Lượng giác toàn tập

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 sñqAM = β vôùi 0 2≤ β ≤ π 
Ñaët k2 ,k Zα = β+ π ∈
Ta ñònh nghóa: 
sin OKα = 
cos OHα = 
sintg
cos
αα = α vôùi cos 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: α vaø −α 
( )sin sin−α = − α 
( )cos cos−α = α 
( ) ( )tg tg−α = − α 
( ) ( )cot g cot g−α = − α 
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b. Buø nhau: α vaø 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 π : vaø α π + 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: 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⎜ ⎟⎝ ⎠
e.Sai nhau 
2
π
: 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⎜ ⎟⎝ ⎠
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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 sinacosb sin bcosa
cos a b cosacosb sinasin b
tga tgbtg a b
1 tgatgb
± = ±
± =
±± =
∓
∓
VI. Coâng thöùc nhaân ñoâi 
=
= − = − =
= −
−=
2 2 2 2
2
2
sin2a 2sinacosa
cos2a cos a sin a 1 2sin a 2cos a 1
2tgatg2a
1 tg a
cot g a 1cot g2a
2cot ga
−
VII. Coâng thöùc nhaân ba: 
3
3
sin3a 3sina 4sin a
cos3a 4cos 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 k2≠ π + π
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22
2
2
2tsina
1 t
1 tcosa
1 t
2ttga
1 t
= +
−= +
= −
X. Coâng thöùc bieán ñoåi toång thaønh tích 
( )
( )
a b a bcosa cosb 2cos cos
2 2
a b a bcosa cosb 2sin sin
2 2
a b a bsina sin b 2cos sin
2 2
a b a bsina sin b 2cos sin
2 2
sin a b
tga tgb
cosacosb
sin b a
cot ga cot gb
sina.sin b
+ −+ =
+ −− = −
+ −+ =
+ −− =
±± =
±± =
XI. Coâng thöùc bieån ñoåi tích thaønh toång 
( ) ( )
( ) ( )
( ) ( )
1cosa.cosb cos a b cos a b
2
1sina.sin b cos a b cos a b
2
1sina.cosb 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
+ − = + − +
= + − −
= − − −
= −
−
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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 
( )2
2
1 cosx1 cosxA 1
sin 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 2cosx 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 sinx 0≠ ) 
Ta coù: 2 2
1 3sin x 1 cos x 1
4 4
= − = − = 
Do: x
2
π 
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. A = − 4 4 2 2 22cos x sin x sin xcos x 3sin x+ +
 b. 
2 cot gx 1
tgx 1 cot gx 1
+
− −B = + 
a. Ta coù: 
4 4 2 2A 2cos x sin x sin xcos x 3sin x= − + + 2 
( ) ( ) ( )
( )
24 2 2 2 2
4 2 4 2 4
A 2cos x 1 cos x 1 cos x cos x 3 1 cos x
A 2cos x 1 2cos x cos x cos x cos x 3 3cos x
⇔ = − − + − + −
⇔ = − − + + − + − 2
A 2⇔ = (khoâng phuï thuoäc x) 
b. Vôùi ñieàu kieän sinx.cosx 0,tgx 1≠ ≠ 
Ta coù: 
2 cot gxB
tgx 1 cot gx 1
1+= +− − 
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1 1
2 2tgxB 1tgx 1 tgx 1 1 tgx1
tgx
+ +⇔ = + = +− −−
1 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 bcot g c
sin c sin b
= − − 
( ) ( )2 2 2 2 2cot g b 1 cot g c 1 cot g b cot g bcot g c 1= + − + − = − (1) 
* 
( )2
2
1 cosa1 cosa 1
2sina sin a
⎡ ⎤−+ −⎢ ⎥⎢ ⎥⎣ ⎦
( )2
2
1 cosa1 cosa 1
2sina 1 cos a
⎡ ⎤−+= −⎢ ⎥−⎢ ⎥⎣ ⎦
1 cosa 1 cosa1
2sina 1 cosa
+ −⎡ ⎤= −⎢ ⎥+⎣ ⎦ 
1 cosa 2cosa. c
2sina 1 cosa
+= =+ ot ga (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= = +
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AÙp duïng baát ñaúng thöùc Cauchy cho ba soá döông tg ta ñöôïc A,tgB,tgC
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 : vaø 
∈
=
\x
y 3Max
∈
=
\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 
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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
2
− − ≥ 0 t 1,1−∀ ∈ 
⇔ ( ) 2g t t 2mt 2 0= + − ≤ [ ]t 1,1−
t
∀ ∈ 
Do ∀ neân g(t) coù 2 nghieäm phaân bieät t2' m 2 0Δ = + > m 1, t2
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 
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Maët khaùc : ( )24 4 2 2 2sin cos sin cos 2sin cosα + α = α + α − α α2 
 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
3π π⎛ ⎞ ⎛= − + −⎜ ⎟ ⎜⎝ ⎠ ⎝
⎞⎟⎠ 
 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 .cos40cos10= 
⇔ ( )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
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⇔ 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 8
4
π = (hieån nhieân ñuùng) 
Baøi :12 : Chöùng minh : 
 a/ 2 2 22 2cos x cos x cos x
3 3
π π⎛ ⎞ ⎛ ⎞ 3
2
+ + + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 
 b/ 1 1 1 1 cot gx cot g16x
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 sinasin b
−− = − = 
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( )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
−− = = 
( ) ( )sin 16x 8x 1cot g8x cot g16x 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 : + =3 0 2 08sin 18 8sin 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 
⇔ 8sin3180 + 8sin2180 – 1 = 0 (do 0 < sin180 < 1) 
Caùch khaùc : 
Chia 2 veá cuûa (1) cho ( sin180 – 1 ) ta coù 
 ( 1 ) ⇔ 8sin2180 ( sin180 + 1 ) – 1 = 0 
Baøi 14 : Chöùng minh : 
 a/ ( )4 4 1sin x cos x 3 cos4x
4
+ = + 
 b/ ( )1sin6x cos6x 5 3cos4x
8
+ = + 
 c/ ( )8 8 1sin x cos x 35 28cos4x cos8x
64
+ = + + 
a/ Ta coù: ( )24 4 2 2 2sin x cos x sin x cos x 2sin x cos x+ = + − 2 
 221 sin 2
4
= − x 
 (  ... ⎪⎩ ⎩
3
3
⇔ =
π⇔ + =
tgB cotgC
B C
2
 ABC⇔ Δ vuoâng taïi A. 
Baøi 213: Cho ABCΔ coù: sin 2A sin 2B 4sin A.sin B+ = 
 Chöùng minh ABCΔ vuoâng. 
Ta coù: + =sin 2A sin 2B 4sin A.sin B
[ ]
[ ]
⇔ + − = − + − −
⇔ + = − + −
2sin(A B) cos(A B) 2 cos(A B) cos(A B)
cos(A B) 1 sin(A B) cos(A B)
[ ]⇔ − = − −cosC 1 sin C cos(A B) 
⇔ − + = − −2cosC(1 sin C) (1 sin C).cos(A B) 
⇔ − + = −2cosC(1 sin C) cos C.cos(A B) 
⇔ = − + = −cosC 0hay (1 sinC) cosC.cos(A B) (*) 
⇔ =cosC 0 
( Do neân sinC 0> (1 sinC) 1− + < −
Maø co .Vaäy (*) voâ nghieäm.) sC.cos(A B) 1− ≥ −
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Do ñoù ABCΔ vuoâng taïi C 
III. TAM GIAÙC CAÂN 
Baøi 214:Chöùng minh neáu ABCΔ coù CtgA tgB 2cotg
2
+ = 
 thì laø tam giaùc caân. 
Ta coù: CtgA tgB 2cotg
2
+ = 
C2cossin(A B) 2
CcosA.cosB sin
2
C2cossinC 2
CcosA.cosB sin
2
C C C2sin cos 2cos
2 2
CcosA cosB sin
2
+⇔ =
⇔ =
⇔ = 2
 ⇔ 2 C Csin cosA.cosB do cos 0
2 2
⎛ ⎞= >⎜ ⎟⎝ ⎠ 
( ) ( ) (
( )
( )
⇔ − = + + −⎡ ⎤⎣ ⎦
⇔ − = − + −
⇔ − =
⇔ =
1 11 cosC cos A B cos A B
2 2
1 cosC cosC cos A B
cos A B 1
)
A B
 ABC⇔ Δ caân taïi C. 
Baøi 215: Chöùng minh ABCΔ caân neáu: 
 3 3A B Bsin .cos sin .cos
2 2 2 2
= A 
Ta coù: 3 3A B Bsin .cos sin .cos
2 2 2 2
= A 
2 2
A Bsin sin1 12 2
A A B Bcos cos cos cos
2 2 2 2
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⇔ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 (do Acos
2
> 0 vaø Bcos
2
>0 ) 
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2 2
3 3
2 2
A A B Btg 1 tg tg 1 tg
2 2 2 2
A B A Btg tg tg tg 0
2 2 2 2
A B A B A Btg tg 1 tg tg tg .tg 0 (*)
2 2 2 2 2 2
⎛ ⎞ ⎛ ⎞⇔ + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⇔ − + − =
⎛ ⎞ ⎡ ⎤⇔ − + + + =⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦
 ⇔ =A Btg tg
2 2
 ( vì 2 2A B A B1 tg tg tg tg 0
2 2 2 2
+ + + > ) 
 ⇔ =A B 
 ABC⇔ Δ caân taïi C 
Baøi 216: Chöùng minh ABCΔ caân neáu: 
 ( )2 2 2 22 2cos A cos B 1 cotg A cotg B (*)sin A sin B 2+ = ++ 
Ta coù: 
 (*) 
2 2
2 2 2 2
cos A cos B 1 1 1 2
sin A sin B 2 sin A sin B
+ ⎛ ⎞⇔ = +⎜ ⎟+ ⎝ ⎠− 
2 2
2 2 2 2
cos A cos B 1 1 11
sin A sin B 2 sin A sin B
+ ⎛ ⎞⇔ + = +⎜ ⎟+ ⎝ ⎠ 
⎛ ⎞⇔ = +⎜ ⎟+ ⎝ ⎠2 2 2 2
2 1 1 1
2sin A sin B sin A sin B
( )⇔ = + 22 2 2 24 sin A sin B sin A sin B 
( )2 20 sin A sin B
sin A sinB
⇔ = −
⇔ =
Vaäy ABCΔ caân taïi C 
Baøi 217: Chöùng minh ABCΔ caân neáu: 
 ( )Ca b tg atgA btgB (*)
2
+ = + 
Ta coù: ( )Ca b tg atgA btgB
2
+ = + 
( )⇔ + = +Ca b cotg atgA btgB
2
⎡ ⎤ ⎡⇔ − + −⎢ ⎥ ⎢⎣ ⎦ ⎣
C Ca tgA cotg b tgB cotg 0
2 2
⎤ =⎥⎦ 
+ +⎡ ⎤ ⎡⇔ − + −⎢ ⎥ ⎢⎣ ⎦ ⎣
⎤ =⎥⎦
A B Aa tgA tg b tgB tg 0
2 2
B 
− −
⇔ ++ + =
A B B Aa sin bsin
2 2 0A B A Bcos A.cos cosB.cos
2 2
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−⇔ = − =A B a bsin 0hay 0
2 cos A cosB
⇔ = =2R sin A 2Rsin BA Bhay
cos A cosB
⇔ = = ⇔ ΔA B hay tgA tgB ABC caân taïi C 
IV. NHAÄN DAÏNG TAM GIAÙC 
Baøi 218: Cho ABCΔ thoûa: a cosB bcosA asin A bsinB (*)− = − 
 Chöùng minh ABCΔ vuoâng hay caân 
 Do ñònh lyù haøm sin: a 2Rsin A,b 2RsinB= = 
 Neân (*) ( )2 22Rsin A cosB 2RsinBcos A 2R sin A sin B⇔ − = − 
( ) ( ) ( )
( ) [ ]
( ) ( ) ( )
( ) ( )
( ) ( )
2 2sin A cosB sinBcosA sin A sin B
1 1sin A B 1 cos2A 1 cos2B
2 2
1sin A B cos2B cos2A
2
sin A B sin A B sin B A
sin A B 1 sin A B 0
sin A B 0 sin A B 1
A B A B
2
⇔ − = −
⇔ − = − − −
⇔ − = −
⇔ − = − + −⎡ ⎤⎣ ⎦
⇔ − − + =⎡ ⎤⎣ ⎦
⇔ − = ∨ + =
π⇔ = ∨ + =
 vaäy ABCΔ vuoâng hay caân taïi C 
Caùch khaùc 
( )
− = −
⇔ − = + −
2 2sin A cosB sin Bcos A sin A sin B
sin A B (sin A sin B) ( sin A sin B)
( ) + − + −⇔ − = A B A B A B A Bsin A B (2sin cos ) (2 cos sin )
2 2 2 2
( ) ( ) ( )
( ) ( )
⇔ − = + −
⇔ − = ∨ + =
π⇔ = ∨ + =
sin A B sin A B sin A B
sin A B 0 sin A B 1
A B A B
2
Baøi 219 ABCΔ laø tam giaùc gì neáu 
 ( ) ( ) ( ) ( )2 2 2 2a b sin A B a b sin A B (*)+ − = − + 
Ta coù: (*) ( ) ( ) ( ) ( )2 2 2 2 2 2 24R sin A 4R sin B sin A B 4R sin A sin B sin A B⇔ + − = − + 
( ) ( ) ( ) ( )2 2sin A sin A B sin A B sin B sin A B sin A B 0⇔ − − + + − + +⎡ ⎤ ⎡⎣ ⎦ ⎣ =⎤⎦
=
( )2 22sin A cosA sin B 2sin Bsin A cosB 0⇔ − + 
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sin A cos A sin BcosB 0⇔ − + = (do sin vaø sin ) A 0> B 0>
sin2A sin2B
2A 2B 2A 2B
A B A B
2
⇔ =
⇔ = ∨ = π −
π⇔ = ∨ + =
Vaäy ABCΔ caân taïi C hay ABCΔ vuoâng taïi C. 
Baøi 220: ABCΔ laø tam giaùc gì neáu: 
2 2a sin2B b sin2A 4abcosA sinB (1)
sin2A sin2B 4sin A sinB (2)
⎧ + =⎨ + =⎩
Ta coù: 
(1) 2 2 2 2 2 24R sin Asin2B 4R sin Bsin2A 16R sin Asin BcosA⇔ + =
( )
2 2 2
2 2
sin A sin2B sin Bsin2A 4sin A sin BcosA
2sin A sinBcosB 2sin A cosA sin B 4sin A sin BcosA
sin A cosB sinBcosA 2sinBcosA (dosin A 0,sinB 0)
sin A cosB sinBcosA 0
sin A B 0
A B
⇔ + =
⇔ + =
⇔ + = >
⇔ − =
⇔ − =
⇔ =
2
>
Thay vaøo (2) ta ñöôïc 
 2sin2A 2sin A=
( )
22sin A cosA 2sin A
cosA sin A dosin A 0
tgA 1
A
4
⇔ =
⇔ = >
⇔ =
π⇔ =
Do ñoù ABCΔ vuoâng caân taïi C 
V. TAM GIAÙC ÑEÀU 
Baøi 221: Chöùng minh ABCΔ ñeàu neáu: 
 ( )bc 3 R 2 b c a (*)= + −⎡ ⎤⎣ ⎦ 
Ta coù:(*) ( ) ( ) ( )2RsinB 2RsinC 3 R 2 2RsinB 2RsinC 2Rsin A⇔ = + −⎡ ⎤⎣ ⎦ 
( ) ( )⇔ = + −2 3 sin BsinC 2 sin B sinC sin B C+ 
( )⇔ = + − −2 3 sin Bsin C 2 sin B sin C sin B cosC sin C cosB 
⎡ ⎤ ⎡⇔ − − + − −⎢ ⎥ ⎢⎣ ⎦ ⎣
1 3 1 32sin B 1 cosC sinC 2sinC 1 cosB sin B 0
2 2 2 2
⎤ =⎥⎦
⎡ π ⎤ ⎡ π ⎤⎛ ⎞ ⎛ ⎞⇔ − − + − − =⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦sin B 1 cos C sinC 1 cos B 0 (1)3 3 
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Do sin vaø B 0> 1 cos C 0
3
π⎛ ⎞− −⎜ ⎟⎝ ⎠ ≥ 
 sin vaø C 0> 1 cos B 0
3
π⎛ ⎞− −⎜ ⎟⎝ ⎠ ≥ 
Neân veá traùi cuûa (1) luoân 0≥
 Do ñoù, (1) 
cos C 1
3
cos B 1
3
⎧ π⎛ ⎞− =⎜ ⎟⎪⎪ ⎝ ⎠⇔ ⎨ π⎛ ⎞⎪ − =⎜ ⎟⎪ ⎝ ⎠⎩
 C B
3
π⇔ = = ⇔ ABCΔ ñeàu. 
Baøi 222: Chöùng minh ABCΔ ñeàu neáu 3 3 3
2
3sinBsinC (1)
4
a b ca (
a b c
⎧ =⎪⎪⎨ − −⎪ =⎪ − −⎩ 2)
Ta coù: (2) 3 2 2 3 3a a b a c a b c⇔ − − = − − 3
 ( )2 3a b c b c⇔ + = + 3 
 ( ) ( ) ( )2 2
2 2 2
a b c b c b bc c
a b bc c
⇔ + = + − +
⇔ = − +
2
c
 (do ñl haøm cosin) 2 2 2 2b c 2bc cosA b c b⇔ + − = + −
⇔ =
π⇔ = ⇔ =
2bc cos A bc
1cos A A
2 3
Ta coù: (1) 4sin BsinC 3⇔ =
( ) ( )⇔ − − +⎡ ⎤⎣ ⎦2 cos B C cos B C 3=
=
( )⇔ − +⎡ ⎤⎣ ⎦2 cos B C cos A 3 
( ) π⎛ ⎞ ⎛ ⎞⇔ − + = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
12 cos B C 2 3 do (1) ta coù A
2 3
( )⇔ − = ⇔ =cos B C 1 B C 
Vaäy töø (1), (2) ta coù ABCΔ ñeàu 
Baøi 223: Chöùng minh ABCΔ ñeàu neáu: 
 sin A sin B sinC sin 2A sin 2B sin 2C+ + = + +
Ta coù: ( ) ( )sin2A sin2B 2sin A B cos A B+ = + − 
 ( )2sinCcos A B 2sinC (1)= − ≤ 
Daáu “=” xaûy ra khi: ( )cos A B 1− = 
Töông töï: sin 2A sin 2C 2sin B+ ≤ (2) 
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Daáu “=” xaûy ra khi: ( )cos A C 1− = 
Töông töï: sin 2B sin 2C 2sin A+ ≤ (3) 
Daáu “=” xaûy ra khi: ( )cos B C 1− = 
Töø (1) (2) (3) ta coù: ( ) ( )2 sin2A sin2B sin2C 2 sinC sinB sinA+ + ≤ + + 
Daáu “=” xaûy ra 
( )
( )
( )
− =⎧⎪⇔ − =⎨⎪ − =⎩
cos A B 1
cos A C 1
cos B C 1
 ⇔ A = =B C 
 ⇔ ABCΔ ñeàu 
Baøi 224: Cho ABCΔ coù: 
 2 2 2
1 1 1 1 (*)
sin 2A sin 2B sin C 2cosA cosBcosC
+ + = 
 Chöùng minh ABCΔ ñeàu 
Ta coù: (*) ⇔ + +2 2 2 2 2 2sin 2B.sin 2C sin 2Asin 2C sin 2Asin 2B
( )
( )
sin2A.sin2B.sin2C sin2A sin2Bsin2C
2cosA cosBcosC
4sin A sinBsinC sin2A sin2Bsin2C
= ⋅
=
Maø: ( ) ( ) (= − − +⎡ ⎤⎣ ⎦4 sin A sin Bsin C 2 cos A B cos A B sin A B)+
)+
( )
( ) (
= − +⎡ ⎤⎣ ⎦
= + −
= + +
2 cos A B cosC sin C
2sin C cosC 2cos A B sin A B
sin 2C sin 2A sin 2B
Do ñoù,vôùi ñieàu kieän ABCΔ khoâng vuoâng ta coù 
 (*) 2 2 2 2 2 2sin 2Bsin 2C sin 2Asin 2C sin 2Asin 2B⇔ + +
( )
( ) ( )
= + +
= + +
⇔ − + −
2 2 2
2 2
sin 2A.sin 2B.sin 2C sin 2A sin 2B sin 2C
sin 2A sin 2Bsin 2C sin 2Bsin 2A sin 2C sin 2Csin 2A sin 2B
1 1sin 2Bsin 2A sin 2Bsin 2C sin 2A sin 2B sin 2A sin 2C
2 2
 ( )21 sin2Csin2A sin2Csin2B 0
2
+ − = 
sin 2Bsin2A sin2Bsin2C
sin2A sin2B sin2A sin2C
sin2A sin2C sin2Csin2B
=⎧⎪⇔ =⎨⎪ =⎩
=⎧⇔ ⎨ =⎩
sin 2A sin 2B
sin 2B sin 2C
A B C⇔ = = ABC⇔ ñeàu 
Baøi 225: Chöùng minh ABCΔ ñeàu neáu: 
a cosA bcosB ccosC 2p (*)
a sinB bsinC csin A 9R
+ + =+ + 
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Ta coù: a cos A bcosB c cosC+ +
( )
( ) ( )
( ) ( )
2Rsin A cosA 2RsinBcosB 2RsinCcosC
R sin2A sin2B sin2C
R 2sin A B cos A B 2sinCcosC
2RsinC cos A B cos A B 4RsinCsin A sinB
= + +
= + +
⎡ ⎤= + − +⎣ ⎦
⎡ ⎤= − − + =⎣ ⎦
Caùch 1: a sin B bsinC csin A+ + 
( )
( )2 2 23
2R sin A sinB sinBsinC sinCsin A
2R sin A sin Bsin C do bñtCauchy
= + +
≥ 
Do ñoù veá traùi : 3a cosA bcosB ccosC 2 sin AsinBsinC
asinB bsinC csin A 3
+ + ≤+ + (1) 
Maø veá phaûi: ( )+ += = + +2p a b c 2 sin A sin B sinC
9R 9R 9
32 sin A sinBsinC
3
≥ (2) 
Töø (1) vaø (2) ta coù 
( * ) ñeàu sin A sin B sinC ABC⇔ = = ⇔ Δ
Caùch 2: Ta coù: (*) 4Rsin AsinBsinC a b c
a sinB bsinC csin A 9R
+ +⇔ =+ + 
a b c4R
a b c2R 2R 2R
b c ca 9Ra b
2R 2R 2R
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ + +⎝ ⎠ ⎝ ⎠ ⎝ ⎠⇔ =⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 ( ) ( )9abc a b c ab bc ca⇔ = + + + + 
Do baát ñaúng thöùc Cauchy ta coù 
3
2 2 23
a b c abc
ab bc ca a b c
+ + ≥
+ + ≥
Do ñoù: ( ) ( )a b c ab bc ca 9abc+ + + + ≥
Daáu = xaûy ra a b c⇔ = = ABC⇔ Δ ñeàu. 
Baøi 226: Chöùng minh ABCΔ ñeàu neáu 
A ( )B Ccot gA cot gB cot gC tg tg tg *
2 2 2
+ + = + + 
Ta coù: 
( )sin A B sinCcot gA cot gB
sin A sinB sin A sinB
++ = = 
 2
sinC
sin A sinB
2
≥ +⎛ ⎞⎜ ⎟⎝ ⎠
 (do bñt Cauchy) 
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2 2 2
C C C2sin cos 2sin
2 2 2
A B A B C Asin .cos cos cos
2 2 2
= = B
2
+ − − 
 C2tg
2
≥ (1) 
Töông töï: Bcot gA cot gC 2tg
2
+ ≥ (2) 
 Acot gB cot gC 2tg
2
+ ≥ (3) 
Töø (1) (2) (3) ta coù 
 ( ) A B C2 cot gA cot gB cot gC 2 tg tg tg
2 2 2
⎛ ⎞+ + ≥ + +⎜ ⎟⎝ ⎠ 
Do ñoù daáu “=” taïi (*) xaûy ra 
− − −⎧ = =⎪⇔ ⎨⎪ = =⎩
=A B A C B Ccos cos cos 1
2 2 2
sin A sin B sin C
A B C
ABC ñeàu.
⇔ = =
⇔ Δ 
BAØI TAÄP 
1. Tính caùc goùc cuûa ABCΔ bieát: 
 a/ = + − 3cos A sin B sinC
2
 (ÑS: 2B C ,A
6 3
π π= = = ) 
 b/ sin 6 (ÑS: A sin 6B sin 6C 0+ + = A B C
3
π= = = ) 
 c/ sin5 A sin5B sin5C 0+ + =
2. Tính goùc C cuûa ABCΔ bieát: 
 a/ ( ) ( )1 cot gA 1 cot gB 2+ + =
 b/ 
2 2 9
A,Bnhoïn
sin A sin B sinC
⎧⎪⎨ + =⎪⎩
3. Cho ABCΔ coù: ⎧ + + <⎨ + + =⎩
2 2 2cos A cos B cos C 1
sin 5A sin 5B sin 5C 0
 Chöùng minh Δ coù ít nhaát moät goùc 36 0. 
4. Bieát . Chöùng minh 2 2 2sin A sin B sin C m+ + =
 a/ m thì 2= ABCΔ vuoâng 
 b/ m thì 2> ABCΔ nhoïn 
 c/ m thì 2< ABCΔ tuø. 
5. Chöùng minh ABCΔ vuoâng neáu: 
 a/ b ccosB cosC
a
++ = 
 b/ b c a
cosB cosC sinBsinC
+ = 
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 c/ sin A sin B sinC 1 cos A cosB cosC+ + = − + +
 d/ 
( ) ( )2
2
2 1 cos B Cb c
b 1 cos2B
⎡ ⎤− −− ⎣ ⎦= − 
6. Chöùng minh ABCΔ caân neáu: 
 a/ 
2 2
1 cosB 2a c
sinB a c
+ += − 
 b/ + + =+ −
sin A sin B sinC A Bcot g .cot g
sin A sin B sinC 2 2
 c/ 2tgA 2tgB tgA.tg B+ =
 d/ C Ca cot g tgA b tgB cot g
2 2
⎛ ⎞ ⎛− = −⎜ ⎟ ⎜⎝ ⎠ ⎝
⎞⎟⎠ 
 e/ ( ) C Bp b cot g ptg
2 2
− = 
 f/ ( )Ca b tg atgA btgB
2
+ = + 
7. ABCΔ laø Δ gì neáu: 
 a/ ( ) A BatgB btgA a b tg
2
++ = + 
 b/ c c cos2B bsin 2B= +
 c/ + +sin 3A sin 3B sin 3C 0=
 d/ ( ) ( )4S a b c a c b= + − + −
8. Chöùng minh ABCΔ ñeàu neáu 
 a/ ( )2 a cos A bcosB c cosC a b c+ + = + +
 b/ ( )= + +2 3 3 33S 2R sin A sin B sin C
 c/ sin A sin B sinC 4sin A sin BsinC+ + =
 d/ a b c
9Rm m m
2
+ + = vôùi laø 3 ñöôøng trung tuyeán a bm ,m ,mc
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