Xem mẫu
- 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 ≤ β ≤ 2π
Ñaët α = β + k2π, k ∈ Z
Ta ñònh nghóa:
sin α = OK
cos α = OH
sin α
tgα = vôùi cos α ≠ 0
cos α
cos α
cot gα = vôùi sin α ≠ 0
sin α
II. Baûng giaù trò löôïng giaùc cuûa moät soá cung (hay goùc) ñaëc bieät
Goùc α ( )
0 0o π
( )
30 o
π
( )
45o
π
( )
60 o
π
( )
90 o
Giaù trò 6 4 3 2
sin α 0 1 2 3 1
2 2 2
cos α 1 3 2 1 0
2 2 2
tgα 0 3 1 3 ||
3
cot gα || 3 1 3 0
3
III. Heä thöùc cô baûn
sin 2 α + cos2 α = 1
1 π
1 + tg 2 α = 2
vôùi α ≠ + kπ ( k ∈ Z )
cos α 2
1
t + cot g 2 = vôùi α ≠ kπ ( k ∈ Z )
sin 2 α
IV. Cung lieân keát (Caùch nhôù: cos ñoái, sin buø, tang sai π ; phuï cheùo)
a. Ñoái nhau: α vaø −α
sin ( −α ) = − sin α
cos ( −α ) = cos α
tg ( −α ) = −tg ( α )
cot g ( −α ) = − cot g ( α )
- b. Buø nhau: α vaø π − α
sin ( π − α ) = sin α
cos ( π − α ) = − cos α
tg ( π − α ) = −tgα
cot g ( π − α ) = − cot gα
c. Sai nhau π : α vaø π + α
sin ( π + α ) = − sin α
cos ( π + α ) = −cosα
tg ( π + α ) = t gα
cot g ( π + α ) = cot gα
π
d. Phuï nhau: α vaø −α
2
⎛π ⎞
sin ⎜ − α ⎟ = cos α
⎝2 ⎠
⎛π ⎞
cos ⎜ − α ⎟ = sin α
⎝2 ⎠
⎛π ⎞
tg ⎜ − α ⎟ = cot gα
⎝2 ⎠
⎛π ⎞
cot g ⎜ − α ⎟ = tgα
⎝2 ⎠
π π
e.Sai nhau : α vaø + α
2 2
⎛π ⎞
sin ⎜ + α ⎟ = cos α
⎝2 ⎠
⎛π ⎞
cos ⎜ + α ⎟ = − sin α
⎝2 ⎠
⎛π ⎞
tg ⎜ + α ⎟ = − cot gα
⎝2 ⎠
⎛π ⎞
cot g ⎜ + α ⎟ = − tgα
⎝2 ⎠
- f.
sin ( x + kπ ) = ( −1) sin x, k ∈ Z
k
cos ( x + kπ ) = ( −1) cos x, k ∈ Z
k
tg ( x + kπ ) = tgx, k ∈ Z
cot g ( x + kπ ) = cot gx
V. Coâng thöùc coäng
sin ( a ± b ) = sin a cos b ± sin b cosa
cos ( a ± b ) = cosa cos b m sin asin b
tga ± tgb
tg ( a ± b ) =
1 m tgatgb
VI. Coâng thöùc nhaân ñoâi
sin 2a = 2sin a cosa
cos2a = cos2 a − sin 2 a = 1 − 2sin 2 a = 2 cos2 a − 1
2tga
tg2a =
1 − tg2 a
cot g2 a − 1
cot g2a =
2 cot ga
VII. Coâng thöùc nhaân ba:
sin 3a = 3sin a − 4sin 3 a
cos3a = 4 cos3 a − 3cosa
VIII. Coâng thöùc haï baäc:
1
sin 2 a = (1 − cos2a )
2
1
cos2 a = (1 + cos2a )
2
1 − cos2a
tg 2 a =
1 + cos2a
IX. Coâng thöùc chia ñoâi
a
Ñaët t = tg (vôùi a ≠ π + k 2 π )
2
- 2t
sin a =
1 + t2
1 − t2
cosa =
1 + t2
2t
tga =
1 − t2
X. Coâng thöùc bieán ñoåi toång thaønh tích
a+ b a−b
cosa + cos b = 2 cos cos
2 2
a+b a−b
cosa − cos b = −2sin sin
2 2
a+ b a−b
sin a + sin b = 2 cos sin
2 2
a+b a−b
sin a − sin b = 2 cos sin
2 2
sin ( a ± b )
tga ± tgb =
cosa cos b
sin ( b ± a )
cot ga ± cot gb =
sin a.sin b
XI. Coâng thöùc bieån ñoåi tích thaønh toång
1
cosa.cos b = ⎡ cos ( a + b ) + cos ( a − b ) ⎦
⎤
2⎣
−1
sin a.sin b = ⎡ cos ( a + b ) − cos ( a − b ) ⎦
⎤
2 ⎣
1
sin a.cos b = ⎡sin ( a + b ) + sin ( a − b ) ⎤
2⎣ ⎦
sin 4 a + cos4 a − 1 2
Baøi 1: Chöùng minh =
sin 6 a + cos6 a − 1 3
Ta coù:
sin 4 a + cos 4 a − 1 = ( sin 2 a + cos2 a ) − 2sin 2 a cos2 a − 1 = −2sin 2 a cos2 a
2
Vaø:
sin 6 a + cos6 a − 1 = ( sin 2 a + cos2 a )( sin 4 a − sin 2 a cos2 a + cos 4 a ) − 1
= sin 4 a + cos 4 a − sin 2 a cos2 a − 1
= (1 − 2sin 2 a cos2 a ) − sin 2 a cos2 a − 1
= −3sin 2 a cos2 a
- sin 4 a + cos4 a − 1 −2sin 2 a cos2 a 2
Do ñoù: = =
sin 6 a + cos6 a − 1 −3sin 2 a cos2 a 3
1 + cos x ⎡ (1 − cos x ) ⎤
2
Baøi 2: Ruùt goïn bieåu thöùc A = = ⎢1 + ⎥
sin x ⎢
⎣ sin 2 x ⎥ ⎦
1 π
Tính giaù trò A neáu cos x = − vaø < x < π
2 2
1 + cos x ⎛ sin x + 1 − 2 cos x + cos2 x ⎞
2
Ta coù: A = ⎜ ⎟
sin x ⎝ sin 2 x ⎠
1 + cos x 2 (1 − cos x )
⇔A= .
sin x sin 2 x
2 (1 − cos2 x ) 2sin 2 x 2
⇔A= = = (vôùi sin x ≠ 0 )
sin 3 x sin 3 x sin x
1 3
Ta coù: sin 2 x = 1 − cos2 x = 1 − =
4 4
π
Do: < x < π neân sin x > 0
2
3
Vaäy sin x =
2
2 4 4 3
Do ñoù A = = =
sin x 3 3
Baøi 3: Chöùng minh caùc bieåu thöùc sau ñaây khoâng phuï thuoäc x:
a. A = 2 cos4 x − sin 4 x + sin 2 x cos2 x + 3sin 2 x
2 cot gx + 1
b. B = +
tgx − 1 cot gx − 1
a. Ta coù:
A = 2 cos4 x − sin 4 x + sin 2 x cos2 x + 3sin 2 x
⇔ A = 2 cos4 x − (1 − cos2 x ) + (1 − cos2 x ) cos2 x + 3 (1 − cos2 x )
2
⇔ A = 2 cos4 x − (1 − 2 cos2 x + cos4 x ) + cos2 x − cos4 x + 3 − 3cos2 x
⇔ A = 2 (khoâng phuï thuoäc x)
b. Vôùi ñieàu kieän sin x.cos x ≠ 0,tgx ≠ 1
2 cot gx + 1
Ta coù: B = +
tgx − 1 cot gx − 1
- 1
+1
2 tgx 2 1 + tgx
⇔B= + = +
tgx − 1 1
− 1 tgx − 1 1 − tgx
tgx
2 − (1 − tgx ) 1 − tgx
⇔ B= = = −1 (khoâng phuï thuoäc vaøo x)
tgx − 1 tgx − 1
Baøi 4: Chöùng minh
1 + cosa ⎡ (1 − cosa ) ⎤ cos2 b − sin 2 c
2
⎢1 − 2
⎥+ 2 2
− cot g 2 b cot g 2 c = cot ga − 1
2sin a ⎢ sin a ⎥ sin bsin c
⎣ ⎦
Ta coù:
cos2 b − sin 2 c
* − cot g 2 b.cot g 2 c
sin b.sin c
2 2
cotg 2 b 1
= − 2 − cot g 2 b cot g 2 c
sin c sin b
2
( ) ( )
= cot g 2 b 1 + cot g 2 c − 1 + cot g 2 b − cot g 2 b cot g 2 c = −1 (1)
1 + cosa ⎡ (1 − cos a ) ⎤
2
* ⎢1 − ⎥
2 sin a ⎢ sin 2 a ⎥
⎣ ⎦
1 + cosa ⎡ (1 − cos a ) ⎤
2
= ⎢1 − ⎥
2 sin a ⎢ 1 − cos2 a ⎥
⎣ ⎦
1 + cosa ⎡ 1 − cosa ⎤
= 1−
2sin a ⎢ 1 + cosa ⎥
⎣ ⎦
1 + cosa 2 cosa
= . = cot ga (2)
2 sin a 1 + cos a
Laáy (1) + (2) ta ñöôïc ñieàu phaûi chöùng minh xong.
Baøi 5: Cho ΔABC tuøy yù vôùi ba goùc ñeàu laø nhoïn.
Tìm giaù trò nhoû nhaát cuûa P = tgA.tgB.tgC
Ta coù: A + B = π − C
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
tgA + tgB + tgC ≥ 3 3 tgA.tgB.tgC
- ⇔ P ≥ 33 P
⇔ 3 P2 ≥ 3
⇔P≥3 3
⎧tgA = tgB = tgC
⎪ π
Daáu “=” xaûy ra ⇔ ⎨ π ⇔ A=B=C=
⎪0 < A,B,C < 2 3
⎩
π
Do ñoù: MinP = 3 3 ⇔ A = B = C =
3
Baøi 6 : Tìm giaù trò lôùn nhaát vaø nhoû nhaát cuûa
a/ y = 2 sin 8 x + cos4 2x
b/ y = 4 sin x − cos x
4
⎛ 1 − cos 2x ⎞
a/ Ta coù : y = 2 ⎜ ⎟ + cos 2x
4
⎝ 2 ⎠
Ñaët t = cos 2x vôùi −1 ≤ t ≤ 1 thì
1 4
y = (1 − t ) + t 4
8
1 3
=> y ' = − (1 − t ) + 4t 3
2
(1 − t ) = 8t 3
3
Ta coù : y ' = 0
⇔ 1 − t = 2t
1
⇔t=
3
1 ⎛1⎞
Ta coù y(1) = 1; y(-1) = 3; y ⎜ ⎟ =
27 ⎝ 3⎠
1
Do ñoù : Max y = 3 vaø Miny =
x∈ x∈ 27
b/ Do ñieàu kieän : sin x ≥ 0 vaø cos x ≥ 0 neân mieàn xaùc ñònh
⎡ π ⎤
D = ⎢ k2π, + k2π ⎥ vôùi k ∈
⎣ 2 ⎦
Ñaët t = cos x vôùi 0 ≤ t ≤ 1 thì t = cos x = 1 − sin x
4 2 2
Neân sin x = 1 − t4
Vaäy y = 1 − t − t treân D ' = [ 0,1]
8 4
−t 3
Thì y ' = − 1 < 0 ∀t ∈ [ 0; 1)
2. (1 − t
8
)
4 7
Neân y giaûm treân [ 0, 1 ]. Vaäy : max y = y ( 0 ) = 1, min y = y (1) = −1
x∈ D x∈ D
Baøi 7: Cho haøm soá y = sin4 x + cos4 x − 2m sin x cos x
Tìm giaù trò m ñeå y xaùc ñònh vôùi moïi x
- Xeùt f (x) = sin 4 x + cos4 x − 2m sin x cos x
f ( x ) = ( sin 2 x + cos2 x ) − m sin 2x − 2 sin 2 x cos2 x
2
1
f ( x) = 1 − sin2 2x − m sin 2x
2
Ñaët : t = sin 2x vôùi t ∈ [ −1, 1]
y xaùc ñònh ∀x ⇔ f ( x ) ≥ 0∀x ∈ R
1 2
⇔ 1− t − mt ≥ 0 ∀t ∈ [ −1,1]
2
⇔ g ( t ) = t 2 + 2mt − 2 ≤ 0 ∀t ∈ [ −1, 1]
Do Δ ' = m2 + 2 > 0 ∀m neân g(t) coù 2 nghieäm phaân bieät t1, t2
Luùc ñoù t t1 t2
g(t) + 0 - 0
Do ñoù : yeâu caàu baøi toaùn ⇔ t1 ≤ −1 < 1 ≤ t 2
⎧1g ( −1) ≤ 0
⎪ ⎧−2m − 1 ≤ 0
⇔⎨ ⇔ ⎨
⎪1g (1) ≤ 0
⎩ ⎩2m − 1 ≤ 0
⎧ −1
⎪m ≥ 2
⎪ 1 1
⇔⎨ ⇔− ≤m≤
⎪m ≤ 1 2 2
⎪
⎩ 2
Caùch khaùc :
g ( t ) = t 2 + 2mt − 2 ≤ 0 ∀t ∈ [ −1,1]
⇔ max g (t ) ≤ 0 ⇔ max { g (−1), g (1)} ≤ 0
t ∈[ −1,1 ]
⎧ −1
⎪m ≥ 2
⎪
⇔ max {−2m − 1),− 2m + 1)} ≤ 0 ⇔ ⎨
⎪m ≤ 1
⎪
⎩ 2
1 1
⇔− ≤m≤
2 2
π 3π 5π 7π 3
Baøi 8 : Chöùng minh A = sin4 + sin4 + sin4 + sin4 =
16 16 16 16 2
7π ⎛π π ⎞ π
Ta coù : sin = sin ⎜ − ⎟ = cos
16 ⎝ 2 16 ⎠ 16
5π ⎛ π 5π ⎞ 3π
sin = cos ⎜ − ⎟ = cos
16 ⎝ 2 16 ⎠ 16
Maët khaùc : sin 4 α + cos4 α = ( sin 2 α + cos2 α ) − 2 sin 2 α cos2 α
2
= 1 − 2sin2 α cos2 α
1
= 1 − sin2 2α
2
- π 7π 3π 5π
Do ñoù : A = sin4 + sin4 + sin4 + sin4
16 16 16 16
⎛ π π ⎞ ⎛ 4 3π 3π ⎞
= ⎜ sin 4 + cos4 ⎟ + ⎜ sin + cos4 ⎟
⎝ 16 16 ⎠ ⎝ 16 16 ⎠
⎛ 1 π⎞ ⎛ 1 3π ⎞
= ⎜ 1 − sin 2 ⎟ + ⎜ 1 − sin 2 ⎟
⎝ 2 8⎠ ⎝ 2 8 ⎠
1⎛ π 3π ⎞
= 2 − ⎜ sin 2 + sin 2 ⎟
2⎝ 8 8 ⎠
1⎛ π π⎞ ⎛ 3π π⎞
= 2 − ⎜ sin 2 + cos2 ⎟ ⎜ do sin = cos ⎟
2⎝ 8 8⎠ ⎝ 8 8⎠
1 3
= 2− =
2 2
Baøi 9 : Chöùng minh : 16 sin 10o .sin 30o .sin 50o .sin 70o = 1
A cos 10o 1
Ta coù : A = = (16sin10ocos10o)sin30o.sin50o.sin70o
cos 10 o
cos 10 o
1 ⎛1⎞
o (
⇔ A= 8 sin 20o ) ⎜ ⎟ cos 40o . cos 20o
cos 10 ⎝2⎠
1
o (
⇔ A= 4 sin 200 cos 20o ) . cos 40o
cos10
1
o (
⇔ A= 2 sin 40o ) cos 40o
cos10
1 cos 10o
⇔ A= sin 80 =o
=1
cos10o cos 10o
A B B C C A
Baøi 10 : Cho ΔABC . Chöùng minh : tg tg + tg tg + tg tg = 1
2 2 2 2 2 2
A+B π C
Ta coù : = −
2 2 2
A+B C
Vaäy : tg = cot g
2 2
A B
tg + tg
⇔ 2 2 = 1
A B C
1 − tg .tg tg
2 2 2
⎡ A B⎤ C A B
⇔ ⎢ tg + tg ⎥ tg = 1 − tg tg
⎣ 2 2⎦ 2 2 2
A C B C A B
⇔ tg tg + tg tg + tg tg = 1
2 2 2 2 2 2
π π π π
Baøi 11 : Chöùng minh : 8 + 4tg + 2tg + tg = cot g ( *)
8 16 32 32
- 1 1 1 1⎡ A B C A B C⎤
+ + = ⎢ tg + tg + tg + cot g + cot g + cot g ⎥
sin A sin B sin C 2 ⎣ 2 2 2 2 2 2⎦
A B C A B C
Ta coù : cot g + cot g + cot g = cot g .cot g .cot g
2 2 2 2 2 2
(Xem chöùng minh baøi 19g )
sin α cos α 2
Maët khaùc : tgα + cot gα = + =
cos α sin α sin 2α
1⎡ A B C A B C⎤
Do ñoù : ⎢ tg + tg + tg + cotg + cotg + cotg ⎥
2⎣ 2 2 2 2 2 2⎦
1⎡ A B C⎤ 1 ⎡ A B C⎤
= ⎢ tg + tg + tg ⎥ + ⎢cotg + cotg + cotg ⎥
2⎣ 2 2 2⎦ 2 ⎣ 2 2 2⎦
1⎡ A A⎤ 1 ⎡ B B⎤ 1 ⎡ C C⎤
= ⎢ tg + cot g ⎥ + ⎢ tg + cot g ⎥ + ⎢ tg + cot g ⎥
2⎣ 2 2⎦ 2⎣ 2 2⎦ 2⎣ 2 2⎦
1 1 1
= + +
sin A sin B sin C
BAØI TAÄP
1. Chöùng minh :
π 2π 1
a/ cos − cos =
5 5 2
cos15 + sin15
o o
b/ = 3
cos15o − sin15o
2π 4π 6π 1
c/ cos + cos + cos =−
7 7 7 2
d/ sin 2x sin 6x + cos 2x.cos 6x = cos3 4x
3 3
e/ tg20o.tg40o.tg60o.tg80o = 3
π 2π 5π π 8 3 π
f/ tg + tg + tg + tg = cos
6 9 18 3 3 9
π 2π 3π 4π 5π 6π 7π 1
g/ cos .cos .cos .cos .cos .cos .cos =
15 15 15 15 15 15 15 27
⎡π ⎤ ⎡π ⎤
h/ tgx.tg ⎢ − x ⎥ .tg ⎢ + x ⎥ = tg3x
⎣3 ⎦ ⎣3 ⎦
k/ tg20o + tg40o + 3tg20o.tg40o = 3
3
e/ sin 20o.sin 40o.sin 80o =
8
m/ tg5 .tg55 .tg65 .tg75 = 1
o o o o
⎧sin x = 2 sin ( x + y )
⎪
2. Chöùng minh raèng neáu ⎨ π
⎪ x + y ≠ ( 2k + 1) ( k ∈ z )
⎩ 2
sin y
thì tg ( x + y ) =
cos y − 2
3. Cho ΔABC coù 3 goùc ñeàu nhoïn vaø A ≥ B ≥ C
- a/ Chöùng minh : tgA + tgB + tgC = tgA.tgB.tgC
b/ Ñaët tgA.tgB = p; tgA.tgC = q
Chöùng minh (p-1)(q-1) ≥ 4
4. Chöùng minh caùc bieåu thöùc khoâng phuï thuoäc x :
a/ A = sin 4 x (1 + sin 2 x ) + cos4 x (1 + cos2 x ) + 5 sin 2 x cos2 x + 1
b/ B = 3 ( sin 8 x − cos8 x ) + 4 ( cos6 x − 2 sin 6 x ) + 6 sin 4 x
c/ C = cos2 ( x − a ) + sin2 ( x − b ) − 2 cos ( x − a ) sin ( x − b ) sin ( a − b )
5. Cho ΔABC , chöùng minh :
cos C cos B
a/ cot gB + = cot gC +
sin B cos A sin C cos A
A B C 3A 3B 3C
b/ sin3 A + sin3 B + sin3 C = 3cos cos cos + cos cos cos
2 2 2 2 2 2
A B−C B A−C
c/ sin A + sin B + sin C = cos .cos + cos .cos
2 2 2 2
C A−B
+ cos .cos
2 2
d/ cotgAcotgB + cotgBcotgC + cotgCcotgA = 1
e/ cos2 A + cos2 B + cos2 C = 1 − 2 cos A cos B cos C
f/ sin3Asin(B- C)+ sin3Bsin(C- A)+ sin3Csin(A- B) = 0
6. Tìm giaù trò nhoû nhaát cuûa :
1 1 π
a/ y = + vôùi 0 < x <
sin x cos x 2
9π
b/ y = 4x + + sin x vôùi 0 < x < ∞
x
c/ y = 2 sin 2 x + 4 sin x cos x + 5
7. Tìm giaù trò lôùn nhaát cuûa :
a/ y = sin x cos x + cos x sin x
b/ y = sinx + 3sin2x
c/ y = cos x + 2 − cos2 x
TT luyện thi đại học CLC Vĩnh Viễn
- Chöông 2: PHÖÔNG TRÌNH LÖÔÏ N G GIAÙ C CÔ BAÛ N
⎡ u = v + k2π
sin u = sin v ⇔ ⎢
⎣ u = π − v + k2π
cos u = cos v ⇔ u = ± v + k2π
⎧ π
⎪u ≠ + kπ
tgu = tgv ⇔ ⎨ 2 ( k, k ' ∈ Z )
⎪u = v + k ' π
⎩
⎧u ≠ kπ
cot gu = cot gv ⇔ ⎨
⎩u = v + k ' π
π
Ñaë c bieä t : sin u = 0 ⇔ u = kπ cos u = 0 ⇔ u = + kπ
2
π
sin u = 1 ⇔ u = + k2π ( k ∈ Z) cos u = 1 ⇔ u = k2π ( k ∈ Z )
2
π
sin u = −1 ⇔ u = − + k2π cos u = −1 ⇔ u = π + k2π
2
Chuù yù : sin u ≠ 0 ⇔ cos u ≠ ±1
cos u ≠ 0 ⇔ sin u ≠ ±1
Baø i 28 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i D, naê m 2002)
Tìm x ∈ [ 0,14 ] nghieä m ñuù ng phöông trình
cos 3x − 4 cos 2x + 3 cos x − 4 = 0 ( * )
Ta coù (*) : ⇔ ( 4 cos3 x − 3 cos x ) − 4 ( 2 cos2 x − 1) + 3 cos x − 4 = 0
⇔ 4 cos3 x − 8 cos2 x = 0 ⇔ 4 cos2 x ( cos x − 2 ) = 0
⇔ cos x = 0 hay cos x = 2 ( loaïi vì cos x ≤ 1)
π
⇔ x= + kπ ( k ∈ Z )
2
π
Ta coù : x ∈ [ 0,14] ⇔ 0 ≤ + kπ ≤ 14
2
π π 1 14 1
⇔ − ≤ kπ ≤ 14 − ⇔ −0, 5 = − ≤ k ≤ − ≈ 3, 9
2 2 2 π 2
⎧ π 3π 5π 7π ⎫
Maø k ∈ Z neâ n k ∈ {0,1, 2, 3} . Do ñoù : x ∈ ⎨ , , , ⎬
⎩2 2 2 2 ⎭
Baø i 29 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i D, naê m 2004)
Giaû i phöông trình :
( 2 cos x − 1)( 2 sin x + cos x ) = sin 2x − sin x ( *)
- Ta coù (*) ⇔ ( 2 cos x − 1)( 2 sin x + cos x ) = sin x ( 2 cos x − 1)
⇔ ( 2 cos x − 1) ⎡( 2 sin x + cos x ) − sin x ⎤ = 0
⎣ ⎦
⇔ ( 2 cos x − 1)( sin x + cos x ) = 0
1
⇔ cos x = ∨ sin x = − cos x
2
π ⎛ π⎞
⇔ cos x = cos ∨ tgx = −1 = tg ⎜ − ⎟
3 ⎝ 4⎠
π π
⇔ x = ± + k2π ∨ x = − + kπ, ( k ∈ Z )
3 4
Baø i 30 : Giaû i phöông trình cos x + cos 2x + cos 3x + cos 4x = 0 (*)
Ta coù (*) ⇔ ( cos x + cos 4x ) + ( cos 2x + cos 3x ) = 0
5x 3x 5x x
⇔ 2 cos .cos + 2 cos .cos = 0
2 2 2 2
5x ⎛ 3x x⎞
⇔ 2 cos ⎜ cos + cos ⎟ = 0
2 ⎝ 2 2⎠
5x x
⇔ 4 cos cos x cos = 0
2 2
5x x
⇔ cos = 0 ∨ cos x = 0 ∨ cos = 0
2 2
5x π π x π
⇔ = + kπ ∨ x = + kπ ∨ = + kπ
2 2 2 2 2
π 2kπ π
⇔ x= + ∨ x = + kπ ∨ x = π + 2π, ( k ∈ Z )
5 5 2
Baø i 31: Giaûi phöông trình sin 2 x + sin 2 3x = cos2 2x + cos2 4x ( * )
1 1 1 1
Ta coù (*) ⇔ (1 − cos 2x ) + (1 − cos 6x ) = (1 + cos 4x ) + (1 + cos 8x )
2 2 2 2
⇔ − ( cos 2x + cos 6x ) = cos 4x + cos 8x
⇔ −2 cos 4x cos 2x = 2 cos 6x cos 2x
⇔ 2 cos 2x ( cos 6x + cos 4x ) = 0
⇔ 4 cos 2x cos 5x cos x = 0
⇔ cos 2x = 0 ∨ cos 5x = 0 ∨ cos x = 0
π π π
⇔ 2x = + kπ ∨ 5x + kπ ∨ x = + kπ, k ∈
2 2 2
π kπ π kπ π
⇔ x= + ∨x= + ∨ x = + kπ , k ∈
4 2 10 5 2
Baø i 32 : Cho phöông trình
⎛π x⎞ 7
sin x.cos 4x − sin 2 2x = 4 sin 2 ⎜ − ⎟ − ( *)
⎝4 2⎠ 2
Tìm caù c nghieä m cuû a phöông trình thoû a : x − 1 < 3
- 1 ⎡ π ⎤ 7
Ta coù : (*)⇔ sin x.cos 4x − (1 − cos 4x ) = 2 ⎢1 − cos ⎛ − x ⎞ ⎥ −
⎜ ⎟
2 ⎣ ⎝2 ⎠⎦ 2
1 1 3
⇔ sin x cos 4x − + cos 4x = − − 2sin x
2 2 2
1
⇔ sin x cos 4x + cos 4x + 1 + 2sin x = 0
2
⎛ 1⎞ ⎛ 1⎞
⇔ cos 4x ⎜ sin x + ⎟ + 2 ⎜ sin x + ⎟ = 0
⎝ 2⎠ ⎝ 2⎠
⎛ 1⎞
⇔ ( cos 4x + 2) ⎜ sin x + ⎟ = 0
⎝ 2⎠
⎡cos 4x = −2 ( loaïi ) ⎡ π
⎢ ⎢ x = − 6 + k 2π
⇔ ⎢sin x = − 1 = sin ⎛ − π ⎞ ⇔ ⎢
⎢ ⎜ ⎟ ⎢ x = 7π + 2hπ
⎣ 2 ⎝ 6⎠ ⎢
⎣ 6
Ta coù : x − 1 < 3 ⇔ −3 < x − 1 < 3 ⇔ −2 < x < 4
π
Vaä y : −2 < − + k2π < 4
6
π π 1 1 2 1
⇔ − 2 < 2kπ < 4 + ⇔ −
- 3
⇔ sin 4x = sin3 4x
4
⇔ 3sin 4x − 4 sin3 4x = 0
⇔ sin12x = 0
kπ
⇔ 12x = kπ ⇔ x= ( k ∈ Z)
12
Baø i 34 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i B, naê m 2002)
Giaû i phöông trình :
sin 2 3x − cos2 4x = sin 2 5x − cos2 6a ( * )
Ta coù : (*)⇔
1 1 1 1
(1 − cos 6x ) − (1 + cos 8x ) = (1 − cos10x ) − (1 + cos12x )
2 2 2 2
⇔ cos 6x + cos 8x = cos10x + cos12x
⇔ 2 cos7x cos x = 2 cos11x cos x
⇔ 2 cos x ( cos 7x − cos11x ) = 0
⇔ cos x = 0 ∨ cos7x = cos11x
π
⇔ x = + kπ ∨ 7x = ±11x + k 2π
2
π kπ kπ
⇔ x = + kπ ∨ x = − ∨x= ,k ∈
2 2 9
Baø i 35 : Giaû i phöông trình
( sin x + sin 3x ) + sin 2x = ( cos x + cos 3x ) + cos 2x
⇔ 2sin 2x cos x + sin 2x = 2 cos 2x cos x + cos 2x
⇔ sin 2x ( 2 cos x + 1) = cos 2x ( 2 cos x + 1)
⇔ ( 2 cos x + 1) ( sin 2x − cos 2x ) = 0
1 2π
⇔ cos x = −= cos ∨ sin 2x = cos 2x
2 3
2π π
⇔ x=± + k2π ∨ tg2x = 1 = tg
3 4
2π π π
⇔ x=± + k2π ∨ x = + k , ( k ∈ Z )
3 8 2
Baø i 36: Giaû i phöông trình
cos 10x + 2 cos2 4x + 6 cos 3x. cos x = cos x + 8 cos x. cos3 3x ( * )
Ta coù : (*)⇔ cos10x + (1 + cos 8x ) = cos x + 2 cos x ( 4 cos3 3x − 3 cos 3x )
⇔ ( cos10x + cos 8x ) + 1 = cos x + 2 cos x.cos 9x
⇔ 2 cos 9x cos x + 1 = cos x + 2 cos x.cos 9x
⇔ cos x = 1
⇔ x = k2π ( k ∈ Z )
Baø i 37 : Giaû i phöông trình
- 4 sin 3 x + 3 cos3 x − 3sin x − sin 2 x cos x = 0 ( * )
Ta coù : (*) ⇔ sin x ( 4 sin 2 x − 3) − cos x ( sin 2 x − 3 cos2 x ) = 0
⇔ sin x ( 4 sin 2 x − 3) − cos x ⎡sin 2 x − 3 (1 − sin 2 x ) ⎤ = 0
⎣ ⎦
⇔ ( 4 sin x − 3) ( sin x − cos x ) = 0
2
⇔ ⎡ 2 (1 − cos 2x ) − 3⎤ ( sin x − cos x ) = 0
⎣ ⎦
⎡ 1 2π
cos 2x = − = cos
⇔ ⎢ 2 3
⎢
⎣sin x = cos x
⎡ π
⎡ 2π ⎢ x = ± + kπ
⇔ ⎢2x = ± 3 + k2π ⇔ ⎢
3
( k ∈ Z)
⎢ ⎢ x = π + kπ
⎣ tgx = 1 ⎢
⎣ 4
Baø i 38 : (Ñeà thi tuyeå n sinh Ñaï i hoï c khoá i B naê m 2005)
Giaû i phöông trình :
sin x + cos x + 1 + sin 2x + cos 2x = 0 ( * )
Ta coù : (*) ⇔ sin x + cos x + 2sin x cos x + 2 cos2 x = 0
⇔ sin x + cos x + 2 cos x ( sin x + cos x ) = 0
⇔ ( sin x + cos x ) (1 + 2 cos x ) = 0
⎡sin x = − cos x
⇔ ⎢
⎢cos 2x = − 1 = cos 2π
⎣ 2 3
⎡ tgx = −1
⇔ ⎢
⎢ x = ± 2π + k 2π
⎣ 3
⎡ π
⎢ x = − 4 + kπ
⇔ ⎢ ( k ∈ Z)
⎢ x = ± 2π + k2π
⎢
⎣ 3
Baø i 39 : Giaû i phöông trình
( 2 sin x + 1)( 3 cos 4x + 2 sin x − 4 ) + 4 cos2 x = 3 ( *)
Ta coù : (*) ⇔ ( 2 sin x + 1)( 3 cos 4x + 2 sin x − 4 ) + 4 (1 − sin 2 x ) − 3 = 0
⇔ ( 2 sin x + 1)( 3 cos 4x + 2 sin x − 4 ) + (1 + 2 sin x )(1 − 2 sin x ) = 0
⇔ ( 2 sin x + 1) ⎡ 3 cos 4x + 2 sin x − 4 + (1 − 2 sin x ) ⎤ = 0
⎣ ⎦
⇔ 3 ( cos 4x − 1)( 2 sin x + 1) = 0
1 ⎛ π⎞
⇔ cos 4x = 1 ∨ sin x = − = sin ⎜ − ⎟
2 ⎝ 6⎠
- π 7π
⇔ 4x = k2π ∨ x = − + k2π ∨ x = + k2π
6 6
kπ π 7π
⇔ x= ∨ x = − + k2π ∨ x = + k2π, ( k ∈ Z)
2 6 6
Baø i 40: Giaû i phöông trình
sin 6 x + cos6 x = 2 ( sin 8 x + cos8 x ) ( * )
Ta coù : (*) ⇔ sin6 x − 2sin8 x + cos6 x − 2 cos8 x = 0
⇔ sin 6 x (1 − 2 sin 2 x ) − cos6 x ( 2 cos2 x − 1) = 0
⇔ sin6 x cos 2x − cos6 x. cos 2x = 0
⇔ cos 2x ( sin 6 x − cos6 x ) = 0
⇔ cos 2x = 0 ∨ sin6 x = cos6 x
⇔ cos 2x = 0 ∨ tg 6 x = 1
π
⇔ 2x = ( 2k + 1) ∨ tgx = ±1
2
π π
⇔ x = ( 2k + 1) ∨ x = ± + kπ
4 4
π kπ
⇔ x= + ,k ∈
4 2
Baø i 41 : Giaû i phöông trình
1
cos x.cos 2x.cos 4x.cos 8x = ( *)
16
Ta thaá y x = kπ khoâ n g laø nghieä m cuû a (*) vì luù c ñoù
cos x = ±1, cos 2x = cos 4x = cos 8x = 1
1
(*) thaøn h : ±1 = voâ nghieä m
16
Nhaâ n 2 veá cuû a (*) cho 16sin x ≠ 0 ta ñöôï c
(*) ⇔ (16 sin x cos x ) cos 2x.cos 4x.cos 8x = sin x vaø sin x ≠ 0
⇔ ( 8 sin 2x cos 2x ) cos 4x.cos 8x = sin x vaø sin x ≠ 0
⇔ ( 4 sin 4x cos 4x ) cos 8x = sin x vaø sin x ≠ 0
⇔ 2sin 8x cos 8x = sin x vaø sin x ≠ 0
⇔ sin16x = sin x vaø sin x ≠ 0
k2π π kπ
⇔x = ∨x= + , ( k ∈ Z)
15 17 17
Do : x = hπ khoâ n g laø nghieä m neâ n k ≠ 15m vaø 2k + 1 ≠ 17n ( n, m ∈ Z )
Baø i 42: Giaû i phöông trình 8cos ⎛ x +
π⎞
= cos 3x ( * )
3
⎜ ⎟
⎝ 3⎠
π π
Ñaët t = x + ⇔x=t−
3 3
- Thì cos 3x = cos ( 3t − π ) = cos ( π − 3t ) = − cos 3t
Vaä y (*) thaø n h 8 cos3 t = − cos 3t
⇔ 8 cos3 t = −4 cos3 t + 3 cos t
⇔ 12 cos3 t − 3 cos t = 0
⇔ 3 cos t ( 4 cos2 t − 1) = 0
⇔ 3 cos t ⎡2 (1 + cos 2t ) − 1⎤ = 0
⎣ ⎦
⇔ cos t ( 2 cos 2t + 1) = 0
1 2π
⇔ cos t = 0 ∨ cos 2t = − = cos
2 3
π 2π
⇔ t = ( 2k + 1) ∨ 2t = ± + k2π
2 3
π π
⇔ t = + kπ ∨ t = ± + kπ
2 3
π
Maø x = t −
3
π 2π
Vaä y (*) ⇔ x = + k2π ∨ x = kπ ∨ x = + kπ, ( vôùik ∈ Z )
6 3
Ghi chuù :
Khi giaû i caù c phöông trình löôï n g giaù c coù chöù a tgu, cotgu, coù aå n ôû maã u , hay
chöù a caê n baä c chaü n ... ta phaû i ñaë t ñieà u kieä n ñeå phöông trình xaù c ñònh. Ta seõ
duø n g caù c caù c h sau ñaâ y ñeå kieå m tra ñieà u kieä n xem coù nhaä n nghieä m hay
khoâ n g.
+ Thay caùc giaù trò x tìm ñöôï c vaø o ñieà u kieä n thöû laï i xem coù thoû a
Hoaë c + Bieå u dieã n caù c ngoï n cung ñieà u kieä n vaø caù c ngoï n cung tìm ñöôïc treâ n cuø n g
moä t ñöôø n g troø n löôï n g giaù c . Ta seõ loaï i boû ngoï n cung cuû a nghieä m khi coù
truø n g vôù i ngoï n cung cuû a ñieà u kieä n .
Hoaë c + So vôi caù c ñieà u kieä n trong quaù trình giaûi phöông trình.
Baø i 43 : Giaû i phöông trình tg 2 x − tgx.tg3x = 2 ( * )
⎧cos x ≠ 0 π hπ
Ñieà u kieä n ⎨ ⇔ cos3x ≠ 0 ⇔ x ≠ +
⎩cos 3x = 4 cos x − 3 cos x ≠ 0
3
6 3
Luù c ñoù ta coù (*) ⇔ tgx ( tgx − tg3x ) = 2
sin x ⎛ sin x sin 3x ⎞
⇔ ⎜ − ⎟=2
cos x ⎝ cos x cos 3x ⎠
⇔ sin x ( sin x cos 3x − cos x sin 3x ) = 2 cos2 x cos 3x
⇔ sin x sin ( −2x ) = 2 cos2 x. cos 3x
⇔ −2 sin2 x cos x = 2 cos2 x cos 3x
⇔ − sin2 x = cos x cos 3x (do cos x ≠ 0 )
1 1
⇔ − (1 − cos 2x ) = ( cos 4x + cos 2x )
2 2
⇔ cos 4x = −1 ⇔ 4x = π + k2π
- ⎛π ⎞
1. Tìm caù c nghieä m treâ n ⎜ , 3π ⎟ cuû a phöông trình:
⎝3 ⎠
⎛ 5π ⎞ ⎛ 7π ⎞
sin ⎜ 2x + ⎟ − 3 cos ⎜ x − ⎟ = 1 + 2 sin x
⎝ 2 ⎠ ⎝ 2 ⎠
⎛ π⎞
2. Tìm caù c nghieä m x treâ n ⎜ 0, ⎟ cuû a phöông trình
⎝ 2⎠
sin 4x − cos 6x = sin (10, 5π + 10x )
2 2
3. Giaû i caù c phöông trình sau:
(
a/ sin 3 x + cos3 x = 2 sin5 x + cos5 x )
sin x + sin 2x + sin 3x
b/ = 3
cos x + cos 2x + cos 3x
1 + cos x
c/ tg 2 x =
1 − sin x
d/ tg2x − tg3x − tg5x = tg2x.tg3x.tg5x
4
e/ cos x = cos2 x
3
⎛ π⎞ 1 1
f/ 2 2 sin ⎜ x + ⎟ = +
⎝ 4 ⎠ sin x cos x
2
i/ 2tgx + cot g2x = 3 +
sin 2x
2
h/ 3tg3x + cot g2x = 2tgx +
sin 4x
2 2 2
k/ sin x + sin 2x + sin 3x = 2
sin 2x
l/ + 2 cos x = 0
1 + sin x
m/ 25 − 4x 2 ( 3sin 2πx + 8 sin πx ) = 0
sin x.cot g5x
n/ =1
cos 9x
2
o/ 3tg6x − = 2tg2x − cot g4x
sin 8x
(
p/ 2 sin 3x 1 − 4 sin 2 x = 1 )
1 + cos x
q/ tg 2 x =
1 − sin x
2
r/ cos3 x cos 3x + sin 3 x sin 3x =
4
⎛x⎞ ⎛x⎞ 5
s/ sin4 ⎜ ⎟ + cos4 ⎜ ⎟ =
⎝ 3⎠ ⎝ 3⎠ 8
t/ cos x − 4 sin x − 3 cos x sin2 x + sin x = 0
3 3
x x
u/ sin4 + cos4 = 1 − 2sin x
2 2
- ⎛ π⎞ ⎛ π⎞
v/ sin ⎜ 3x − ⎟ = sin 2x.sin ⎜ x + ⎟
⎝ 4⎠ ⎝ 4⎠
4
w/ tg x + 1 =
( 2 − sin x ) sin 3x
2
cos4 x
⎛ x ⎞
y/ tgx + cos x − cos2 x = sin x ⎜ 1 + tg tgx ⎟
⎝ 2 ⎠
4. Cho phöông trình: ( 2 sin x − 1)( 2 cos 2x + 2 sin x + m ) = 3 − 4 cos2 x (1)
a/ Giaû i phöông trình khi m = 1
b/ Tìm m ñeå (1) coù ñuù n g 2 nghieä m treâ n [ 0, π ]
( ÑS: m = 0 ∨ m < −1 ∨ m > 3 )
5. Cho phöông trình:
4 cos5 x sin x − 4 sin5 x.cos x = sin2 4x + m (1)
Bieá t raè n g x = π laø moä t nghieä m cuû a (1). Haõ y giaû i phöông trình trong tröôø n g
hôï p ñoù .
Th.S Phạm Hồng Danh
TT luyện thi Đại học CLC Vĩnh Viễn
nguon tai.lieu . vn
