . .
< sin-------- --------- = sin = => (pcm).
4 3 2Cch khc: du xy ra khi A = B = c = nn ta lp ghp ri
3dng bt ng thc Csi nh sau:
sinA + sinB + sinC = -=(sin A + sin B. ) + cos B + SJ COS
A).& 2 2 v3 V3 /__ ' A B O 3/3
Bi 162: Chng minh vi ABC l tam gic bt k: COS + COS+ COS < -2
2 2 2
Gii:~ n ^ ^ n c o s x + c o s y ^ ____X + y
rng vi < X, y < thi ---------- 0 .Du bng xy ra ch khi sinA
= sin B - sinC = 1 : v l.nn sinA + sinB + sinC sinA sinB sinB sinC
sinC sinA > 0.V t ( sinA)(] sinB)(l sinC) > 0 suy ra1 sinA -
sinB sinC + sinA sin B + sinB sinC + sinC sinA
+ sinA sinB sinC > 0V sinA, sinB, sinC > 0 nn c:sinA +
sinB + sinC - sinAsinB - sinBsinC sinCsinA
< 1 sinA sinB sinC < 1 .Bi 164: Chng minh vi ABC l tam gic
bt k:
31 < cosA + cosB + cosC < .
2
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
38/341
Gii:_ A B CTa c: cosA + cosB + cosC 1 + 4sin - s i n - sin >
1 .
2
2
23
v: cosA + cosB + cosC <
A + B A B 2f + 3 2 c o s ------ r----- c o s
-------------------2 c o s -------- ------ + 1 <
2 2 ^ 2 22f A + A + B _ A B 1 .
COS -------------- COS-------------- COS------- -------- + >;
0V 2 J 2 2 4
A + B 1 _ ' A - B2 1 A - ~ , [cos-------------- COS---------- ]
+ sin2 ----- ------ > 0: ng
2 2 2 4 { 2 J
Cch khc: du xy ra khi A = B = c = nn ta lp ghp rio
dng bt ng thc Csi nh sau:cosA + cosB + cosC = (cosA + cosB). 1 +
sin A.sin B - COS A cosB.
Bi 165: Cho tam gic ABC c 3 gc nhn.Chng minh rng: tanA + tanB +
tanC ^ 3 s.
Gii:Ta c h thc: tanA + tanB + tanC = tanA tanB tanC.Vi tam gic
nhn nn tan A, tan B, tan c > 0, p dng bt ng thc Csi:tan A + tanB
+ tanC > 3 tan A tan B tan cnn (tanA + tanB + tanC)3> 27 tanA
tanB tanC = 27(tanA + tanB + tanC)o (tanA + tanB + tanC)2> 27
=i> (pcm).
Bi 166: Chng minh rng, trong tam gic ABC bt k ta c:
A - B B - C C - ACOS----------- + c o s ---------- + c o s
-----------2 2 2^ i f A *0 1-o 71 1< cos A + cos B + cos^r c
3 \ 3 J 3J 3^3 ,Gii:
_ I , n ^ , A - C ^ , 7 r A A C -, < --------- < => 0
< cos ---------< 1;
4 4 4 4
0< B - | < - => cos 1B - > COS IB ~ I > 03 3 - 4
3 2 3 3 4 3
, 1 , A - B , B - c , _ A - C A + C - 2Bne n ta CO (c o s
----------- + COS------- ) = COS-----------
COS-----------------------2 2 2 4 4_ A - C 3, 71 -QN_ A - c 3 n I=
cos---------COS B) = cos---------cos B _
4 4 3 4 4 3
< cos |B I= cos (B )3 3 3 3 .
37
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
39/341
_ \ , B - C , c - A. , rc.Tng t: (cos - COS ) ^ cos^-(C-r)
2 2 ^ 3 3 '
-- (cos + COS ) ^ COS ( A ) = > pcm.2 2 2 3 3
Tng t
c o s ( A - | ) < | c o s ( A c o s | ( C - | ) < - ~ c o
s ( C - I ) =>(pcm)
Bi X67: Chng minh rng nu tam gic ABC tha iu kin
cos2A + cos2B + cos2C 1
th sinA + sinB + sinC < 1+y2 .
Gii:
Ta c cos2A + COS2B + COS2C = 1 4cosAcosBcosC
nn bt ng thc bi suy ra cosAcosBcosC < 0 do tam gic ABC
khng nhn, gi s A > .2
Ta c: sinA + sinB + sinC = sinA + 2cos COS_ ^ < sinA + 2c os
.2 2 2
Xt hm s f(x) = sinx + 2COS-; < X < l2 2
f (x) = cosx sin = (1 - 2sin )(1 + s in ) < 0,2 2 2
nn f(x) nghch bin trn [ ; r)9o
f(A) < f(-^ ) nn sinA + 2cos < 1 + y2 =>pcm.2 2
Bi 168: Chng minh rng trong tam gic ABC ta c
ab(a + b - 2c) + bc(b + c 2a) + ca(c +- a - 2b) > 0.
Gii:Bt ng thc bi tng ng:
a + b - 2 c , b + c - 2 a c + a - 2 b----- ------- +
------------ + ------- ----- 0c a b
( + ) + ( + ) + ( + b ) a 6 .c a c b b a
p dng bt ng thc Csi th c pcm.
B 169: Cho tam gic ABC, chng minh a4+ b4+ c42:16S 2.
38
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
40/341
Gi:Ta c bt ng thc hin nhin
a 2 > a 2 ( b - c ) 2 = ( a 4- b - c ) ( a b + c )
b2 > b2 - (c - a)2 = (b + c - a.) ct - c + a) ,
c2s c2(a b) 2= (c + a - b)(c - a + b)=> a 2b 2c 2 > (a + b
- c ) 2(b + c - a ) 2(c + a - b ) 2
=> ab c> (a + h - c)(b + c a)(c +a b)Mt khc, p ng bt ng hc
Csi2 (a 2b c + a b 2c + ab c2) < a 4 + b2c2 + b 4 + c 2a 2 + c4
+ a 2b 2
< a 4 -- b 4 + d 1+ (b 4 + c 4 + c 4 + a 4 + a 4 + b 4)2
= 2(a4+ b4+ C*)nn a4+ b4+ c4: abc(a + b + c) do :
a4+ b4+ c4> (a + b + c)(a + b c)(b + c a)(c + a b) = 16S2Bi
170: Trong cc tam gic ABC, tm gi tr b nht ca
T = cotA + cotB + cotCGii:
Ta chng minh: cotA + cotB + co te > V3
(cotA + cotB + cotC)2> 3cot2A + cot2B + cot2C + 2(cotAcotB +-
cotB cotC + cotC cot A) > 3 (cotA - cotB ) 2+ (cotB - cotC) 2+
(cotC - cotA)2> 0
Du ng thc xy ra khi cogA = cotA = cotB = coC A = B ~ c.Vy
min(cotA + cotB + cotC) = y/s.
Bi 171: Trong cc tam gic ABC khng vung, c anA, anB anC theoth t
lp thnh cp s cng. Tm gi tr b nht ca gc B.
Gii:Ta c h thc anA + tanB + tanC - tanA tanB tanC, v anA, tanB,
tandlp thnh cp s cng nn 2tanB = tanA + anCdo 3tanB = tanA tanB tanC
suy ra tanAtanC = 3 > 0 .Vy tanA, tanB, tanC > 0 nn tam gic'
ABC nhn.
Theo bt ng thc Csin _.___ . ( tan A + taaC _ A 2tt,3 = tanA tanC
< ! ------------- I= tan B
=> tanB > y/s=> < B < .Vy min B = .3 2 3
B i 172: Cho tam gic ABC bt k. Tm gi ln nht ca biu thc
p = \3cosB + 3(cosA + cosC).
39
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
41/341
Gii:
p = V cosB + 6sin cos < \/s cosB + 6sin2 2 2
- 73 (1 2sin2) + 6sin = _ 273 (sin < t .2 2 2 2 2 2
rv: M- A - c _ . , . B _ ^3Du = xy ra khi COS- = 1v sin = 2 2
2
A = c = v B = 2 . Vy max p =6 3
2_ 5 7 3
2
3. I LUYN TPBi 173: Chng minh bt ng thc vi ABC l tam gic bt
k:
cos2A + cos2B + cos2C > .4 ,,
HO: Bin i tcmg ng
A 0 / 1 ^ ' ^ sin 2A + sin 2B + sin 2C! Bi 174: Cho tam gic ABC
tu . Chng m in h ---- -------------------- 3.cos A + COS B + cos
c
HJD: Bin i tng ngBi 175: Trong mi tam gic ABC, chng minh
/ s i n A + %/s inB + / s c
J A B , r 7 c3 cos + 3 COS + ?/COS~V 2 V 2 V 2
< 1.
HI): Ta chng minh mi X, y > 0 t h i x + y < ^/4(x3 + y3)
.
' A B C 3V3B 176: Chng minh vi ABC l tam gic bt k: COS COS COS
< 2 2 2 8
HI): Dng bt ng thc Csi.
A B C ^/3Bi 177: Chng minh vi ABC tam gic bt k: tan tantan <
.2 2 2 9
i n * A B * c _ c + AHD: tan---tan + tan-tan + tan tan = 1.2 2 2
2 2 2
Bi 78:Chng minh vi ABC l tam gic bt k:
X2
1 + > c o s B + (co s A + c o s C ) x ,V x 2HI): Dng tam thc
bc hai.
179: Chng minh vi ABC 1
HD: Dng bin i tng ng
cBi 179: Chng minh vi ABC l tam gic bt k: sinA. sinB < COS2
.
40
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
42/341
Bi 180: Chng minh vi ABG l tam gic bt k:
(1 cosA) (1 cosB) (1 - cosC) .8
HD: Dng bt ng thc Csi.Bi 181:Chng minh vi ABC l tam gic bt
k:
1 1 1 1 1 - f - 1 ^ --
sin A sin B sinC A BCOS - cos - cos2 2 2
HD: Ghp cp ri nh gi.Bi 182: Chng minh vi ABC l tam gic nhn:
Vtan A + VtanB + VtanC s Jc ot +./cot + JcotV 2 V 2 V 2
HD: Dng bt ng thc Csi.Bi 183: Chng minh vi ABC l tam gic bt
k:
1 1 + -------7 > 1 2 .
sin2 t f J sin [ f j sin
TITT* 2 A _ 1 A\ - 1 /1 b 2 + c 2 - a 2 .HD: sin ~ = (1 - cosA)
( 1 ------------------)2 2 2 2bc
Bi 184:Chng minh vi ABC l tmgic bt k:
a2(l V3cotA) + b2(l y cotB) + c2( l y/s cotC) > 0
HD: Dng bt ng thc Csi,
Bi 185:Chng minh vi ABC l tam gic bt k: ma mb mc___ 3 2 , 1-2 ,
2HD: Chng minh a.ma -----* -----
Bi 186: Chng minh vi ABCl tam gic bt k: + 2 + 2a b c 4r
HD: Dng so snh.Bi 187: Chng minh vi ABC l tam gic bt k:
3 (a + b + c ) < m a + m b + m c < a + b + c4
HD: Dng hnh bnh hnh AB CD tm M.Bi 188: Chng minh vi ABC l tam
gic bt k:
a2(p - b)(p - c) + b2(p - e)(p a) + c2(p - a)(p - b) < p^
2
HD: Dng tam thc bc hai theo X = p a.
41
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
43/341
Bi 189: Chng minh vi ABC l tam gic bt k:y cosA + 2cosB + 2
/3cosC < 4 .
HD: Dng bt ng thc tng ng
(sinB *!?> sinC)2 + (cosB + V3 cosC 2)2 > 02 2
Bi 190:Chiig minh vi ABC l tam gie bt k: < +- k+ - < 22 b
+ c c + a a + b
HD: Dng so snh v bt ng thc Csi.Bi 191:Chng minh vi ABC l tam gic
nhn:
tan A + tannB + tannC > 3^/(3V3)nHD: Dng bt ng thc Csi.
Bi 192: Chng minh v i ABC l tam gic bt k:Nu c sin A + sin 2B +
sin 2C th anA.tanB < 1HD: sin 2A + s in 2B + s n 2C2 +2cosAcosB
cosC.
Bi 193: Hai gc A, B ca tam gic ABC tha mn iu kin tan + tan = .2
2
3 c Chng minh rang < tan < 1.4 2
HD: Dng bt ng thc Csi.B i 194: Chng minh vi ABC l tam gic
nhn:
2 _ 1 _ _ (sinA + sinB + sinC) + (tanA + tanB + anC) > 7r3
3
HD: Xt hm s f(x) = sinx + tanx X vi 0 < X <
w 3 3 2Bi 195: Chng minh vi ABC l tam gc khng c gc t:sin A + sin
B + sinC _ \2-----------------------------> 1 + cosA + cosB +
cosC 2
c2 COS + sinC
0D : VT > --------------------- v dng o hm.2sin + COSc
2Bi 196: Trong cc tam gic AB C, tm gi tr ln nht ca biu thc
E = cosA cosB cosC.
BS: max E = .8
Bi 197: Trong cc tam gic ABC, tm gi tr b nht ca biu thc:
+ ~ X l + ~ h sn sin sn-r2 2 2
S: min N = 27.
42
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
44/341
Bi 198: Trong cc tam gic ABC, tim gi tr b nht ca biu thc
R = tan2 + tan2 + tan22 2 2
HX>: Dng bt ng thc Csi.Bi 199:Trong cc tam gic ABC, tm gi tr
ln nht ca biu thc
M = 3cosA + 2 (cosB + cosC).
BS: max M = ~3
Bi 200: Trong tt c tam gic ABC ni tip ng trn (0;R) cho tnrc,tm
tam gic c din tch ln nht.
S: tam gic ABC u .
-DNG3.-.---'. : , :D^ G'.TAM:G 1C ________________
1. L THUYT V PHNG PHP GII TOM ________v':'i;Mt;ani gide thm cc
iu kin: v cnir hoc v gQ th c dng cbf:;/tm"gic'-vungi'tm'-gic cn,
tam gic.u, tam gic nhn,...hoc c gc3 0,6Q ,2Q , ..7 ^
X c p h m r a g p h p I s |b i i : .
Bin i lng gic trc tip a v phng trnh lng gic c bn^ Lp hiu so 2 v
ri bin i iih gi bt ng th ci s v ng gic G ithitcho ubng ca bt ng
thc
. Ch : ^ _ '1) Phi iifp cc bin iliicmg gic v cc nh l trong tam
gic.
.' 2) Cc bi ton nh dng th phi gii theo bin i tang
ucmg._______
2. CC BI TON IN HNH
Bi 201: Cho tam gic ABC tha mn = 2cos A.sinB
Chng minh tam gic ABC cn. Gii:
Ta c sin ^ = 2cos A nn sinC = 2cosAsinB sinB
=> sin(A + B) = 2sinB cosA=> sinA cosB + sinB cosA = 2
sinB cosA=> sinA cosB sinB cosA = 0 => sin(A B) = 0=> A B
kt. V A, B l cc gc tam gic nn k = 0 o A = B.Vy tam gic tam gic ABC
cn ti c .
43
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
45/341
Bi 202: Chng t rng nu trong tam gic ABC ta c:c
tanA + tanB = 2cot th tam gic ABC cn.2
Gii:
ng thc cho tong ng vi sin~^ = 2tan~ cos A cos B 2
A + BA + B A + B _ rts 2 A T ,
2 s i n ----- c o s - = 2 -------- T m-n c o s A c o s B2 2 __A
+ B
cos - 2
CC)S2 f ^ | = cosAcosB
-- 1 + cos(A + B)] = [cos(A - B) + cos(A + B)] cos(A - B) = 1.i'
2
T d suy ra tam gic cho l tam gic cn.
Bi 203: Chng minh rng tam gic ABC tha mn iu kin:cos2A + cos2B +
cos2C + 1 = 0
th tam gic l tam gic vung.Gii:
Ta chng minh ng thc:cos2A + cos2B + cos2C = 14cosA cosB cosC
Vy ang thc bi tng ng vicosA oosB cosC = 0o cosA = 0hoc cosB = 0
hoc cosC = 0.Vy lim gic cho l tam gic vung.
b2 + c2 < a2Bi 204: Xt dng tam gic ABC tho mn: 0 nn (b +
c)2< 2(b2+ c2) < 2a2
=> b + c < a /2 = > a + b + c < a y/2 + a = a (y/2 +
1)=> sinA + sinB + sinC < (V2 + l)snA < V2 + 1 .
Du "=="xy ra: ----- v c o s 3B c o s 2B >
--------------------------------- .3 12 3 12
=> (cos3A cos2A) + (cossB cos2B) > - 3 3 6
Do ng thc xy ra nn cosA = v cosB = o A = B = 60.2 2
Vy tam gic ABC u.Bi 208: Cho tam gic ABC khng t, tha mn iu
kin
COS2A +2\2COSB+2\2COSc = 3. Tnh ba gc ca tam gic ABC.
( H khi A nm 204)Gii
t gi thit tam gic AB C khng t th c. _ A. _ B C , ______2 A A
s i n - r - > 0,COS - l , c o s A < COSA nn2 2
c o s 2 A + 2^/2 c o s B + 2-J2COSc - 3
= 2 cos2A 1 + 2V2 .2 cos- .cos-B - 32 2
^ 2 cos2A 1 + 2%/22c o s - 32
< 2cos A 1 + 2-42.2sin - 3 = -2{s2.sin - l)2 02 2
Do du "" xy ra nn:
B - C ,COS- = 2
COS = 1 . Vy A = 90, B = c = 45.
\Z2.sin = 12 / .
Bi 209: Cc gc A, B, c ca tam gic ABC tha.mnA B c
sinA + sinB + sinC - 2sin sin =2 sin .2 2 2
Chng minh rng c = 120.
Gii
Ta chng minh h thc sinA + sinB -- sinC = 4cos ^ COS COS nn t2 2
2
g i t h i t t h c :
A _ B c , A , B _ A + B4COS- co s COS 2 si n -^ si n . = 2
COS------r------
2 2 2 2 2 2
46
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
48/341
A B c A . B A B A . c x4c os -r cos co s- - 2 s m - s i n
------2(cos-C O S-------s i n - f s i n - O
2 2 2 2 2 2 2 2 S:
li n c o s COS ( 2 c o s - 1) - 0 .Zi
Vi cos cos > 0 nn COS = => = 60 c ~ 120.2 2 2 2 2
Bi 210: Tnh gc c ca tam gic ABC bit rng: (1 + cotA)(l + cotB) =
2.
Ta chng minh h thc cotAcotB + cotB cote + cotC cotA 1
nn t gi thit
cotA + cotB + cotAcotB = 1cotA + cotB + cotAcotB = cotAcotB --
cotBcotC + cotCcotA
nn cotA + cotB = cotB cotC + cotCcotA = cotC (cotA + cotB)Gi s
cotA + cotB = 0=> cotA = cot(B) =^A + B = 7T:V
nn cote = 1 => c = 45.
(A > B > c
Bi 211: Tnh cc gc ca tam gic ABC, bit rng
T iu kin th nht suy ra A > ; > B > 0; > c > 0
3 2 3_ 3tc ^ 3A _ TE 3jc 3B ^ A n . 3C ^ . . 3B ^ A . 3C .
n=> > > 41; 0; -Z >> 0nn s in - > 0; sin > 0.2
2 2 4 2 2 2 2 2
, A v chon A = .
2 3 3. , , 5^ 5g 5(3 /
T iu kin th ba bin i thnh COS COS COS = 02 2 2
V cos 0 v do B > c nn > c > 0 CGS-^ > 0.2 6 2
do cos = 0 => B = +v chon B= . T c = .2 5 5 5 15
Bi 212; Cho tam gic ABC tho-^- = Tnh gc A. K b c
47
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
49/341
T- 1 _ 1 1 ~ 1 (1 . 1 1 _ 1 1Ta CO - = + --------*A b e 2 c o s
' c ' b c
2
cos = + k2r A = 4- k47t2 2 2 3 3
27TVi A l gc mt tam gic nn A = .
33. BI LUYN TPBi 213: Cho tam gic ABC tha:
cosC(sinA + sinB) = sinC.cos(A B). Hy tnh cosA + cosB.S: cosA +
cosB = 1 .
Bi 214: Xt dng tam gic ABC tha mn iu kin:A 3 B . B 3 A
s i n c o s = s i n COS z~ .2 2 2 2
SrTam gic ABC cn ti c.
Bi 215: Cho tam gic ABC tha mn co t = + c .2 bChng minh rng tam
gic ABC l tam gic vung.
S: Tam gic ABC vung ti A hay c .
Bi 216: Cho tam gic ABC tho mn sin(B + C) + sin(C + A) + cos(A +
B) = .
Chng minh tam gic cn.S: tam gic ABC cn ti c.
Bi 217: Xt dng ca tam gic ABC tha mn iu kinsin A + sinB + s in C
___. A
= c o t - c o t .sin A + s inB - s inC 2 2
S: tam gic ABC cn.Bi 218: Chng minh nu tam gic ABC tho mn iu
kin
QCOS A cos B = sin 2 th tam gic cn.
2HD: Dng cng thc h bc.
Bi 219: Xt dng ca tam gic ABC tha mn iu kin0
a 2 s i n 2 B + b 2 s i n 2 A = c 2 c o t 2
BS: tam gic ABC cnB 220:C hng minh rng tam gic AB C vung hoc cn
khi v ch khi:
a cosB b cosA = a sinA b sinB.HD: Dng nh l sin
48
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
50/341
j___, ____., .r,/-,*, - cosA cosB cosC 5Bi 221: Xt dng tam gic
ABC thoa- + I- - - = 3 4 5 1 2
HD: Dng tng bnh phcmg hoc vct.
Bi 222: Xc inh dang ca tam gic ABC nu s = (a + b - c)(a - b +
c).4
S: Tam gic ABC vung ti A.
Bi 223: Chng minh rng nu ong tam gic ABC c_ (a j- b)(b + c -
a)(a + c - b) +U
HD: Dng nh l cosin
______ _ D^D + c - a x a + c - d; . , . ,cosB =
------------------------------th tam gic l vung.
2abc
Bi 224: Xc inh dang ca tam gic ABC nu 7.cosB cosC sin B sin
c
S: tam gic ABC vung
Bi 225: X t dng tam gic ABC tho sin = J ------ .2 V 2cL
S: Tam gic ABC vung.
Bi 226: Chng minh rng, nu ta c be V3 = R(2(b +c) - a) th tam gic
l u.
H: snB[>/3 sinC + cosC 2] + siC[n/3 sinB + cosB 2] = 0
B i 227:Tam gic ABC tha mn 3S = 2R2(sin3A + sin3B + sin 3C).
Chng minh rng ABC l tam gic u.
HD: Dng bt ng thc Csi.___ , 9R
Bi 228: Chng minh nu tam gic ABC tho m a + mb + mc = th tam
gic ABC u.
HD: Dng nh l trung tuyn.
Bi 229: Xt dng tam gic ABC tho ab sin + be s in + ca s in^- = 2S
V3 .2 2 2
1 2SHD: s = bc sinA Sy ra bc = : r~ -
2 sin A
Bi 230: Xt dng tam gic ABC tho
A . B . B . c . c . A 5 _rs i n s i n + s i n s i n + s i n s m
- +
2 2 2 2 2 2 8 4R
S: tam gic u
49
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
51/341
Bi 231: C tn ti hay khng tam gic ABC tho mn iu kin
s = a2 - (b - c)2
sin2 A + sin2 B + sin 2 c = cot A + cotB + cotC
BS: khng tn ti
Bi 232: Cc cnh tam gic ABC tho mn a4 = b4 4- c4.Chng t rng tam
gi c ABC nhn v 2s in2A - tanB tanC.HX>: So snh cnh gc trong tam
gic.
I Bi 233: Tnh 3 gc ca tam gic ABC tho A B C
cosA + cosB + cosC = sin + sin + sin 2 2 2
BS: A = B = c = 60.
Bi 234: Gho tam gic ABCc 3 cnh: a = x2 + x + l ; b = 2 x + l ; c
X2 1.
Tnh gc ln nht. , .1 BS: A = 120.
Bi 235: Tnh 3 gc catam gic ABCtho mn: S = (a + b + c)2
B S: A = B = c = 60.
Bi 236: Tnh 3 gc ca tam gic ABC tho: b + c = +
HD: Dng nh l sin v din tch.Bi 237: Tnh cc gc ca tam gic ABC, bit
chng tha mn
3 .sin(B A) sinC + sinA + cosB =2
S A = 30; B = 60; c = 90.'Tr,
1 + cosC 2a + b
Bi 238: Tnh 3 gc ca tam gic ABC tho mn: snC ^4a2 ~ b2
a 2(b + c a) = b3 + c3 a 3
BS: A = B = c = 60
Bi 239: Cho tam gic ABC tho s in ^ - - CQ- anA. Tnh gc A.sin B +
cos ABS: A = 90.
Bi 240: Cho tam gic ABC c gc A, B nhn : sin2A + sin2B = %/sin c
.
Tnh gc c.
BS: c = 90.
50
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
52/341
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
53/341
Phng php: V- \ ''Xt cosx .= 0, xt cosx 7* Or chia 2 ye chp
,cos"x a v phng trinhtheo t = tanx. :; ^ ;V.;>. V ;' ^-:;Ncu
chia sinnx th v phng teirii Aeo"t3= cofcc. : " . : - Ch : Bc tng
/gim ixrTig i cua lng g ic . ; . ;sin2x = 2sinxcosx; 1 = s in 2x .+
cos2x; cos2x;= cos^x-- sin 2 :..
c = C(sin2x + cos2x) = C(sin2x + cosSc)"* I:;vPhcrag trnh bc nh
t theo siiij: S (G in):Phcrag trnh bc nht theo siii es (G in):
;Dng: a. sinu + b: cosu:= c . : J /',iu kin c nghim: 2 + b 2 >
c2 : ;
Phng php: Chia 2 v ch Va2 + b2 th c: . : V
a ' : ' .. b i : .' ' V C V= - ; s i n a + ; ; _. .COS a 7r;
.2 . 1.2 - / 7. -1 5? . ~ . . 9. ,
52
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
54/341
2 _g q b i t o n i n h n h{Bai 24 l Gii phng trnh: 2cos2x c o s
x - 1 = 0
-----^ Gii:
2 c o s 2x c o sx 1 = 0 c o s x 1 hay c o s x = 2
o x - k2n hay X = + k271X - z .3 3B i 242: Gii phng nh: c o s 23
x c o s 2 x c o s 2x = 0
Gii:(H khi A nm 2003)
PT (1 + cos6x)cos2x (1 + cos2x) = 0 cos6xcos2x 1 = 0cos8x +
cos4x 2 = 0 2cos24x + cos4x 3 = 0
cos4x = l 3 cos4x = 1 X = , k z .
cos4x = - (lo i) 22
/ Bi 243: Gii phng trnh: 5snx 2 = 3(1 sinx)tan2x(H khi B nm
2004)G:
iu kin: cosx 0 X = + k7, k
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
55/341
Gii:
Ta c: sin2x + sin2x 2cos2x = *
o s i n 2 x + s in 2 x 2 c o s 2X = ( s in 2x + c o s 2x )2
sin2x + 4s inxcosx 5cos2x = 0Xt cosx = 0 phng trinh: sinx = 0
=>sin ^ + cos^ = 0 (loi)Xt cosx 5* 0. Chia 2 v cho coszx: ta n2x
+ 4tan x 5 = 0V a + b + c = 0 nn tanx = 1 hay tanx = 5 = ta na
V y n g h i m X = + k t, X = a + kt, k s Z .4
Bi 246: Gii phng trnh: s in3x ~ 3 sin 2xcosx H- 4cos2x . sin x =
0Gii:
Phng trnh: sin3x 3sin2xcosx -- 4cos2x . sinx = 0Xt cosx = 0 th
sin 3x = 0 => sinx 0=> sin2x + coszx = 0: V l.Xt csx ^ 0. Chi
2 v cho cos3x:t a n 3x 3 t a n 2x + 4 t a n x = 0 o t a n x ( t a n
2x ~ 3 t a n x + 4 ) = 0
t a n X = 0
tan X 3t an x + 4 = 0 (v nghim v A < 0)
tanx = 0.Vy nghim X = k7,k Z .
Bi 247; Gii phng trnh: 6s in x -+- 8cosx = 5Gii:
Chia 2 v cho V36 + 6410.PT: . sin X + 4- COS X =5 5. 2
V ' l Y T i T - 1 - _ 3 -_ 4V + = 1 nn tCOS cp= , sin (p= \ 5 J
W ' 5 5
X 1 7TPT: COSC) . sinx +sincp. cosx = sin x + cp) = = sin 2 2
6
X = tp + k2tVy: 6
SuX P -cp + k2E, k e z
.. . 6
Bi 248: Gii phong trinh: c 0 sx : 2 s ln ^c0 sx = s
Bi 245: Gii phng trnh: sinzx + sin 2x - 2cos X =
2 cos X + sin X 1
54
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
56/341
G:>iu kin 2cos2x + sinx i 5*0cosx-2sinxcosx _ ^ COS X - sin
2x ^
2 cos2 X + sin X - 1 cos 2x + sin X
"J3cos2x + sin2x = cosx \3 .sinx COS 2x I= coss X K e ) { s
)
x = + k 2 7 t ( l o a i )
X = k - , k
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
57/341
Phng trnh tr thnh: 6u +l - u ' + 6 0
12u + 1 u2 + 12 = 0 u2 12u 13 0
y/2 c o s X + = -1 cos X + ] - l=r{ 4J l 4 j 4
u = - l
U = 13>V2
n _ 37 1 o _X + = 4 - k 2 7 T4 4t _ 371 *
X + = + k2z4 4
Vy nghim:X = +k2t , rr
2 k e Z
X = 7+ k2jt
Bi 252: Gii phng trnh: 2sinx(l + cos2x) + sin2x = 1 + 2cosx(H
khi D nm. 2008)
Gii: .Phng trnh tng ng: 2s inx.2cos2X + sin2x = 1 + 2cosx 2sin
2xcosx + sin2x = 1 + 2cosxs in2x (2cosx + 1) = + 2cosx (2cosx + l)
(s in 2x 1) = 0
c o s x = -----
2s i n 2 x = 1
x = - ^ + k27t3
X = + k i , k e Z 4
Bi 253: Gii phng trnh: cos3x + cos2x cosx 1 = 0(Hkh D nm
2006)
G:
PT: cos3x cosx + cos2x 1 = 0 2sinJ|x.siii2x 2sin 2x = 0 s in x (
s n 2 x + s in x ) = 0 s i n 2x (2 c o s x + 1 ) = 0
osin X = 0
1 ^ cosx = 2
X = k !
x = +k2t, k e Z3
Bi 254: Gii phng trnh: (2cosx l)(2 sin x + cosx) sn2x sinx(H khi
D nm 2004)
Gi:PT (2cosx l)(s inx + cosx) = 0
To V _ 1_ n 1 X = + k2I COS X - 1 = 0 cos X = 3s i n X 4- c o s
x = 0 . _ _ , n , _ _
t a n X = -1 x = - + K7T, k e Z L 4
Bi 255: Gii phng trnh: 2sir?t+ sin7x 1 = sinx(H Kh i B nm
2007)
56
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
58/341
PT: 2snz2x 1 + sin7x sinx = 0 o cos4x 2cos4xsin3x = 0Gii:
co s4 x( l 2sin 3x) = 0 cos4x = 0
1 sin 3x - 2
71 - nX = + k
8 47t T 2 n
X " -r k 18 3
.251X " + k1, kZ18 3'Bai 256: Gii phng trnh: (1 + sin2x)cosx +
(1 + cos2x)sinx = 1 + sin2x
v (H kh i A nm 2007)Gii:
PT: sinx + cosx + sinxcosx(sinx + cosx) = (sinx + cosx)2(sin x +
cosx)(l + sinxcosx sinx cosx) = 0 (sinx + cosx)(l sinx)'(l cosx) =
0
s i n X + cosX = 0 s i n X = 1
COS X = 1
X = 4- k t
4X = + k2r
2X = k27t, k e Z
Bi 258: Gii phng trnh: sin X + - + COS X + *s i n X
3. BI LUYN TP
Bi 257: Gii phng trnh: sin23x cos24x = sin25x cos26x
HD: Cng thc h bc
_____
10cos X 3
HD: t t = sinx + cosx,
Bi 259: Gii phcmg trnh: 2sin 2x + 3cos2x = 5sin 2x + 6 HD: PT: 7
sin2x + lp sinxcosx + 3cos2x = 0
B i 260: Gii phng trnh: sin x + cosx + 5sin2x = 1
H: t y = sinx + cosx \/2s in (x + ), ly < V24
/2Bi 261: Gii phng trnh: s in (2 x -1 5 ) = vi 120 < X <
90
S:-105, 30,75
Bi 262: Phng trnh: tan(3x + 2) =yfz vi < X < c bao nhiu
nghim.2 2
S: 3 nghim.
57
I?;>
WWW.FACEBOOK.COM/DAYKEM.QU
WWW.FACEBOOK.COM/BOIDUONGHOAHOCQU
B
I
D
N
GT
O
N
-
L
-
H
A
C
P
2
3
1
0
0
0
B
T
R
N
H
N
G
O
T
P
.
Q
U
Y
N
H
N
W.DAYKEMQUYNHON.UCOZ.COM
ng gp PDF bi GV. Nguyn Thanh T
-
8/10/2019 1234 BI TP T LUN IN HNH HNH HC-LNG GIC - L HONH PH
59/341
S:18+-k72Bi 264: Gii phng trnh: sin22x + cos23x =1
HD: S dng cng thc h bc.Bi 265: Gii phng trnh: tan 2x tan 3x tanx
= tan2x.tan3x.tan5x
HD: a v tan 5x = tan X. , , sin x + s in 2 x + sin3x nr
Bi 266: Gii phng trnh: ------------- ----- ------- = V3cos X +
cos 2x + cos 3x
HD: a v tan 2x = V3Bi 267: Gii phng nh: cos2xcosx + sinxcos3x =
sin2xsinx sin3xcos X
KD: a v COS 3x = sin 4x
Bi 268: Gii phng trnh: s in 4X - s in 4 (x + ) = 4sin . COS
.cosx
Bi 263: Gii phng trnh: sin3x = cos2x
7r , 7T
+ k-r8 2Bi 269: Gii phng trnh: 3sinx + 2cosx = 2 + 3tan X
HD: a v (cosx l)(3tanx + 2) = 0.
. cot(x + )Bi 270: Gii phng trnh: x _