chuyen de bat dang thuc luong giac (2)

8/14/2019 Chuyen de Bat Dang Thuc Luong Giac (2)

1/35

Trng THPT chuyn L T Trng Cn Th Bt ng thc lng gicChng 2 Cc phngphp chng minh

The Inequalities Trigonometry 31

Chng 2 :

Cc phng php chng minh

Chng minh bt ng thc i hi k nng v kinh nghim. Khng th khi khi m tam u vo chng minh khi gp mtbi bt ng thc. Ta s xem xt n thuc dngbino, nn dng phngphp no chng minh. Lc vic chng minh bt ng thcmi thnh cng c.

Nh vy, c th ng u vi cc bt ng thc lng gic,bn c cn nm vngcc phngphp chng minh. s l kim ch nam cho ccbi bt ng thc. Nhngphngphp cng rt phongph v a dng : tng hp, phn tch, quy c ng, clng non gi, i bin, chn phn t cc tr Nhng theo kin ch quan ca mnh,nhng phng php tht s cn thit v thng dng s c tc gi gii thiu trongchng 2 : Cc phng php chng minh.

Mc lc :2.1. Bin i lng gic tng ng ... 322.2. S dng cc bc u cs... 382.3. a v vector v tch v hng .. 462.4. Kt hp cc bt ng thc c in .. 482.5. Tn dng tnh n diu ca hm s 572.6. Bi tp . 64


2/35



2.1. Bin i lng gic tng ng :

C th ni phngphp ny l mt phngphp xa nhTrit.N s dng cccng thc lng gic v sbin i qua li gia cc btng thc. c ths dng

tt phngphp ny bn c cn trang b cho mnh nhng kin thc cn thit vbin ilng gic (bn c c th tham kho thm phn 1.2. Cc ng thc,bt ng thctrong tamgic).

Thng thng th vi phng php ny, tas a btng thc cn chng minh vdng btng thc ng hay quen thuc. Ngoi ra, ta cng c ths dng hai ktququen thuc 1cos;1sin xx .

V d 2.1.1.

CMR :7

cos3

14sin2

14sin1

>

Li gii :

Ta c :

( )17

3cos

7

2cos

7cos

14sin2

14sin1

7

3cos

7

2cos

7

cos

14

sin2

14

5sin

14

7sin

14

3sin

14

5sin

14sin

14

3sin

14sin1

++=

++=

++=

Mt khc ta c :

( )27

cos7

3cos

7

3cos

7

2cos

7

2cos

7cos

7

2cos

7

4cos

7cos

7

5cos

7

3cos

7cos

2

1

7cos

++=

+++++=

t7

3cos;

7

2cos;

7cos

=== zyx

Khi t ( ) ( )2,1 ta c bt ng thc cn chng minh tng ng vi :

( ) ( )33 zxyzxyzyx ++>++

m 0,, >zyx nn :

( ) ( ) ( ) ( ) ( )403 222 >++ xzzyyx


3/35



V zyx ,, i mt khc nhau nn ( )4 ng pcm.

Nhv y, vi cc btng thc nh trn th vic bin i lng gic l quytnhsng cn vi vic chng minh btng thc. Sau khi s dng cc bin i th vicgiiquyt btng thc trnn d dng thm ch l hin nhin (!).

V d 2.1.2.

CMR : ( )xbcxcaxabcba sin2cos3sin2222 +++

Li gii :

Bt ng thc cn chng minh tng ng vi :( ) ( ) ( )

( )( )

( ) ( ) 0cos2sinsin2cos

0coscos2sin22sin

sin22cos2sin2cos2sin2cossin22cos2

cos2sin2cossin2cossin2cos2sin

22

2222

22222

2222222

+

++

+++

+

++++++

xbxacxbxa

xbxxabxa

xbcxcaxxabcxbxaxbcxca

xxxxabcxxbxxa

Bt ng thc cui cng lun ng nn ta c pcm.

V d 2.1.3.

CMR vi ABC btk ta c :

4

9sinsinsin 222 ++ CBA

Li gii :

Bt ng thc cn chng minh tng ng vi :

( )

( )

( )( ) 0sin

4

1

2

coscos

04

1coscoscos

04

12cos2cos

2

1cos

4

9

2

2cos1

2

2cos1cos1

2

2

2

2

2

+

+

+++

+

+

CBCB

A

CBAA

CBA

CBA

pcm.ng thc xy ra khi v ch khi ABC u.


4/35



V d 2.1.4.

Cho ( )Zkk +

2

,, l bagc tha 1sinsinsin 222 =++ . CMR :

222

2

tantantan213

tantantantantantan

++

Li gii :

Ta c :

222222222

222

222

222

tantantan21tantantantantantan

2tan1

1

tan1

1

tan1

1

2coscoscos

1sinsinsin

=++

=+

++

++

=++

=++

Khi bt ng thc cn chng minh tng ng vi :

( ) ( ) ( ) 0tantantantantantantantantantantantan

tantantantantantan3

tantantantantantan

222

222222

2

++

++

++

pcm.

ng thc xy ra

tantantan

tantantantan

tantantantan

tantantantan

==

=

=

=

V d 2.1.5.

CMR trong ABC btk ta c :

++++

2tan

2tan

2tan3

2cot

2cot

2cot

CBACBA

Li gii :

Ta c :

2cot

2cot

2cot

2cot

2cot

2cot

CBACBA=++

t2

cot;2

cot;2

cotC

zB

yA

x === th

=++

>

xyzzyx

zyx 0,,



5/35



( )( )

( ) ( )( ) ( ) ( ) 0

3

3

1113

222

2

++

++++

++++

++++

xzzyyx

zxyzxyzyx

xyz

zxyzxyzyx

zyxzyx

pcm.ng thc xy ra CBA cotcotcot ==

CBA ==

ABC u.

V d 2.1.6.

CMR :xxx cos2

2sin31

sin31

+

+

+

Li gii :

V 1sin1 x v 1cos x nn :0sin3;0sin3 >>+ xx v 0cos2 >+

Khi bt ng thc cn chng minh tng ng vi :( ) ( )

( )

( )( ) 02cos1cos

04cos6cos2

cos1218cos612

sin92cos26

2

2

2

+

+

+

xx

xx

xx

xx

do 1cos x nn bt ng thc cui cng lun ng pcm.

V d 2.1.7.

CMR2

;3


6/35



do

zyx Khi theo AM GM th :

( )( )( ) ( )( )( )( )( )( )bacacbcbaxyz

zxyzxyxzzyyxabc +++==

+++=

8

222

8

( )3 ng pcm.

2.3 a v vector v tch v hng :

Phng php ny lun a ra cho bn c nhng li gii bt ng v th v. N ctrng cho skt hp hon gia i s v hnh hc. Nhng tnh chtca vectorli mang

n ligii thtsngsa v p mt. Nhng slng cc bi ton ca phngphp nykhng nhiu.

V d 2.3.1.


17/35



A

BC

e

e

e

1

2

3

O

A

B C

CMR trong mi tamgic ta c :

2

3coscoscos ++ CBA

Li gii :

Ly cc vector n v 321 ,, eee ln lt trn cc cnh CABCAB ,, .

Hin nhin ta c :

( )( ) ( ) ( )

( )

2

3coscoscos

0coscoscos23

0,cos2,cos2,cos23

0

133221

2

321

++

++

+++

++

CBA

CBA

eeeeee

eee

pcm.

V d 2.3.2.

Cho ABC nhn. CMR :

2

32cos2cos2cos ++ CBA

Li gii :

Gi O, G ln lt l tm ng trn ngoi tip v trng tm ABC .

Ta c : OGOCOBOA 3=++ Hin nhin :

( )( ) ( ) ( )[ ]

( )

2

32cos2cos2cos

02cos2cos2cos23

0,cos,cos,cos23

0

22

22

2

++

+++

+++

++

CBA

BACRR

OAOCOCOBOBOARR

OCOBOA

pcm.

ng thc xy ra ABCGOOGOCOBOA ==++ 00 u.

V d 2.3.3.


18/35



O

A

B C

Cho ABC nhn. CMR Rzyx ,, ta c :

( )2222

12cos2cos2cos zyxCxyBzxAyz ++++

Li gii :

Gi O l tm ng trn ngoi tip ABC .Ta c :

( )

( )222

222

222

2

2

12cos2cos2cos

02cos22cos22cos2

0.2.2.2

0

zyxCxyBzxAyz

BzxAyzCxyzyx

OAOCzxOCOByzOBOAxyzyx

OCzOByOAx

++++

+++++

+++++

++

pcm.

2.4. Kt hp cc bt ng thc c in :

Vni dung cng nhcch thc s dng cc btng thc chng ta bn chng1: Cc bc u cs. V th phn ny, taskhng nhc li m xt thm mt s vdphc tp hn, th v hn.

V d 2.4.1.

CMR ABC ta c :

2

39

2cot

2cot

2cot

2sin

2sin

2sin

++

++

CBACBA

Li gii :

Theo AM GM ta c :

3

2

sin

2

sin

2

sin

3

2sin

2sin

2sin

CBA

CBA

++

Mt khc :

2sin

2sin

2sin

2cos

2cos

2cos

2cot

2cot

2cot

2cot

2cot

2cot

CBA

CBA

CBACBA==++


19/35



( )

2sin

2sin

2sin

2cos2sin2cos2sin2cos2sin

2

3

2sin

2sin

2sin2

2cos

2sin

2cos

2sin

2cos

2sin

2sin

2sin

2sin

sinsinsin4

1

3

CBA

CCBBAA

CBA

CCBBAA

CBA

CBA

++

=

++

=

Suy ra :

( )12

cot2

cot2

cot2

92

sin

2

sin

2

sin

2cos

2sin

2cos

2sin

2cos

2sin

2sin

2sin

2sin

2

9

2cot

2cot

2cot

2sin

2sin

2sin

3

3

CBA

CBA

CCBBAACBA

CBACBA

=

++

++

m ta cng c : 332

cot2

cot2

cot CBA

( )22

3933

2

9

2cot

2cot

2cot

2

9 33 =CBA

T ( )1 v ( )2 :

2

39

2

cot

2

cot

2

cot

2

sin

2

sin

2

sin

++

++

CBACBA

pcm.

V d 2.4.2.

Cho ABC nhn. CMR :

( )( )2

39tantantancoscoscos ++++ CBACBA

Li gii :V ABC nhn nn CBACBA tan,tan,tan,cos,cos,cos u dng.

Theo AM GM ta c : 3 coscoscos3

coscoscosCBA

CBA

++

CBA

CBACBACBA

coscoscos

sinsinsintantantantantantan ==++


20/35



( )

CBA

CCBBAA

CBA

CCBBAA

CBA

CBA

coscoscos2

cossincossincossin

2

3

coscoscos2

cossincossincossin

coscoscos

2sin2sin2sin4

1

3

++=

++

=

Suy ra :

( )( )

( )1tantantan2

9

coscoscos

cossincossincossincoscoscos

2

9tantantancoscoscos

3

3

CBA

CBA

CCBBAACBACBACBA

=

++++

Mt khc : 33tantantan CBA

( )22

3933

2

9tantantan

2

9 33= CBA

T ( )1 v ( )2 suy ra :

( )( )2

39tantantancoscoscos ++++ CBACBA

pcm.

V d 2.4.3.

Cho ABC ty. CMR :

34

2tan

1

2tan

2tan

1

2tan

2tan

1

2tan

++

++

+C

C

B

B

A

A

Li gii :

Xt ( )

=

2;0tan

xxxf

Khi : ( ) =xf ''

Theo Jensen th : ( )132

tan2

tan2

tan ++CBA

Xt ( )

= 2;0cot

xxxg

V ( ) ( )

>+=

2;00cotcot12'' 2

xxxxg

Theo Jensen th : ( )2332

cot2

cot2

cot ++CBA

Vy ( ) ( )+ 21 pcm.


21/35



V d 2.4.4.

CMR trong mi tamgic ta c :3

3

21

sin

11

sin

11

sin

11

+

+

+

+

CBA

Li gii :

Ta s dng b sau :B : Cho 0,, >zyx v Szyx ++ th :

( )12

11

11

11

1

3

+

+

+

+

Szyx

Chng minh b :Ta c :

( ) ( )2111111111 xyzzxyzxyzyxVT +

+++

+++=

Theo AM GM ta c :

( )399111

Szyxzyx

++++

Du bng xy ra trong ( )3

3S

zyx ===

Tip tc theo AM GM th :33 xyzzyxS ++

( )4271

27 3

3

Sxyzxyz

S

Du bng trong ( )4 xy ra3

Szyx ===

Vn theo AM GM ta li c :

( )51

3111

3

2

++

xyzzxyzxy

Du bng trong ( )5 xy ra3

Szyx ===

T ( ) ( )54 suy ra :

( )627111

2Szxyzxy++

Du bng trong ( )6 xy ra ng thi c du bng trong ( ) ( )3

54S

zyx ===

T ( )( )( )( )6432 ta c :


22/35



( )3

32

31

2727911

+=+++

SSSSVT

B c chng minh. Du bng xy ra ng thi c du bng trong ( )( )( )643

3

Szyx ===

p dng vi 0sin,0sin,0sin >=>=>= CzByAx

m ta c2

33sinsinsin ++ CBA vy y

2

33=S

Theo b suy ra ngay :3

3

21

sin

11

sin

11

sin

11

+

+

+

+

CBA

Du bng xy ra2

3sinsinsin === CBA

ABC

u.

V d 2.4.5.

CMR trong mi tamgic ta c :

3plll cba ++

Li gii :

Ta c : ( ) ( ) ( )1222cos2 appcb

bc

bc

app

cb

bc

cb

A

bcla

+=

+=

+=

Theo AM GM ta c 12

+ cb

bcnn t ( )1 suy ra :

( ) ( )2appla

Du bng trong ( )2 xy ra cb = Hon ton tng t ta c :

( ) ( )

( ) ( )4

3

cppl

bppl

c

b

Du bng trong ( ) ( )43 tng ng xy ra cba ==

T ( )( ) ( )432 suy ra :

( ) ( )5cpbpapplll cba ++++ Du bng trong ( )5 xy ra ng thi c du bng trong ( )( ) ( ) cba ==432p dng BCS ta c :


23/35



( ) ( )( )63

332

pcpbpap

cbapcpbpap

++

++

Du bng trong ( )6 xy ra cba ==

T ( ) ( )65 ta c : ( )73plll cba ++

ng thc trong ( )7 xy ra ng thi c du bng trong ( ) ( ) cba ==65ABC u.

V d 2.4.6.

Cho ABC btk. CMR :

R

r

abc

cba 24

333

++

Li gii :

Ta c : ( )( )( )cpbpappprR

abcS ===

4

( )( )( ) ( )( )( )

( )( )( )abc

abccbacaacbccbabba

abc

cbabacacb

abc

cpbpap

pabc

cpbpapp

pabc

S

R

r

2

222222882

333222222

2

+++++=

+++=

=

==

abccba

ca

ac

bc

cb

ab

ba

abccba

Rr

333333

624++

++++++

++=

pcm.

V d 2.4.7.

Cho ABC nhn. CMR :

abcbA

a

C

ca

C

c

B

bc

B

b

A

a27

coscoscoscoscoscos

+

+

+

Li gii :


CBABA

A

C

CA

C

C

B

BC

B

B

A

Asinsinsin27sin

cos

sin

cos

sinsin

cos

sin

cos

sinsin

cos

sin

cos

sin

+

+

+


24/35



27coscos

coscos1

coscos

coscos1

coscos

coscos1

sinsinsin27sincoscos

sinsin

coscos

sinsin

coscos

sin

AC

AC

CB

CB

BA

BA

CBABAC

BA

CB

AC

BA

C

t

+

=

+

=

+

=


25/35



Li gii :

Bt ng thc cn chng minh tng dng vi :

( )

( ) ( )cba

abccbacba

cba

abccbacba

+++++++

+++

++++

72935

2

435

36

2222

2

222

Theo BCS th : ( ) ( )2222 3 cbacba ++++ ( ) ( ) ( )1279 2222 cbacba ++++

Li c :

++

++

3 222222

3

3

3

cbacba

abccba

( )( )( )( )

( ) ( )2728

7289

222

222

222

cba

abccba

abccbacbaabccbacba

++++

++++

++++

Ly ( )1 cng ( )2 ta c :

( ) ( ) ( )

( ) ( )cba

abccbacba

cba

abccbacbacba

+++++++

++++++++++

72935

729827

2222

2222222

pcm.

V d 2.4.9.

CMR trong ABC ta c :

6

2sin

2cos

2sin

2cos

2sin

2cos

+

+

C

BA

B

AC

A

CB

Li gii :

Theo AM GM ta c :

( )1

2sin

2cos

2sin

2cos

2sin

2cos

3

2sin

2cos

2sin

2cos

2sin

2cos

3C

BA

B

AC

A

CB

C

BA

B

AC

A

CB

+

+


26/35



m :

( )( )( )CBA

BAACCB

CC

BABA

BB

ACAC

AA

CBCB

C

BA

B

AC

A

CB

sinsinsinsinsinsinsinsinsin

2sin

2cos2

2cos

2sin2

2sin

2cos2

2cos

2sin2

2sin

2cos2

2cos

2sin2

2sin

2cos

2sin

2cos

2sin

2cos

+++=

+

+

+

=

Li theo AM GM ta c :

+

+

+

ACAC

CBCB

BABA

sinsin2sinsin

sinsin2sinsin

sinsin2sinsin

( )( )( )

( )( )( )( )28

sinsinsin

sinsinsinsinsinsin

sinsinsin8sinsinsinsinsinsin

+++

+++

CBA

BAACCB

CBABAACCB

T ( )( )21 suy ra :

683

2sin

2cos

2sin

2cos

2sin

2cos

3=

+

+

C

BA

B

AC

A

CB

pcm.

V d 2.4.10.

CMR trong mi ABC ta c : 29sinsinsinsinsinsin

++

R

rACCBBA

Li gii :


2

2

2

36

9222222

9sinsinsinsinsinsin

rcabcab

raccbba

rACRCBRBAR

++

++

++

Theo cng thc hnh chiu :

+=

+=

+=

a

BArc

a

ACrb

a

CBra cot

2cot;cot

2cot;cot

2cot


27/35



+

++

+

+

++

+

+=++

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2

22

CBBAr

BAACr

ACCBrcabcab

Theo AM GM ta c :

( )1cotcotcot42

cot2

cot22

cot2

cot22

cot2

cot2

cot2

cot 2 BACACCBACCB

=

+

+

Tng t :

( )

( )3cotcotcot42

cot2

cot2

cot2

cot

2cotcotcot42

cot2

cot2

cot2

cot

2

2

ACBCBBA

CBABAAC

+

+

+

+

T ( )( )( )321 suy ra :

( )42

cot2

cot2

cot122

cot2

cot2

cot2

cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

2cot

3222 CBABAAC

BAACBAAC

+

++

+

+

++

+

+

Mt khc ta c : ( )5272

cot2

cot2

cot332

cot2

cot2

cot 222 CBACBA

T ( ) ( )54 suy ra : ( )6363.122

cot2

cot2

cot123 222 =CBA

T ( ) ( )64 suy ra pcm.

2.5. Tn dng tnh n iu ca hm s :

Chng ny khi c th bn c cn c kin thc cbn v o hm, khosthm sca chng trnh 12 THPT. Phngphp ny thc s c hiu qu trong cc bi btngthc lnggic. c ths dng tt phngphp ny th bn c cn n nhng kinhnghimgii ton cc phngphp nu cc phn trc.

V d 2.5.1.

CMR :

xx

2sin > vi

2;0

x

Li gii :


28/35



Xt ( )

2sin=

x

xxf vi

2;0

x

( )2

sincos'

x

xxxxf

=

Xt ( ) xxxxg sincos = vi

2;0

x

( ) ( )xgxxxxg

vi

2;0

Li gii :


( )

( ) 0cossin

cossin

3

1

3

1

>

>

xx

x

x

Xt ( ) ( ) xxxxf =

3

1

cossin vi

2;0 x

Ta c : ( ) ( ) ( ) 1cossin3

1cos' 3

42

3

2

=

xxxxf

( ) ( ) ( ) ( )

>+=

2;00cossin

9

4sin1cos

3

2'' 4

73

3

1 xxxxxxf

( )xf ' ng bin trong khong ( ) ( ) 00'' => fxf

( )xf cng ng bin trong khong ( ) ( ) => 00fxf pcm.

V d 2.5.3.

CMR nu a l gc nhn hay 0=a th ta c :1tansin 222 ++ aaa

Li gii :


29/35



p dng AM GM cho hai s dng asin2 v atan2 ta c :aaaaaa tansintansintansin

2222222+

=+

Nh vy ta ch cn chng minh : aaa 2tansin >+ vi2

0

+=

+=+=

2;00

cos

cos1cos1cos1

cos

1cos2cos2

cos

1cos'

22

23

2

x

x

xxx

x

xx

xxxf

( )xf ng bin trn khong ( ) ( )0faf > vi aaaa 2tansin2

;0 >+

12tansin22222

++= aaaa

1tansin 222 ++ aaa (khi 0=a ta c du ng thc xy ra).

V d 2.5.4.

CMR trong mi tamgic ta u c :

( ) CBACBABABABA coscoscoscoscoscos12

13coscoscoscoscoscos1 ++++++

Li gii :

Bt ng thc cn chng minh tng ng vi :( ) ( )CBABABABACBA coscoscos

6

131coscoscoscoscoscos2coscoscos21 ++++++

( ) ( CBABABABACBA coscoscos6

131coscoscoscoscoscos2coscoscos 222 ++++++++

( ) ( )CBACBA coscoscos6

131coscoscos

2+++++

6

13

coscoscos

1coscoscos

+++++

CBACBA

t

2

31coscoscos =

2

3;10

11'

2ng bin trn khong .

( ) =

6

13

2

3fxf pcm.


30/35



V d 2.5.5.

Cho ABC c chu vi bng 3. CMR :

( )2

222

4

13sinsinsin8sinsinsin3

RCBARCBA +++

Li gii :

Bt ng thc cn chng minh tng ng vi :( )( )( ) 13sin2sin2sin24sin4.3sin4.3sin4.3 222222 +++ CRBRARCRBRAR

134333222

+++ abccba

Do vai tr ca cba ,, l nh nhau nn ta c th gi s cba

Theo gi thit :2

3133 >+=++ ccccbacba

Ta bin i :

( )( )[ ]( )

( ) ( )

( ) ( )cabcc

cabcc

ababccc

abccabba

abccba

abccbaT

232333

322333

64333

4323

433

4333

22

22

22

22

222

222

+=

++=

++=

+++=

+++=

+++=

v 0230322

3>


31/35



Li gii :

Ta c :

( )( )

( )

( )( )

( )

( )( )

( )

p

cp

p

bp

p

apCBA

cpp

bpapC

bpp

apcpB

app

cpbpA

=

=

=

=

2tan

2tan

2tan

2tan

2tan

2tan

v( )( )( )

p

cp

p

bp

p

ap

p

cpbpapp

p

S

S

r

=

==

22

2

Do :2

tan2

tan2

tan2 CBA

S

r=

Mt khc :

( ) ( )

( )

( )

2cot

2cot

2cot

2cos

2sin

2sin

2sin

2cos

2cos

2cos

2cos

2tansinsinsin2

sinsinsin2

2tan

2tan2

CBA

A

A

CBA

CBA

AACBR

CBAR

Aacb

cba

Aap

cba

r

p

==

+

++=

+

++=

++=


33

28

2cot

2cot

2cot

2cot

2cot

2cot

1

33

28

2cot

2cot

2cot

2tan

2tan

2tan

+

+

CBA

CBA

CBACBA

t 332

cot2

cot2

cot = tCBA

t

Xt ( )t

ttf1

+= vi 33t

( ) 3301

1'2

>= tt

tf

( ) ( ) =+==33

28

33

13333min ftf pcm.

V d 2.5.7.


32/35



CMR vi mi ABC ta c :

( )( )( ) 233

38222 eRcRbRaR


33/35



( ) ( ) ( )[ ]

( ) ( ) 1coscoscoscos69

2cos2cos22

1coscos69

22

22

+++=

++++=

BACBAC

BABABAC

do ( ) 1cos BA

( )22

cos3coscos69 CCCP =+ m 0cos >C

( ) ( ) ( )CCCP 222 cos1cos3cos1 ++

Mt khc ta c :2

1cos600 0 < CC

Xt ( ) ( ) ( )22 13 xxxf += vi

1;

2

1x

( ) ( )( )( )

= 1;

2

1012132' xxxxxf

( )xf ng bin trn khong .( ) ( )( )( ) +++=

16

125cos1cos1cos1

16

125

2

1 222 CBAfxf pcm.

V d 2.5.9.

Cho ABC bt k. CMR :

( ) 32cotcot

sin

1

sin

12 +

+ CB

CB

Li gii :

Xt ( ) xx

xf cotsin

2= vi ( );0x

( ) ( )3

0'sin

cos21

sin

1

sin

cos2'

222

==

=+= xxf

x

x

xx

xxf

( ) 3cotsin

23

3max =

= x

xfxf

Thay x bi CB, trong bt ng thc trn ta c :

3cotsin

2

3cotsin

2

CC

BB pcm.


34/35



V d 2.5.10.

CMR :20

720sin

3

1 0


35/35


2.6.1. ( )2

5cos2cos2cos3 + BCA

2.6.2. 42cos322cos22cos3 ++ CBA

2.6.3. ( )( ) ( ) 542cos532cos2cos15 ++++ CBA

2.6.4. 342

tan2

tan2

tan ++ CBA vi ABC c mt gc 32

2.6.5.2222 4

1111

rcba++

2.6.6.cba r

c

r

b

r

a

r

abc 333++

2.6.7.( )( )( )

23

chuyen de bat dang thuc luong giac (2)

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