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CHNG 1: CONG THC LNG GIAC

CHÖÔNG 1: COÂNG THÖÙC LÖÔÏNG GIAÙC I. Ñònh nghóa

Treân maët phaúng Oxy cho ñöôøng troøn löôïng giaùc taâm O baùn kính R=1 vaø ñieåm M treân ñöôøng troøn löôïng giaùc maø sñ �AM = β vôùi 0 2≤ β ≤ π Ñaët k2 ,k Zα = β+ π ∈ Ta ñònh nghóa: sin OKα = cos OHα =

sintg cos

αα = α vôùi co s 0α ≠ coscot g sin

αα = α vôùi sin 0α ≠ II. Baûng giaù trò löôïng giaùc cuûa moät soá cung (hay goùc) ñaëc bieät

Goùc α Giaù trò

( )o0 0 ( )o306π ( )o454π ( )o603π ( )o902π sinα 0 1

2 2

2

3 2

1

cosα 1 3 2

2

2

1 2

0

tgα 0 3 3

1 3 ||

cot gα || 3 1 3 3

0

III. Heä thöùc cô baûn

2 2sin cos 1α + α = 2

2

11 tg cos

+ α = α vôùi ( )k k Z2 πα ≠ + π ∈

2 2

1t cot g sin

+ = α vôùi ( )k k Zα ≠ π ∈ IV. Cung lieân keát (Caùch nhôù: cos ñoái, sin buø, tang sai π ; phuï cheùo)

a. Ñoái nhau: vaø −α α ( )sin sin−α = − α ( )cos cos−α = α

( ) ( )tg tg−α = − α ( ) ( )cot g cot g−α = − α

b. Buø nhau: vaø α π −α

( ) ( )

( ) ( )

sin sin

cos cos

tg tg

cot g cot g

π −α = α π−α = − α π−α = − α π−α = − α

c. Sai nhau : vaø π + π α α ( ) ( )

( ) ( )

sin sin

cos cos

tg t g

cot g cot g

π+ α = − α π+α = − α π+α = α π+α = α

d. Phuï nhau: vaø α 2 π −α

sin cos 2

cos sin 2

tg cot g 2

cot g tg 2

π⎛ ⎞− α = α⎜ ⎟⎝ ⎠ π⎛ ⎞− α = α⎜ ⎟⎝ ⎠ π⎛ ⎞− α = α⎜ ⎟⎝ ⎠

π⎛ ⎞− α = α⎜ ⎟⎝ ⎠

e.Sai nhau 2 π

: α vaø 2 π + α

sin cos 2

cos sin 2

tg cot g 2

cot g tg 2

π⎛ ⎞+ α = α⎜ ⎟⎝ ⎠ π⎛ ⎞+ α = − α⎜ ⎟⎝ ⎠ π⎛ ⎞+ α = − α⎜ ⎟⎝ ⎠

π⎛ ⎞+ α = − α⎜ ⎟⎝ ⎠

f.

( ) ( ) ( ) ( )

( ) ( )

+ π = − ∈ + π = − ∈

+ π = ∈ + π =

k

k

sin x k 1 sin x,k Z

cos x k 1 cosx,k Z

tg x k tgx,k Z

cot g x k cot gx

V. Coâng thöùc coäng

( ) ( )

( )

sin a b sin acos b sin b cosa

cos a b cosacos b sin asin b

tga tgbtg a b 1 tgatgb

± = ± ± =

±± = m

m

VI. Coâng thöùc nhaân ñoâi

= = − = − =

= − −=

2 2 2 2

2

2

sin2a 2sin acosa

cos2a cos a sin a 1 2sin a 2 cos a 1 2tgatg2a

1 tg a

cot g a 1cot g2a 2 cot ga

VII. Coâng thöùc nhaân ba:

3

3

sin3a 3sina 4sin a

cos3a 4 cos a 3cosa

= − = −

VIII. Coâng thöùc haï baäc:

( ) ( )

2

2

2

1sin a 1 cos2a 2 1cos a 1 cos2a 2

1 cos2atg a 1 cos2a

= −

= + −= +

IX. Coâng thöùc chia ñoâi

Ñaët at tg 2

= (vôùi a k ) 2≠ π + π

2

2

2

2

2tsin a 1 t 1 tcosa 1 t 2ttga

1 t

= + −= +

= −

X. Coâng thöùc bieán ñoåi toång thaønh tích

( ) ( )

a b a bcosa cos b 2cos cos 2 2

a b a bcosa cos b 2sin sin 2 2

a b a bsina sin b 2cos sin 2 2

a b a bsina sin b 2 cos sin 2 2

sin a b tga tgb

cosacos b sin b a

cot ga cot gb sina.sin b

+ −+ = + −− = −

+ −+ = + −− =

±± = ±± =

XI. Coâng thöùc bieån ñoåi tích thaønh toång

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

1cosa.cos b cos a b cos a b 2 1sina.sin b cos a b cos a b

2 1sina.cos b sin a b sin a b 2

= ⎡ + + − ⎤⎣ ⎦ −= ⎡ + − −⎣ ⎦

= ⎡ + + − ⎤⎣ ⎦

Baøi 1: Chöùng minh 4 4

6 6

sin a cos a 1 2 sin a cos a 1 3

+ − =+ − Ta coù:

( )24 4 2 2 2 2 2sin a cos a 1 sin a cos a 2sin acos a 1 2sin acos a+ − = + − − = − 2 Vaø: ( )( )

( )

6 6 2 2 4 2 2 4

4 4 2 2

2 2 2 2

2 2

sin a cos a 1 sin a cos a sin a sin acos a cos a 1

sin a cos a sin acos a 1

1 2sin acos a sin acos a 1

3sin acos a

+ − = + − + = + − − = − − − = −

Do ñoù: 4 4 2 2

6 6 2 2

sin a cos a 1 2sin acos a 2 sin a cos a 1 3sin acos a 3

+ − −= =+ − −

Baøi 2: Ruùt goïn bieåu thöùc ( )221 cosx1 cosxA 1sin x sin x ⎡ ⎤−+= = +⎢ ⎥⎢ ⎥⎣ ⎦

Tính giaù trò A neáu 1cosx 2

= − vaø x 2 π < < π

Ta coù: 2 2

2

1 cosx sin x 1 2 cosx cos xA sin x sin x

⎛ ⎞+ + − += ⎜ ⎟⎝ ⎠

( ) 2

2 1 cosx1 cosxA . sin x sin x

−+⇔ = ( )2 2

3 3

2 1 cos x 2sin x 2A sin x sin x sin x

−⇔ = = = (vôùi sin x 0≠ )

Ta coù: 2 2 1 3sin x 1 cos x 1 4 4

= − = − =

Do: x 2 π < < π neân sin x 0>

Vaäy 3sin x

2 =

Do ñoù 2 4 4A

sin x 33 = = = 3

Baøi 3: Chöùng minh caùc bieåu thöùc sau ñaây khoâng phuï thuoäc x: a. 4 4 2 2A 2cos x sin x sin x cos x 3sin x= − + + 2 b.

2 cot gxB tgx 1 cot gx 1

+= +− − 1

a. Ta coù:

4 4 2 2A 2cos x sin x sin x cos x 3sin x= − + + 2 ( ) ( ) ( ) ( )

24 2 2 2 2

4 2 4 2 4

A 2 cos x 1 cos x 1 cos x cos x 3 1 cos x

A 2 cos x 1 2 cos x cos x cos x cos x 3 3cos x

⇔ = − − + − + − ⇔ = − − + + − + − 2

A 2⇔ = (khoâng phuï thuoäc x) b. Vôùi ñieàu kieän sin x.cosx 0,tgx 1≠ ≠ Ta coù:

2 cot gxB tgx 1 cot gx 1

1+= +− −

1 1 2 2 1 tgxtgxB 1tgx 1 tgx 1 1 tgx1

tgx

+ +⇔ = + = +− −− −

( )2 1 tgx 1 tgxB 1

tgx 1 tgx 1 − − −⇔ = = = −− − (khoâng phuï thuoäc vaøo x)

Baøi 4: Chöùng minh

( )2 2 2 2 2 2 2 2

1 cosa1 cosa cos b sin c1 cot g bcot g c cot ga 1 2sina sin a sin bsin c

⎡ ⎤−+ −− + − =⎢ ⎥⎢ ⎥⎣ ⎦ −

Ta coù:

* 2 2

2 2 2 2

cos b sin c cot g b.cot g c sin b.sin c

− − 2

2 2 2 2

cotg b 1 cot g b cot g c sin c sin b

= − − ( ) ( )2 2 2 2 2cot g b 1 cot g c 1 cot g b cot g b cot g c= + − + − 1= − (1)

* ( )2

2

1 cosa1 cosa 1 2sin a sin a

⎡ ⎤−+ −⎢ ⎥⎢ ⎥⎣ ⎦

( )2 2

1 cosa1 cosa 1 2sin a 1 cos a

⎡ ⎤−+= −⎢ ⎥−⎢ ⎥⎣ ⎦

1 cosa 1 cosa1 2sin a 1 cosa + −⎡ ⎤= −⎢ ⎥+⎣ ⎦

1 cosa 2 cosa. cot ga 2sin a 1 cosa += =+ (2)

Laáy (1) + (2) ta ñöôïc ñieàu phaûi chöùng minh xong.

Baøi 5: Cho tuøy yù vôùi ba goùc ñeàu laø nhoïn. ABCΔ Tìm giaù trò nhoû nhaát cuûa P tgA.tgB.tgC=

Ta coù: A B C+ = π − Neân: ( )tg A B tgC+ = −

tgA tgB tgC 1 tgA.tgB

+⇔ =− − tgA tgB tgC tgA.tgB.tgC⇔ + = − +

Vaäy: P tgA.tgB.tgC tgA tgB tgC= = + + AÙp duïng baát ñaúng thöùc Cauchy cho ba soá döông tgA,tgB,tgC ta ñöôïc

3tgA tgB tgC 3 tgA.tgB.tgC+ + ≥

3P 3 P⇔ ≥ 3 2P 3

P 3 3

⇔ ≥ ⇔ ≥

Daáu “=” xaûy ra = =⎧ π⎪⇔ ⇔ =⎨ π< <⎪⎩

tgA tgB tgC A B C

30 A,B,C 2

= =

Do ñoù: MinP 3 3 A B C 3 π= ⇔ = = =

Baøi 6 : Tìm giaù trò lôùn nhaát vaø nhoû nhaát cuûa a/ 8 4y 2sin x cos 2x= + b/ 4y sin x cos= − x

a/ Ta coù : 4

41 cos2xy 2 cos 2x 2

−⎛ ⎞= +⎜ ⎟⎝ ⎠ Ñaët vôùi thì t cos2x= 1 t 1− ≤ ≤

( )4 41y 1 t 8

= − + t

=> ( )3 31y ' 1 t 4t 2

= − − + Ta coù : Ù ( ) y ' 0= 3 31 t 8t− = ⇔ 1 t 2t− = ⇔ 1t

3 =

Ta coù y(1) = 1; y(-1) = 3; 1 1y 3 2

⎛ ⎞ =⎜ ⎟⎝ ⎠ 7

Do ñoù : ∈

= �x y 3Max vaø

∈ =

�x

1yMin 27

b/ Do ñieàu kieän : sin vaø co neân mieàn xaùc ñònh x 0≥ s x 0≥

π⎡ ⎤= π + π⎢ ⎥⎣ ⎦D k2 , k22 vôùi ∈ �k Ñaët t cos= x x vôùi thì 0 t 1≤ ≤ 4 2 2t cos x 1 sin= = − Neân 4sin x 1 t= − Vaäy 8 4y 1 t= − − t treân [ ]D' 0,1= Thì ( )

−= − < − 3

748

ty ' 1 0 2. 1 t

[ )t 0; 1∀ ∈ Neân y giaûm treân [ 0, 1 ]. Vaäy : ( )∈ = =x Dmax y y 0 1, ( )∈ = = −x Dmin y y 1 1

Baøi 7: Cho haøm soá 4 4y sin x cos x 2msin x cos= + − x Tìm giaù trò m ñeå y xaùc ñònh vôùi moïi x

Xeùt 4 4f (x) sin x cos x 2msin x cos x= + − ( ) ( )22 2 2f x sin x cos x msin 2x 2sin x cos x= + − − 2 ( ) 21f x 1 sin 2x msin2x

2 = − −

Ñaët : vôùi t sin 2x= [ ]t 1,∈ − 1 y xaùc ñònh ⇔ x∀ ( )f x 0 x R≥ ∀ ∈ ⇔ 211 t mt 0

2 − − ≥ [ ]t 1,1−∀ ∈

⇔ ( ) 2g t t 2mt 2 0= + − ≤ [ ]t 1,∀ ∈ − 1

t

Do neân g(t) coù 2 nghieäm phaân bieät t1, t2 2' m 2 0Δ = + > m∀ Luùc ñoù t t1 t2

g(t) + 0 - 0 Do ñoù : yeâu caàu baøi toaùn ⇔ 1 2t 1 1≤ − < ≤ ⇔ ⇔ ( )( )

1g 1 0 1g 1 0

− ≤⎧⎪⎨ ≤⎪⎩ 2m 1 0

2m 1 0 − − ≤⎧⎨ − ≤⎩

⇔ 1m 2 1m 2

−⎧ ≥⎪⎪⎨⎪ ≤⎪⎩ ⇔ 1 1m

2 2 − ≤ ≤

Caùch khaùc : g t ( ) 2t 2mt 2 0= + − ≤ [ ]t 1,1−∀ ∈ { }

[ , ] max ( ) max ( ), ( ) t

g t g g ∈ −

⇔ ≤ ⇔ − ≤ 11

0 1 1 0

{ }max ), )m m⇔ − − − + ≤2 1 2 1 0⇔ 1m 2 1m 2

−⎧ ≥⎪⎪⎨ ⎪ ≤⎪⎩ m⇔− ≤ ≤1 1

2 2

Baøi 8 : Chöùng minh 4 4 4 43 5 7A sin sin sin sin 16 16 16 16 2 π π π π= + + + 3=

Ta coù : 7sin sin cos 16 2 16 16 π π π π⎛ ⎞= − =⎜ ⎟⎝ ⎠

π π π⎛ ⎞= − =⎜ ⎟⎝ ⎠ 5 5sin cos cos 16 2 16 16

π3 Maët khaùc : ( )24 4 2 2 2 2cos sin cos 2sin cosα + α = α + α − α αsin 2 21 2sin cos= − α α 211 sin 2

2 = − α

Do ñoù : 4 4 4 47 3A sin sin sin sin 16 16 16 16 π π π π= + + + 5

4 4 4 43 3sin cos sin cos 16 16 16 16 π π π⎛ ⎞ ⎛= + + +⎜ ⎟ ⎜⎝ ⎠ ⎝

π ⎞⎟⎠

2 21 11 sin 1 sin 2 8 2 8

π π⎛ ⎞ ⎛= − + −⎜ ⎟ ⎜⎝ ⎠ ⎝ 3 ⎞⎟⎠

2 21 32 sin sin 2 8 8

π π⎛ ⎞= − +⎜ ⎟⎝ ⎠

2 212 sin cos 2 8 8

π π⎛ ⎞= − +⎜ ⎟⎝ ⎠ π π=⎝ ⎠

3do sin cos 8 8

⎛ ⎞⎜ ⎟

1 32 2 2

= − =

Baøi 9 : Chöùng minh : o o o o16sin10 .sin 30 .sin50 .sin70 1= Ta coù :

o

o

A cos10 1A cos10 cos10

= = o (16sin10ocos10o)sin30o.sin50o.sin70o

⇔ ( )o oo1 1 oA 8sin 20 cos 40 .cos 202cos10 ⎛ ⎞= ⎜ ⎟⎝ ⎠ ⇔ ( )0 oo1 oA 4 sin 20 cos20 .cos 40cos10= ⇔ ( )o oo1A 2sin 40 cos40cos10= ⇔

o o

o o

1 cos10A sin 80 1 cos10 cos10

= = =

Baøi 10 : Cho ABCΔ . Chöùng minh : A B B C C Atg tg tg tg tg tg 1 2 2 2 2 2 2

+ + =

Ta coù : A B C 2 2 + π

2 = −

Vaäy : A B Ctg cot g 2 2 + =

⇔ A Btg tg 12 2

A B C1 tg .tg tg 2 2 2

+ =

⇔ A B C Atg tg tg 1 tg tg 2 2 2 2

⎡ ⎤+ = −⎢ ⎥⎣ ⎦ B 2

⇔ A C B C A Btg tg tg tg tg tg 1 2 2 2 2 2 2

+ + =

Baøi 11 : Chöùng minh : ( )π π π π+ + + =8 4tg 2tg tg cot g * 8 16 32 32

Ta coù : (*) ⇔ 8 cot g tg 2tg 4tg 32 32 16 8 π π π= − − − π

Maø :

2 2cosa sina cos a sin acot ga tga sina cosa sina cosa

−− = − = cos2a 2cot g2a1 sin2a 2

= =

Do ñoù :

cot g tg 2tg 4tg 8 32 32 16 8

π⎡⎢ π π π⎤− − − =⎥⎣ ⎦ (*) ⇔

2cot g 2tg 4tg 8 16 16 8 π π π⎡ ⎤− −⎢ ⎥⎣ ⎦ ⇔ =

4cot g 4tg 8⇔ 8 8 π π = −

8cot g 8π⇔ = (hieån nhieân ñuùng) 4

Baøi :12 : Chöùng minh :

2 2 22 2cos x cos x cos x 3 3 π π⎛ ⎞ ⎛ ⎞+ + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3 2

= a/ 1 1 1 1 cot gx cot g16x b/

sin2x sin4x sin8x sin16x + + + = −

a/ Ta coù : 2 2 22 2cos x cos x cos x 3 3 π π⎛ ⎞ ⎛+ + + −⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠ ( )1 1 4 1 41 cos2x 1 cos 2x 1 cos 2x

2 2 3 2 3 ⎡ π ⎤ ⎡ π ⎤⎛ ⎞ ⎛ ⎞= + + + + + + −⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

3 1 4 4cos2x cos 2x cos 2x 2 2 3 3

⎡ π π ⎤⎛ ⎞ ⎛ ⎞= + + + + −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ 3 1 4cos2x 2cos2x cos 2 2 3

π⎡ ⎤= + +⎢ ⎥⎣ ⎦ 3 1 1cos2x 2cos2x 2 2 2

⎡ ⎤⎛ ⎞= + + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ 3= 2

b/ Ta coù : cosa cosb sin bcosa sina cosbcot ga cot gb sina sin b sina sin b

−− = − = ( )sin b a

sina sin b −=

Do ñoù : ( ) ( )sin 2x x 1cot gx cot g2x 1

sin xsin2x sin2x −− = =

( ) ( )sin 4x 2x 1cot g2x cot g4x 2 sin2xsin4x sin4x

−− = =

( ) ( )sin 8x 4x 1cot g4x cot g8x 3 sin4xsin8x sin8x

−− = = ( ) ( )sincot g8x cot g16x− = 16x 8x 1 4

sin16xsin8x sin16x − =

Laáy (1) + (2) + (3) + (4) ta ñöôïc 1 1 1 1cot gx cot g16x

sin2x sin4x sin8x sin16x − = + + +

Baøi 13 : Chöùng minh : 38sin 18 + =0 2 08sin 18 1

Ta coù: sin180 = cos720 ⇔ sin180 = 2cos2360 - 1 ⇔ sin180 = 2(1 – 2sin2180)2 – 1 ⇔ sin180 = 2(1 – 4sin2180+4sin4180)-1 ⇔ 8sin4180 – 8sin2180 – sin180 + 1 = 0 (1 ) ⇔ (sin180 – 1)(8sin3180 + 8sin2180 – 1) = 0

0 < 1)

Chia 2 veá cuûa (1) cho ( sin180 – 1 ) ta coù ( sin180 + 1 ) – 1 = 0

Baøi 14 :

⇔ 8sin3180 + 8sin2180 – 1 = 0 (do 0 < sin18 Caùch khaùc :

( 1 ) ⇔ 8sin2180

Chöùng minh :

( ) a/ 4 4si + = 1n x cos x 3 cos4x 4

+

b/ ( )1sin6x cos6x 5 3cos4x 8

+ = +

c/ ( )8 8 1sin x cos x 35 28cos4x cos8x 64

+ = + + ( )24 4 2 2 2sin x cos x sin x cos x 2sin x cos x+ = + − 2a/ Ta coù:

221 sin 2 4

= − x

( )11 1 cos4 4

= − − x

3 1 cos4x 4 4

= + b/ Ta coù : sin6x + cos6x )( ) (2 2 4 2 2 4sin x cos x sin x sin x cos x cos x= + − +

( )4 4 21sin x cos x sin 2x4= + − ( )3 1 1cos4x 1 cos4x

4 4 8 ⎛ ⎞= + − −⎜ ⎟⎝ ⎠ ( do keát quaû caâu a ) 3 5cos4x 8 8

= + ( )+ = + −28 8 4 4 4sin x cos x sin x cos x 2sin x cos x 4c/ Ta coù :

( )= + −2 41 23 cos4x sin 2x 16 16

( ) ( )⎡ ⎤= + + − −⎢ ⎥⎣ ⎦ 2

21 1 19 6cos4x cos 4x 1 cos4x 16 8 2

( ) ( )29 3 1 1cos4x 1 cos8x 1 2cos4x cos 4x16 8 32 32= + + + − − + ( )= + + + − +9 3 1 1 1cos4x cos8x cos4x 1 cos8x

16 8 32 16 64

35 7 1cos4x cos8x 64 16

= + 64

+ Baøi 15 : Chöùng minh : 3 3 3sin3x.sin x cos3x.cos x cos 2x+ =

Caùch 1: Ta coù : 33 3sin3x.sin x cos3x.cos x cos 2x+ = ( ) ( )3 3 3 33sin x 4sin x sin x 4 cos x 3cos x cos x= − + −

4 6 6 4s3sin x 4sin x 4cos x 3co x= − + − ( ) ( )4 4 6 63 sin x cos x 4 sin x cos x= − − − ( ) ( )2 2 2 23 sin x cos x sin x cos x= − + ( ) ( )2 2 4 2 2 44 sin x cos x sin x sin x cos x cos x− − + +

2 23cos2x 4 cos2x 1 sin x cos x⎡ ⎤= − + −⎣ ⎦ 213cos2x 4 cos2x 1 sin 2x

4 ⎛ ⎞= − + −⎜ ⎟⎝ ⎠

21cos2x 3 4 1 sin 2x 4

⎡ ⎤⎛ ⎞= − + −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ ( )2cos2x 1 sin 2x= − 3cos 2x=

Caùch 2 : Ta coù : 3 3sin3x.sin x cos3x.cos x+

3sin x sin3x 3cos x cos3xsin3x cos3x 4 4 − +⎛ ⎞ ⎛= +⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠ ( ) ( )2 23 1sin3xsin x cos3x cos x cos 3x sin 3x4 4= + + −

( )3 1cos 3x x cos6x 4 4

= − +

(1 3cos2x cos3.2x 4

= + ) ( )= + −31 3cos2x 4cos 2x 3cos2x ( boû doøng naøy cuõng ñöôïc) 4 3cos 2x=

o o o o o 3 1cos12 cos18 4 cos15 .coBaøi 16 : s21 cos24 2 ++ − = − Chöùng minh :

( )o o o ocos12 cos o8 4 cos15 cos21 cos24+ − 1Ta coù : ( )o o o o2cos15 cos3 2cos15 cos45 cos3= − + o

os3 2cos15 cos45 2cos15 cos3= − −

− +

o o o o o o2cos15 c o o2cos15 cos45= − ( )o ocos60 cos30=

3 1 2

= − +

Baøi 17 : Tính o 2 o 2 oP sin 50 sin 70 cos50 cos70= + − ( ) ( ) ( )= − + − − +o o o1 1 1P 1 cos100 1 cos140 cos120 cos202 2 2 oTa coù :

( )o o1 1 1P 1 cos100 cos140 cos202 2 2⎛ ⎞= − + − − +⎜ ⎟⎝ ⎠ o ( )o o 1 1P 1 cos120 cos20 cos204 2= − + − o

o o5 1P cos2 1 50 cos20 4 2 2 4

= + − =

Baøi 18 : Chöùng minh : o o o o 8 3tg30 tg40 tg50 tg60 cos20 3

+ + + = o

( )sin a btga tgb cosa cos b

++ = AÙp duïng : Ta coù : )o( ) (o o otg50 tg40 tg30 tg60+ + +

o o

o o o sin90 sin90

cos50 cos40 cos30 cos60 = + o

o o o

1 1 1sin40 cos40 cos30 2

= +

o o 2 2

sin80 cos30 = +

o o 1 12

cos10 cos30 ⎛ ⎞= +⎜ ⎟⎝ ⎠

o o

o o cos30 cos102 cos10 cos30

⎛ ⎞+= ⎜ ⎟⎝ ⎠

p o

o o s20 cos10 co4

cos10 cos30 =

o8 3 cos20 3

= Baøi 19 : Cho ABCΔ , Chöùng minh :

a/ A B CsinA sinB sinC 4cos cos cos 2 2

+ + = 2

A b/ B CcA cosB cosC 1 4sin sin sin 2 2 2

+ + = + so c/ sin 2A sin 2B sin 2C 4sin A sinBsinC+ + = d/ 2 2A 2cos cos B cos C 2cosA cosBcosC+ + = − e/ tgA tgB tgC tgA.tgB.tgC+ + = f/ =cot gA.cot gB cot gB.cot gC cot gC.cot gA 1+ + g/ + + =A B C A Bcot g cot g cot g cot g .cot g .cot g

2 2 2 C

2 2

2

a/ Ta coù : ( )A B A BsinA sinB sinC 2sin cos sin A B 2 2 + −+ + = + +

A B A B A B2sin= cos cos 2 2 2 + − +⎛ ⎞+⎜ ⎟⎝ ⎠

+ π⎛ ⎞= =⎜ ⎟⎝ ⎠ C A B A B C4cos cos cos do 2 2 2 2 2 2

b/ Ta coù :

( )A B A BcosA cosB cosC 2cos cos cos A B 2 2 + −+ + = − +

2A B A B A B2cos cos 2cos 1 2 2 2 + − +⎛ ⎞= − ⎜ ⎟⎝ ⎠ −

A B A B A B2cos cos cos 1 2 2 2 + − +⎡ ⎤= −⎢ ⎥⎣ ⎦ + A B A B4cos sin sin 1 2 2 2 + ⎛ ⎞− +⎜ ⎟⎝ ⎠ = −

C A B4sin sin sin 1 2 2 2

= + ( ) ( )sin2A sin2B sin2C 2sin A B cos A B 2sinCcosC+ = + − + c/

= − +2sinCcos(A B) 2sinCcosC = − −2sinC[cos(A B) cos(A B) ] +

d/ 2

= − −4sinCsinAsin( B) = 4sinCsin A sinB

+ +2 2cos A cos B cos C ( ) 211 cos2A cos2B cos C

2 = + + +

( ) ( ) 21 cos A B cos A B cos C= + + − + ( )1 B= cosC cos A− −⎡ ⎤⎣ ⎦ do ( )( )cos A B cosC+ = − cosC− ( ) ( )1 cosC cos A B cos A B= − − + +⎡ ⎤⎣ ⎦

1 2cosC.cosA.cosB= − e/ Do neân ta coù

g A B tgC+ = − a b C+ = π − ( ) t

tgA tgB tgC 1 tgAtgB

+ = −− ⇔ ⇔ tgC tgA tgB tgC tgAtgB+ = − + ⇔

a coù : cotg(A+B) = - cotgC tgA tgB tgC tgAtgBtgC+ + =

f/ T 1 tgAtgB cot gC⇔ tgA + tgB − = −

⇔ cot gA cot gB 1 cot gC cot gB cot gA

− = −+ (nhaân töû vaø maãu cho cotgA.cotgB)

⇔ = g/ Ta coù :

cot gA cot gB 1 cot gCcot gB cot gA cot gC− = − − ⇔ cot gA cot gB cot gBcot gC cot gA cot gC 1+ +

A B Ctg cot g 2 2 + =

⇔ A Btg tg C2 2 cot gA B 21 tg tg

2 2

+ =

A Bcot g cot g C2 2 cot gA B 2cot g .cot g 1 2 2

+ =

− .cotg B

2 A 2

⇔ (nhaân töû vaø maãu cho cotg )

⇔ A B A B C Ccot g 2 + cot g cot g cot g cot g cot g

2 2 2 2 2 = −

A B C A B⇔ C.cot g .cot g 2 2 2

Baøi 20 :

cot g cot g cot g cot g 2 2 2 + + =

ABC . Chöùng minh : Cho Δ cos2A + cos2B + cos 2C + 4cosAcosBcosC + 1 = 0

Ta coù : (cos2A + cos2B) + (cos2C + 1)

= 2 cos (A + B)cos(A - B) + 2cos2C = - 2cosCcos(A - B) + 2cos2C = - 2cosC[cos(A – B) + cos(A + B)] = - 4cosAcosBcosC

Do ñoù : cos2A + cos2B + cos2C + 1 + 4cosAcosBcosC = 0 Baøi 21 : ABCΔ Cho . Chöùng minh :

3A 3B 3C4sin sin sin 2 2

cos3A + cos3B + cos3C = 1 - 2

Ta coù : (cos3A + cos3B) + cos3C 23 32cos (A B)cos (A B) 1 2sin

2 2 = + − + − 3C

2

Maø : A B C+ = π − neân ( )3 3A B 2 2

+ = π − 3C 2

=> ( )3cos A B cos+ = 3 3C 2 2 2

π⎛ ⎞−⎜ ⎟⎝ ⎠ 3Ccos

2 2 π⎛ ⎞= − −⎜ ⎟⎝ ⎠ 3Csin 2

= − Do ñoù : cos3A + cos3B + cos3C

( ) 23 A B3C 3C2sin cos 2sin 1 2 2 2

−= − − + ( )3 A B3C 3C2sin cos sin 1

2 2 2 −⎡ ⎤= − + +⎢ ⎥⎣ ⎦

( ) ( )3 A B3C 32sin cos cos A B 1 2 2 2

= − − +⎢⎣ −⎡ ⎤ +⎥⎦

−= +3C 3A 3B4sin sin sin( ) 1 2 2 2

3C 3A 3B4sin sin sin 1 2 2 2

= − +

Baøi 22 : A, B, C laø ba goùc cuûa moät tam giaùc. Chöùng minh : sinA sinB sinC A B Ctg tg cot g

cosA cosB cosC 1 2 2 2 + − =+ − +

2

A B A B C C2sin cos 2sin cossinA sinB sinC 2 2 2 A B A B CcosA cosB cosC 1 2cos cos 2sin 2 2 2

2 + − −+ − = + −+ − + +

Ta coù :

C A B C A B A2cos cos sin cos cosC2 2 2 2 2cot g .

B

A B AC A B C 2 cos cos2sin cos sin 2 22 2 2

−⎡ ⎤

B

− +− −⎢ ⎥⎣ ⎦= = − +−⎡ ⎤ ++⎢ ⎥⎣ ⎦

A B2sin C 2 2

− .sin cot g . A B2 2cos .cos

2 2

⎛ ⎞−⎜ ⎟⎝ ⎠=

C A Bcot g .tg .tg 2 2

= 2

Baøi 23 : Cho ABCΔ h : . Chöùng min

A B C B C A C A Bsin cos cos sin cos cos sin cos cos 2 2 2 2 2 2 2 2 2

+ +

( )A B C A B B C A Csin sin sin gtg tg tg t tg tg * 2 2 2 2 2 2 2 2 2

= + + +

ATa coù : B C 2 2 2 + π= − vaäy A B Ctg cot g

2 2 2 ⎛ ⎞+ =⎜ ⎟⎝ ⎠

⇔ A Btg tg 12 2

A B C1 tg tg tg 2 2 2

+ =

⇔ A B C Atg tg tg 1 tg tg 2 2 2 2

⎡ ⎤+ = −⎢ ⎥⎣ ⎦ B 2

⇔ ( )A C B C A Btg tg tg tg tg tg 1 1 2 2 2 2 2 2

+ + =

A B C B C Ac sin cos cos C A Bsin os cos sin cos cos 2 2 2 2 2 2 2 2 2

+ + Do ñoù : (*) Ù A B Csin sin sin 1 2 2 2

= + (do (1)) A B C B C A B C C Bsin 2

⇔ cos cos sin sin cos sin cos sin cos 1 2 2 2 2 2 2 2 2 2

⎡ ⎤ ⎡ ⎤− + + =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⇔ A B C A B Csin cos cos sin 1 2 2 2 2

+ ++ =

⇔ A B Csin 1 2

+ + = π⇔ =sin 1 2

( hieån nhieân ñuùng)

Baøi 24 : ( )A B C 3 cosA cosB cosCtg tg tg * 2 2 2 sinA sinB sinC

+ + ++ + = + + Chöùng minh : Ta coù :

2A B A B CcosA cosB cosC 3 2cos cos 1⎡ 2sin 3 2 2 2 + − ⎤+ + = + +⎥⎣ ⎦ −⎢+

2C A B2sin cos 4 2s C 2 2 2

− in= + − C A B C2sin cos sin 4 2 2 2

−⎡ ⎤− +⎢ ⎥⎣ ⎦ = C A B A B2sin cos cos 4 2 2 2

− +⎡ ⎤− +⎢ ⎥⎣ ⎦ = C A Bin4sin sin .s 4 2 2 2

+ (1) = A B A BsinA sinB sinC 2sin cos sinC 2 2 + −+ + = +

C A B C2cos cos 2sin cos 2 2 2

C 2

−= + C A B A B2cos cos cos 2 2 2

− +⎡ ⎤= +⎢ ⎥⎣ ⎦ C A B

Töø (1) vaø (2) ta coù :

4cos cos cos 2 2 2

= (2)

(*) ⇔ A B C A B Csin sin sin sin sin sin 1 2 2 2 2 2 2 A B C A B Ccos cos cos cos co

+ s cos

2 2 2 2 2 2

+ + =

A B C B A C C A Bsin cos cos sin cos cos sin cos cos 2 2 2 2 2 2 2 2 2 ⎡ ⎤ ⎡ ⎤ ⎡+ +⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

⎤⎥⎦

A B Csin sin sin 1 2 2 2

= +

⇔ A B C B C A B C C Bsin cos cos sin sin cos sin cos sin cos 1 2 2 2 2 2 2 2 2 2 2 ⎡ ⎤ ⎡− + +⎢ ⎥ ⎢⎣ ⎦ ⎣

⎤ =⎥⎦ ⇔ A B C A+ B Csin .cos cos sin 1

2 2 2 2 ++ =

A⇔ B C 1 2

+ + ⎤ =⎢ ⎥⎣ ⎦ sin ⎡

⇔ sin π 1 2 = ( hieån nhieân ñuùng)

Baøi 25 : . Chöùng minh:

A B Csin sin sin 2 2 2 2B C C A A Bcos cos cos cos cos cos

2 2 2 2 2 2

+ + = ABCΔ Cho

Caùch 1 :

Ta coù :

A B A A Bsin sin sin cos sin cos 2 2 2 2 2

B C C A

B 2

B Ccos cos cos cos cos cos cos 2 2 2 2 2 2 2

+ + =

A

A B Asin cos BsinA sinB 2 2 +

1 A B C A B C2 cos cos cos cos cos cos 2 2 2 2 2 2

− + = =

−⎛ ⎞− ⎜ ⎟⎝ ⎠= = A BC A B coscos .cos 22 2

A B C Acos .cos .cos cos cos 2 2 2 2

B 2

Do ñoù : Veá traùi

A B C A B Acos sin cos cos2 2 2 A B A B A Bcos cos cos cos cos cos 2 2 2 2

B 2

2

−⎛ ⎞ − ++⎜ ⎟⎝ ⎠= + = 2

A B2cos cos 2 2 2A Bcos cos 2 2

= =

Caùch 2 :

B C A C A Bcos cos cos 2 2

B C C A A Bcos cos cos cos cos cos 2 2 2 2 2

+ + + = + + 2

2

Ta coù veá traùi

B C B C A C A Ccos cos sin sin cos cos sin sin 2 2 2 2 2 2 2

B C C Acos cos cos cos 2 2 2 2

− − = + 2

A B Acos cos sin sin 2 2 2

A Bcos cos 2 2

− +

B 2

B C A C A B3 g tg tg tg tg tg t 2 2 2 2 2 2

⎡ ⎤= − + +⎢ ⎥⎣ ⎦ Maø : A B B C A Btg tg tg tg tg tg 1

2 2 2 2 2 2 + + =

(ñaõ chöùng minh taïi b Do ñoù : Veá traùi = 3 – 1 = 2

Baøi 26 :

aøi 10 )

. Coù A B Ccot g ,cot g ,cot g 2 2

ABCΔ Cho 2

theo töù töï taïo caáp soá coäng.

A Ccot g .cot g 3 2 2

= Chöùng minh A B Ccot g ,cot g ,cot g 2 2

Ta coù : 2

laø caáp soá coäng

⇔ A C Bcot g cot g 2cot g 2 2 + =

2

⇔ +

= A Csin 2cos 2 2

B

A C Bsin sin sin 2 2 2

⇔ Bcos 2cos 2 2

B

A C Bsin sin sin 2 2 2

=

neân Bcos 0 2 >⇔ = +

1 2 A C A Csin sin cos 2 2 2

(do 0

⇔ A C A Ccos cos sin sin 2 2 2 2 2A Csin .sin

2 2

− ⇔ A Ccot g cot g 3=

2 2 =

Baøi 27 : ABCΔ Cho . Chöùng minh :

1 t+ 1 1 1 A B C A B Ctg tg tg cot g co g cot g sin A sinB sinC 2 2 2 2 2 2 2

⎡ ⎤+ = + + + + +⎢ ⎥⎣ ⎦ A B C A Bcot g cot g cot g cot g .cot g .cot g 2 2 2 2 2 + + = Ta coù : C

2 (Xem chöùng minh baøi 19g )

Maët khaùc : sin cos 2tg cot g cos sin sin2

α αα + α = + =α α α 1 A B C A B Ctg tg tg cotg cotg cotg 2 2 2 2 2 2 2 ⎡ ⎤+ + + + +⎢ ⎥⎣ ⎦ Do ñoù : 1 A B C 1 Acotg⎡ +⎢

B Ctg tg tg cotg cotg 2 2 2 2 2 2

⎡ ⎤ ⎤= + + + +⎢ ⎥ ⎥⎣ ⎦ ⎣ ⎦ 2 2 1 A A 1 B B 1 C Ctg cot g tg cot g tg cot g 2 2 2 2 2 2 2 2 2 ⎡ ⎤ ⎡ ⎤ ⎡= + + + + +⎢ ⎥ ⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎦ ⎣

⎤⎥⎦ 1 1 1

sinA sinB sinC = + +

BAØI TAÄP

1. Chöùng minh :

a/ 2 1cos cos 5 5 π π− =

2

b/ o o

o o cos15 sin15 3 cos15 sin15

+ =− 2 4 6cos cos cos 7 7 7 π π π+ + = c/ 1

2 −

d/ 3+ =3 3sin 2xsin6x cos 2x.cos6x cos 4x o o o otg20 .tg40 .tg60 .tg80 3= e/

π π π π+ + + =2 5 π3tg tg tg cos 6 9 18 3 3 9

8tgf/

7 2 3 4 5 6 7 1os .cos .cos .cos .cos .cos .cos

15 15 15 15 15 15 15 2 π π π π π π = c πg/

h/ tgx.tg x .tgπ⎡ ⎤−⎢ ⎥ x tg3x3 3 π⎡ ⎤+ =⎢ ⎥⎣ ⎦ ⎣ ⎦

k/ o o o otg20 tg40 3tg20 .tg40 3+ + = o o o 3sin 20 .sin 40 .sin 80e/

8 =

m/ o o o otg5 .tg55 .tg65 .tg75 1= ( )

2. Chöùng minh raèng neáu ( ) (x y 2k 1 k z 2 π+ ≠ + ∈⎪⎩ )

x y+

thì

sin x 2sin=⎧⎪⎨

sin( ) cos

ytg x y y

+ = − 2 3. Cho coù 3 goùc ñeàu nhoïn vaø A B C≥ ≥ ABCΔ

a/ Chöùng minh : tgA + tgB + tgC = tgA.tgB.tgC b/ Ñ Chöùng minh (p-1)(q-1)

aët tgA.tgB = p; tgA.tgC = q 4

4. Chöùng minh caùc bieåu thöùc khoâng phuï thuoäc x : a/

( ) ( )4 2 4 2 2 2A sin x 1 sin x cos x 1 cos x 5sin x cos x 1= + + + + + ( ) ( )8 8 6 6B 3 sin x cos x 4 cos x 2sin x 6sin x= − + − + b/ 4

c/ ( ) ( ) ( ) ( ) (2 2C cos x a sin x b 2cos x a sin x b sin a b= − + − − − − − ) 5. Cho , chöùng minh : ABCΔ

cosC cosBcota/ gB cot gC sinBcosA sinCcosA

+ = +

b/ 3 3 3 A B CC 3cos cos cos co 3A 3B 3Cs cos cos 2 2 2 2 2 2

= + sin A sin B sin+ + A B C B A CsinA sinB sic/ nC scos .co cos .cos 2 2 2 2

− −+ + + = C Acos .co B 2 2

−s+ otgAcotgB + cotgBcotgC + cotgC otgA = 1

s C 1 2cosA cosBcosC= − in3Asin(B- C)+ sin3Bsin(C- A)+ sin3Csin(A- B) = 0

6. Tìm giaù trò nhoû nhaát cuûa :

d/ c c e/ 2 2cos A cos B co+ + 2 f/ s

1 1y sin x cos x

= + vôùi 0 x 2 π< < a/

π= + +9y 4x sin x x

vôùi 0 x< < ∞ b/ 2y 2sin x 4sin x cos x 5= + + c/

7. Tìm giaù trò lôùn nhaát cuûa : a/ y sin x cos x cos x sin x= + b/ y = sinx + 3sin2x c/ 2y cos x 2 cos x= + −

TT luyện thi đại học CLC Vĩnh Viễn

Đơn vị chủ quản: CÔNG TY TNHH THƯƠNG MẠI ĐIỆN TỬ THIÊN THI
372/10 Điện Biên Phủ, Phường 17, Q.Bình Thạnh, HCM
giấy phép MXH: 102/GXN - TTĐT
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