Chương 1: Công thức lượng giác

ậy : ( )∈ = =x Dmax y y 0 1, ( )∈ = = −x Dmin y y 1 1 
Bài 7: Cho hàm số 4 4y sin x cos x 2msin x cos= + − x 
Tìm giá trị m để y xác định với mọi x 
Xét 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
= − − 
Đặt : với t sin 2x= [ ]t 1,∈ − 1 
y xác định ⇔ 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 nên g(t) có 2 nghiệm phân biệt t1, t2 2' m 2 0Δ = + > m∀
Lúc đó t t1 t2 
 g(t) + 0 - 0 
Do đó : yêu cầu bài toán ⇔ 1 2t 1 1≤ − < ≤ 
⇔ ⇔ ( )( )
1g 1 0
1g 1 0
− ≤⎧⎪⎨ ≤⎪⎩
2m 1 0
2m 1 0
− − ≤⎧⎨ − ≤⎩
⇔ 
1m
2
1m
2
−⎧ ≥⎪⎪⎨⎪ ≤⎪⎩
 ⇔ 1 1m
2 2
− ≤ ≤ 
Cách khác : 
 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
Bài 8 : Chứng minh 4 4 4 43 5 7A sin sin sin sin
16 16 16 16 2
π π π π= + + + 3= 
Ta có : 7sin sin cos
16 2 16 16
π π π π⎛ ⎞= − =⎜ ⎟⎝ ⎠
π π π⎛ ⎞= − =⎜ ⎟⎝ ⎠
5 5sin cos cos
16 2 16 16
π3 
Mặt khác : ( )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 đó : 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
= − = 
Bài 9 : Chứng minh : o o o o16sin10 .sin 30 .sin50 .sin70 1= 
Ta có : 
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
= = = 
Bài 10 : Cho ABCΔ . Chứng minh : A B B C C Atg tg tg tg tg tg 1
2 2 2 2 2 2
+ + = 
Ta có : A B C
2 2
+ π
2
= − 
Vậy : 
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
+ + = 
Bài 11 : Chứng minh : ( )π π π π+ + + =8 4tg 2tg tg cot g *
8 16 32 32
Ta có : (*) ⇔ 8 cot g tg 2tg 4tg
32 32 16 8
π π π= − − − π
Mà : 
2 2cosa sina cos a sin acot ga tga
sina cosa sina cosa
−− = − = 
cos2a 2cot g2a1 sin2a
2
= = 
Do đó : 
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⇔ = (hiển nhiên đúng) 
4
Bài :12 : Chứng 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 có : 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 có : cosa cosb sin bcosa sina cosbcot ga cot gb
sina sin b sina sin b
−− = − = 
( )sin b a
sina sin b
−= 
Do đó : 
( ) ( )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
− = 
Lấy (1) + (2) + (3) + (4) ta được 
1 1 1 1cot gx cot g16x
sin2x sin4x sin8x sin16x
− = + + + 
Bài 13 : Chứng minh : 38sin 18 + =0 2 08sin 18 1 
Ta có: 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 vế của (1) cho ( sin180 – 1 ) ta có 
 ( sin180 + 1 ) – 1 = 0 
Bài 14 :
⇔ 8sin3180 + 8sin2180 – 1 = 0 (do 0 < sin18
Cách khác : 
 ( 1 ) ⇔ 8sin2180
 Chứng 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 có:
221 sin 2
4
= − x 
( )11 1 cos4
4
= − − x
 3 1 cos4x
4 4
= + 
b/ Ta có : 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 kết quả câu a ) 
3 5cos4x
8 8
= + 
( )+ = + −28 8 4 4 4sin x cos x sin x cos x 2sin x cos x 4c/ Ta có : 
( )= + −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
+ 
Bài 15 : Chứng minh : 3 3 3sin3x.sin x cos3x.cos x cos 2x+ = 
Cách 1: 
Ta có : 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= 
Cách 2 : 
Ta có : 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 ( bỏ dòng này cũng được) 4
3cos 2x= 
o o o o o 3 1cos12 cos18 4 cos15 .coBài 16 : s21 cos24
2
++ − = − Chứng minh : 
( )o o o ocos12 cos o8 4 cos15 cos21 cos24+ − 1Ta có : 
( )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
= − + 
Bài 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 có : 
( )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
= + − = 
 Bài 18 : Chứng minh : o o o o 8 3tg30 tg40 tg50 tg60 cos20
3
+ + + = o
( )sin a btga tgb
cosa cos b
++ = Áp dụng : 
Ta có : )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
= 
Bài 19 : Cho ABCΔ , Chứng 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 có : ( )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 có : 
( )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 nên ta có 
g A B tgC+ = − 
 a b C+ = π −
( ) t
tgA tgB tgC
1 tgAtgB
+ = −− ⇔ 
⇔ tgC tgA tgB tgC tgAtgB+ = − +
⇔
a có : 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
− = −+ (nhân tử và mẫu cho cotgA.cotgB) 
⇔ =
g/ Ta có : 
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
⇔ (nhân tử và mẫu 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
Bài 20 :
cot g cot g cot g cot g
2 2 2
+ + =
ABC . Chứng minh : Cho Δ
cos2A + cos2B + cos 2C + 4cosAcosBcosC + 1 = 0 
Ta có : (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 đó : cos2A + cos2B + cos2C + 1 + 4cosAcosBcosC = 0 
Bài 21 : ABCΔ Cho . Chứng minh : 
3A 3B 3C4sin sin sin
2 2
 cos3A + cos3B + cos3C = 1 - 
2
Ta có : (cos3A + cos3B) + cos3C 
23 32cos (A B)cos (A B) 1 2sin
2 2
= + − + − 3C
2
Mà : A B C+ = π − nên ( )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 đó : 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