Mô tả:
Lượng giác toàn tập
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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
vôùi cos α ≠ 0
tgα =
cos α
cos α
vôùi sin α ≠ 0
cot gα =
sin α
II. Baûng giaù trò löôïng giaùc cuûa moät soá cung (hay goùc) ñaëc bieät
Goùc α
Giaù trò
( )
0 0o
sin α
0
cos α
1
tgα
0
cot gα
||
π
30o
6
1
2
( )
3
2
3
3
3
π
45o
4
2
2
2
2
π
60o
3
3
2
1
2
π
90o
2
1
3
||
1
3
3
0
( )
( )
III. Heä thöùc cô baûn
sin 2 α + cos2 α = 1
1
π
vôùi α ≠ + kπ ( k ∈ Z )
1 + tg2α =
2
cos α
2
1
vôùi α ≠ kπ ( k ∈ Z )
t + cot g2 =
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 )
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( )
1
0
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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
d. Phuï nhau: α vaø
π
−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
: α vaø + α
2
2
⎛π
⎞
sin ⎜ + α ⎟ = cos α
⎝2
⎠
⎛π
⎞
cos ⎜ + α ⎟ = − sin α
⎝2
⎠
e.Sai nhau
⎛π
⎞
tg ⎜ + α ⎟ = − cot gα
⎝2
⎠
⎛π
⎞
cot g ⎜ + α ⎟ = − tgα
⎝2
⎠
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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 ∓ sin asin b
tg ( a ± b ) =
tga ± tgb
1 ∓ 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 − tg2a
cot g2a − 1
cot g2a =
2 cot ga
VII. Coâng thöùc nhaân ba:
sin3a = 3sin a − 4sin3 a
cos3a = 4 cos3 a − 3cosa
VIII. Coâng thöùc haï baäc:
1
(1 − cos2a )
2
1
cos2 a = (1 + cos2a )
2
1 − cos2a
tg2a =
1 + cos2a
sin 2 a =
IX. Coâng thöùc chia ñoâi
Ñaët t = tg
a
(vôùi a ≠ π + k2 π )
2
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2t
1 + t2
1 − t2
cosa =
1 + t2
2t
tga =
1 − t2
sin a =
X. Coâng thöùc bieán ñoåi toång thaønh tích
a+b
a−b
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
cosa + cos b = 2 cos
XI. Coâng thöùc bieån ñoåi tích thaønh toång
1
⎡ 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
cosa.cos b =
Baøi 1: Chöùng minh
sin 4 a + cos4 a − 1 2
=
sin 6 a + cos6 a − 1 3
Ta coù:
sin 4 a + cos4 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 + cos4 a ) − 1
= sin 4 a + cos4 a − sin 2 a cos2 a − 1
= (1 − 2sin 2 a cos2 a ) − sin 2 a cos2 a − 1
= −3sin 2 a cos2 a
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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
2
1 + cos x ⎡ (1 − cos x ) ⎤
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 2 x + 1 − 2 cos x + cos2 x ⎞
Ta coù: A =
⎜
⎟
sin x ⎝
sin 2 x
⎠
1 + cos x 2 (1 − cos x )
.
sin x
sin 2 x
2 (1 − cos2 x ) 2sin 2 x
2
(vôùi sin x ≠ 0 )
⇔A=
=
=
3
3
sin x
sin 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
⇔A=
Baøi 3: Chöùng minh caùc bieåu thöùc sau ñaây khoâng phuï thuoäc x:
a. A = 2 cos4 x − sin 4 x + sin2 x cos2 x + 3sin 2 x
b. B =
2
cot gx + 1
+
tgx − 1 cot gx − 1
a. Ta coù:
A = 2 cos4 x − sin 4 x + sin2 x cos2 x + 3sin2 x
2
⇔ A = 2 cos4 x − (1 − cos2 x ) + (1 − cos2 x ) cos2 x + 3 (1 − cos2 x )
⇔ 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.cosx ≠ 0,tgx ≠ 1
Ta coù: B =
2
cot gx + 1
+
tgx − 1 cot gx − 1
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1
+1
2
2
1 + tgx
tgx
⇔ B=
+
=
+
tgx − 1 1 − 1 tgx − 1 1 − tgx
tgx
⇔ B=
2 − (1 − tgx ) 1 − tgx
=
= −1 (khoâng phuï thuoäc vaøo x)
tgx − 1
tgx − 1
Baøi 4: Chöùng minh
2
1 + cosa ⎡ (1 − cosa ) ⎤ cos2 b − sin 2 c
− cot g2 b cot g2 c = cot ga − 1
⎢1 −
⎥+
2
2
2
2sin a ⎢
sin a ⎥ sin bsin c
⎣
⎦
Ta coù:
cos2 b − sin 2 c
*
− cot g2 b.cot g2 c
2
2
sin b.sin c
cotg2 b
1
=
− 2 − cot g2 b cot g2 c
2
sin c sin b
= cot g2 b 1 + cot g2 c − 1 + cot g2 b − cot g 2 b cot g2 c = −1 (1)
(
) (
)
2
1 + cosa ⎡ (1 − cosa ) ⎤
*
⎢1 −
⎥
2sin a ⎢
sin 2 a ⎥
⎣
⎦
2
1 + cosa ⎡ (1 − cosa ) ⎤
=
⎢1 −
⎥
2sin a ⎢
1 − cos2 a ⎥
⎣
⎦
1 + cosa ⎡ 1 − cosa ⎤
1−
=
2sin a ⎢⎣ 1 + cosa ⎥⎦
1 + cosa 2 cosa
=
= cot ga (2)
.
2sin a 1 + cosa
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
⇔
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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=
3
⎪⎩ 0 < A,B,C < 2
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 ⎞
4
a/ Ta coù : y = 2 ⎜
⎟ + cos 2x
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
3
Ta coù : y ' = 0 Ù (1 − t ) = 8t 3
⇔ 1 − t = 2t
1
⇔t=
3
⎛1⎞
⎝ 3⎠
1
27
1
Do ñoù : Max y = 3 vaø Miny =
x∈
27
x∈
Ta coù y(1) = 1; y(-1) = 3; y ⎜ ⎟ =
b/ Do ñieàu kieän : sin x ≥ 0 vaø cos x ≥ 0 neân mieàn xaùc ñònh
π
⎡
⎤
D = ⎢ k2π, + k2π ⎥ vôùi k ∈
2
⎣
⎦
4
2
2
Ñaët t = cos x vôùi 0 ≤ t ≤ 1 thì t = cos x = 1 − sin x
Neân sin x =
1 − t4
4
Vaäy y = 1 − t − t treân D ' = [ 0,1]
8
Thì y ' =
−t 3
2. (1 − t
8
)
4 7
− 1 < 0 ∀t ∈ [ 0; 1)
Neân y giaûm treân [ 0, 1 ]. Vaäy : max y = y ( 0 ) = 1, min y = y (1) = −1
x∈ D
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x∈ D
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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
sin2 2x − m sin 2x
2
Ñaët : t = sin 2x vôùi t ∈ [ −1, 1]
f ( x) = 1 −
y xaùc ñònh ∀x ⇔ f ( x ) ≥ 0∀x ∈ R
1 2
t − mt ≥ 0 ∀t ∈ [ −1,1]
2
⇔ g ( t ) = t 2 + 2mt − 2 ≤ 0 ∀t ∈ [ −1,1]
⇔ 1−
Do Δ ' = m2 + 2 > 0 ∀m neân g(t) coù 2 nghieäm phaân bieät t1, t2
Luùc ñoù
t
t1
t2
g(t)
Do ñoù : yeâu caàu baøi toaùn
+
0
-
⇔ t1 ≤ −1 < 1 ≤ t 2
0
⎧⎪1g ( −1) ≤ 0
⎧−2m − 1 ≤ 0
⇔⎨
⇔ ⎨
⎩2m − 1 ≤ 0
⎪⎩1g (1) ≤ 0
−1
⎧
⎪⎪ m ≥ 2
1
1
⇔⎨
⇔− ≤m≤
2
2
⎪m ≤ 1
⎪⎩
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
+ sin4
+ sin4
+ sin4
=
16
16
16
16 2
7π
π
⎛π π ⎞
Ta coù : sin
= sin ⎜ −
⎟ = cos
16
16
⎝ 2 16 ⎠
3π
5π
⎛ π 5π ⎞
= cos ⎜ −
sin
⎟ = cos
16
16
⎝ 2 16 ⎠
Baøi 8 : Chöùng minh A = sin4
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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
+ sin4
+ sin4
+ sin4
Do ñoù : A = sin4
16
16
16
16
π
π ⎞ ⎛
3π ⎞
⎛
4 3π
= ⎜ sin 4
+ cos4
+ cos4
⎟ + ⎜ sin
⎟
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⎞
= cos ⎟
= 2 − ⎜ sin 2 + cos2 ⎟ ⎜ do sin
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
o
o
cos 10
cos 10
1
⎛1⎞
⇔ A=
8 sin 20o ) ⎜ ⎟ cos 40o . cos 20o
o (
cos 10
⎝2⎠
1
4 sin 200 cos 20o ) . cos 40o
⇔ A=
o (
cos10
1
2 sin 40o ) cos 40o
⇔ A=
o (
cos10
1
cos 10o
o
sin
80
=
=1
⇔ A=
cos10o
cos 10o
Baøi 10 : Cho ΔABC . Chöùng minh : tg
A B
B C
C A
tg + tg tg + tg tg = 1
2
2
2
2
2
2
A+B π C
= −
2
2 2
A+B
C
= cot g
Vaäy : tg
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
Ta coù :
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⇔ tg
A C
B C
A B
tg + tg tg + tg tg = 1
2
2
2
2
2
2
π
π
π
π
+ 2tg
+ tg
= cot g
( *)
8
16
32
32
π
π
π
π
− tg
− 2tg
− 4tg
Ta coù : (*) ⇔ 8 = cot g
32
32
16
8
2
2
cos a sin a cos a − sin a
−
=
Maø : cot ga − tga =
sin a cos a
sin a cos a
cos 2a
=
= 2 cot g2a
1
sin 2a
2
Do ñoù :
π
π⎤
π
π
⎡
(*) ⇔ ⎢ cot g
− tg ⎥ − 2tg
− 4tg = 8
32
32 ⎦
16
8
⎣
π
π⎤
π
⎡
⇔ ⎢ 2 cot g
− 2tg ⎥ − 4tg = 8
16
16 ⎦
8
⎣
π
π
⇔ 4 cot g − 4tg = 8
8
8
π
⇔ 8 cot g = 8 (hieån nhieân ñuùng)
4
Baøi 11 : Chöùng minh : 8 + 4tg
Baøi :12 : Chöùng minh :
⎛ 2π
⎞
⎛ 2π
⎞ 3
a/ cos2 x + cos2 ⎜
+ x ⎟ + cos2 ⎜
− x⎟ =
⎝ 3
⎠
⎝ 3
⎠ 2
1
1
1
1
+
+
+
= cot gx − cot g16x
b/
sin 2x sin 4x sin 8x sin16x
⎞
⎛ 2π
⎞
⎛ 2π
a/ Ta coù : cos2 x + cos2 ⎜
+ x ⎟ + cos2 ⎜
− x⎟
⎠
⎝ 3
⎠
⎝ 3
1
1⎡
4π ⎞ ⎤ 1 ⎡
⎞⎤
⎛
⎛ 4π
= (1 + cos 2x ) + ⎢1 + cos ⎜ 2x +
− 2x ⎟ ⎥
⎟ ⎥ + ⎢1 + cos ⎜
2
2⎣
3 ⎠⎦ 2 ⎣
⎝
⎝ 3
⎠⎦
3
+
2
3
= +
2
=
1⎡
4π ⎞
⎞⎤
⎛
⎛ 4π
cos 2x + cos ⎜ 2x +
− 2x ⎟ ⎥
⎟ + cos ⎜
⎢
2⎣
3 ⎠
⎠⎦
⎝
⎝ 3
1⎡
4π ⎤
cos 2x + 2 cos 2x cos ⎥
⎢
2⎣
3⎦
3 1⎡
⎛ 1 ⎞⎤
+ ⎢ cos 2x + 2 cos 2x ⎜ − ⎟ ⎥
2 2⎣
⎝ 2 ⎠⎦
3
=
2
cos a cos b sin b cos a − sin a cos b
−
=
b/ Ta coù : cot ga − cot gb =
sin a sin b
sin a sin b
=
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=
sin ( b − a )
sin a sin b
sin ( 2x − x )
1
=
(1 )
sin x sin 2x sin 2x
sin ( 4x − 2x )
1
=
cot g2x − cot g4x =
( 2)
sin 2x sin 4x sin 4x
sin ( 8x − 4x )
1
cot g4x − cot g8x =
=
( 3)
sin 4x sin 8x sin 8x
sin (16x − 8x )
1
cot g8x − cot g16x =
=
(4)
sin16x sin 8x sin16x
Laáy (1) + (2) + (3) + (4) ta ñöôïc
1
1
1
1
cot gx − cot g16x =
+
+
+
sin 2x sin 4x sin 8x sin16x
Do ñoù : cot gx − cot g2x =
Baøi 13 : Chöùng minh : 8sin3 180 + 8sin2 180 = 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 :
1
( 3 + cos 4x )
4
1
b/ sin 6x + cos 6x = ( 5 + 3cos 4x )
8
1
c/ sin8 x + cos8 x =
( 35 + 28 cos 4x + cos 8x )
64
a/ sin4 x + cos4 x =
a/ Ta coù: sin 4 x + cos4 x = ( sin 2 x + cos2 x ) − 2 sin 2 x cos2 x
2
2
sin2 2x
4
1
= 1 − (1 − cos 4 x )
4
3 1
= + cos 4x
4 4
=1−
b/ Ta coù : sin6x + cos6x
= ( sin 2 x + cos2 x )( sin 4 x − sin 2 x cos2 x + cos4 x )
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= ( sin4 x + cos4 x ) −
1
sin2 2x
4
⎛3 1
⎞ 1
= ⎜ + cos 4x ⎟ − (1 − cos 4x )
⎝4 4
⎠ 8
3
5
= cos 4x +
8
8
( do keát quaû caâu a )
c/ Ta coù : sin 8 x + cos8 x = ( sin 4 x + cos4 x ) − 2 sin 4 x cos4 x
2
1
2
2
( 3 + cos 4x ) − sin4 2x
16
16
2
1
1 ⎡1
⎤
2
=
( 9 + 6 cos 4x + cos 4x ) − 8 ⎢⎣ 2 (1 − cos 4x )⎥⎦
16
9 3
1
1
=
+ cos 4x +
(1 + cos 8x ) − (1 − 2 cos 4x + cos2 4x )
16 8
32
32
9 3
1
1
1
=
+ cos 4x +
cos 8x +
cos 4x −
(1 + cos 8x )
16 8
32
16
64
35 7
1
=
+
cos 4x +
cos 8x
64 16
64
=
Baøi 15 : Chöùng minh : sin 3x.sin3 x + cos 3x.cos3 x = cos3 2x
Caùch 1:
Ta coù : sin 3x.sin3 x + cos 3x.cos3 x = cos3 2x
= ( 3sin x − 4 sin 3 x ) sin 3 x + ( 4 cos3 x − 3 cos x ) cos3 x
= 3sin4 x − 4 sin6 x + 4 cos6 x − 3cos4 x
= 3 ( sin 4 x − cos4 x ) − 4 ( sin 6 x − cos6 x )
= 3 ( sin 2 x − cos2 x )( sin 2 x + cos2 x )
−4 ( sin 2 x − cos2 x )( sin 4 x + sin 2 x cos2 x + cos4 x )
= −3 cos 2x + 4 cos 2x ⎡⎣1 − sin 2 x cos2 x ⎤⎦
1
⎛
⎞
= −3 cos 2x + 4 cos 2x ⎜ 1 − sin 2 2x ⎟
4
⎝
⎠
1
⎡
⎛
⎞⎤
= cos 2x ⎢ −3 + 4 ⎜ 1 − sin 2 2x ⎟ ⎥
4
⎝
⎠⎦
⎣
= cos 2x (1 − sin 2 2x )
= cos3 2x
Caùch 2 :
Ta coù : sin 3x.sin3 x + cos 3x.cos3 x
⎛ 3sin x − sin 3x ⎞
⎛ 3 cos x + cos 3x ⎞
= sin 3x ⎜
⎟
⎟ + cos 3x ⎜
4
4
⎠
⎝
⎠
⎝
3
1
= ( sin 3x sin x + cos 3x cos x ) + ( cos2 3x − sin2 3x )
4
4
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3
1
cos ( 3x − x ) + cos 6x
4
4
1
= ( 3cos 2x + cos 3.2x )
4
1
= ( 3cos 2x + 4 cos3 2x − 3cos 2x ) ( boû doøng naøy cuõng ñöôïc)
4
= cos3 2x
3 +1
Baøi 16 : Chöùng minh : cos12o + cos18o − 4 cos15o.cos 21o cos 24 o = −
2
o
o
o
o
o
Ta coù : cos12 + cos18 − 4 cos15 ( cos 21 cos 24 )
=
= 2 cos15o cos 3o − 2 cos15o ( cos 45o + cos 3o )
= 2 cos15o cos 3o − 2 cos15o cos 45o − 2 cos15o cos 3o
= −2 cos15o cos 45o
= − ( cos 60o + cos 30o )
=−
3 +1
2
Baøi 17 : Tính P = sin2 50o + sin2 70 − cos 50o cos70o
1
1
1
Ta coù : P = (1 − cos100o ) + (1 − cos140o ) − ( cos120o + cos 20o )
2
2
2
1
1
1
⎛
⎞
P = 1 − ( cos100o + cos140o ) − ⎜ − + cos 20o ⎟
2
2⎝ 2
⎠
1 1
P = 1 − ( cos120o cos 20o ) + − cos 20o
4 2
5 1
1
5
P = + cos 20o − cos 20o =
4 2
2
4
Baøi 18 : Chöùng minh : tg30o + tg40o + tg50o + tg60o =
sin ( a + b )
cos a cos b
o
o
Ta coù : ( tg50 + tg40 ) + ( tg30o + tg60o )
AÙp duïng : tga + tgb =
sin 90o
sin 90o
=
+
cos 50o cos 40o cos 30o cos 60o
1
1
=
+
o
o
1
sin 40 cos 40
cos 30o
2
2
2
=
+
o
sin 80
cos 30o
1 ⎞
⎛ 1
= 2⎜
+
⎟
o
cos 30o ⎠
⎝ cos10
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8 3
cos 20o
3
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⎛ cos 30o + cos10o ⎞
= 2⎜
o
o ⎟
⎝ cos10 cos 30 ⎠
cos 20p cos10o
=4
cos10o cos 30o
8 3
=
cos 20o
3
Baøi 19 : Cho ΔABC , Chöùng minh :
A
B
C
cos cos
2
2
2
A
B
C
b/ socA + cos B + cos C = 1 + 4 sin sin sin
2
2
2
c/ sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
d/ cos2 A + cos2 B + cos2 C = −2 cos A cos B cos C
e/ tgA + tgB + tgC = tgA.tgB.tgC
f/ cot gA.cot gB + cot gB.cot gC + cot gC.cot gA = 1
A
B
C
A
B
C
g/ cot g + cot g + cot g = cot g .cot g .cot g
2
2
2
2
2
2
a/ sin A + sin B + sin C = 4 cos
a/ Ta coù : sin A + sin B + sin C = 2sin
A+B
A−B
cos
+ sin ( A + B )
2
2
A + B⎛
A−B
A + B⎞
+ cos
⎜ cos
⎟
2 ⎝
2
2 ⎠
C
A
B
⎛ A + B π C⎞
= 4 cos cos cos
= − ⎟
⎜ do
2
2
2
2
2 2⎠
⎝
A+B
A−B
cos
− cos ( A + B )
b/ Ta coù : cos A + cos B + cos C = 2 cos
2
2
A+B
A−B ⎛
A+B
⎞
= 2 cos
− ⎜ 2 cos2
− 1⎟
cos
2
2
2
⎝
⎠
A+B⎡
A−B
A + B⎤
+1
= 2 cos
cos
− cos
⎢
2 ⎣
2
2 ⎥⎦
A+B
A
⎛ B⎞
= −4 cos
sin sin ⎜ − ⎟ + 1
2
2
⎝ 2⎠
C
A
B
= 4 sin sin sin + 1
2
2
2
c/ sin 2A sin 2B + sin 2C = 2 sin ( A + B ) cos ( A − B ) + 2 sin C cos C
= 2 sin
= 2 sin C cos(A − B) + 2 sin C cos C
= 2sin C[cos(A − B) − cos(A + B) ]
= −4 sin Csin A sin( − B)
= 4 sin C sin A sin B
d/ cos2 A + cos2 B + cos2 C
1
= 1 + ( cos 2A + cos 2B ) + cos2 C
2
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= 1 + cos ( A + B ) cos ( A − B ) + cos2 C
= 1 − cos C ⎡⎣cos ( A − B ) − cos C ⎤⎦ do ( cos ( A + B ) = − cos C )
= 1 − cos C ⎡⎣cos ( A − B ) + cos ( A + B ) ⎤⎦
= 1 − 2 cos C.cos A.cos B
e/ Do a + b = π − C neân ta coù
tg ( A + B ) = −tgC
tgA + tgB
= −tgC
1 − tgAtgB
⇔ tgA + tgB = −tgC + tgAtgBtgC
⇔ tgA + tgB + tgC = tgAtgBtgC
f/ Ta coù : cotg(A+B) = - cotgC
1 − tgAtgB
= − cot gC
⇔
tgA + tgB
cot gA cot gB − 1
= − cot gC (nhaân töû vaø maãu cho cotgA.cotgB)
⇔
cot gB + cot gA
⇔ cot gA cot gB − 1 = − cot gC cot gB − cot gA cot gC
⇔ cot gA cot gB + cot gB cot gC + cot gA cot gC = 1
A+B
C
= cot g
g/ Ta coù : tg
2
2
A
B
tg + tg
2
2 = cot g C
⇔
A B
2
1 − tg tg
2
2
A
B
cot g + cot g
2
2 = cot g C (nhaân töû vaø maãu cho cotg A .cotg B )
⇔
A
B
2
2
2
cot g .cot g − 1
2
2
A
B
A
B
C
C
⇔ cot g + cot g = cot g cot g cot g − cot g
2
2
2
2
2
2
A
B
C
A
B
C
⇔ cot g + cot g + cot g = cot g .cot g .cot g
2
2
2
2
2
2
⇔
Baøi 20 : Cho ΔABC . Chöùng minh :
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
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Baøi 21 : Cho ΔABC . Chöùng minh :
cos3A + cos3B + cos3C = 1 - 4 sin
3A
3B
3C
sin
sin
2
2
2
Ta coù : (cos3A + cos3B) + cos3C
3
3
3C
= 2 cos (A + B) cos (A − B) + 1 − 2sin2
2
2
2
3
3
3C
Maø :
A + B = π − C neân ( A + B ) = π −
2
2
2
3
⎛ 3π 3C ⎞
=> cos ( A + B ) = cos ⎜
−
⎟
2
2 ⎠
⎝ 2
⎛ π 3C ⎞
= − cos ⎜ −
⎟
2 ⎠
⎝2
3C
= − sin
2
Do ñoù : cos3A + cos3B + cos3C
3 ( A − B)
3C
3C
= −2 sin
cos
− 2sin 2
+1
2
2
2
3 ( A − B)
3C ⎡
3C ⎤
= −2 sin
+ sin
⎢cos
⎥ +1
2 ⎣
2
2 ⎦
3 ( A − B)
⎤
3C ⎡
3
= −2 sin
− cos ( A + B ) ⎥ + 1
⎢cos
2 ⎣
2
2
⎦
−3B
3C
3A
sin
sin(
) +1
2
2
2
3C
3A
3B
= −4 sin
sin
sin
+1
2
2
2
= 4 sin
Baøi 22 : A, B, C laø ba goùc cuûa moät tam giaùc. Chöùng minh :
sin A + sin B − sin C
A B
C
= tg tg cot g
cos A + cos B − cos C + 1
2
2
2
A+B
A−B
C
C
2 sin
cos
− 2 sin cos
sin A + sin B − sin C
2
2
2
2
=
Ta coù :
A+B
A−B
cos A + cos B − cos C + 1
2 C
+ 2 sin
2 cos
cos
2
2
2
C⎡
A−B
C⎤
A−B
A+B
2 cos ⎢cos
− sin ⎥
cos
− cos
C
2⎣
2
2⎦
2
2
=
= cot g .
C⎡
A−B
C⎤
2 cos A − B + cos A + B
2 sin ⎢cos
+ sin ⎥
2
2
2⎣
2
2⎦
= cot g
C
.
2
A
⎛ B⎞
.sin ⎜ − ⎟
2
⎝ 2⎠
A
B
2 cos .cos
2
2
−2 sin
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= cot g
C
A
B
.tg .tg
2
2
2
Baøi 23 : Cho ΔABC . Chöùng minh :
A
B
C
B
C
A
C
A
B
sin 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 C
= sin sin sin + tg tg + tg tg + tg tg ( *)
2
2
2
2
2
2
2
2
2
A+B π C
C
⎛ A B⎞
= − vaäy tg ⎜ + ⎟ = cot g
2
2 2
2
⎝ 2 2⎠
A
B
tg + tg
2
2 = 1
⇔
A B
C
1 − tg tg
tg
2
2
2
B⎤ C
A B
⎡ A
⇔ ⎢ tg + tg ⎥ tg = 1 − tg tg
2⎦ 2
2
2
⎣ 2
A C
B C
A B
⇔ tg tg + tg tg + tg tg = 1 (1)
2
2
2
2
2
2
A
B
C
B
C
A
C
A
B
Do ñoù : (*) Ù sin cos cos + sin cos cos + sin cos cos
2
2
2
2
2
2
2
2
2
A
B
C
= sin sin sin + 1 (do (1))
2
2
2
A⎡
B
C
B
C⎤
A⎡
B
C
C
B⎤
⇔ sin ⎢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+C
+ cos sin
=1
⇔ sin cos
2
2
2
2
A+B+C
π
= 1 ⇔ sin = 1 ( hieån nhieân ñuùng)
⇔ sin
2
2
Ta coù :
Baøi 24 : Chöùng minh : tg
A
B
C 3 + cos A + cos B + cos C
+ tg + tg =
( *)
2
2
2
sin A + sin B + sin C
Ta coù :
A+B
A−B ⎡
C⎤
+ ⎢1 − 2 sin 2 ⎥ + 3
cos
2
2
2⎦
⎣
C
A−B
C
2sin cos
+ 4 − 2sin2
2
2
2
C⎡
A−B
C⎤
− sin ⎥ + 4
2 sin ⎢cos
2⎣
2
2⎦
C⎡
A−B
A + B⎤
− cos
+4
2 sin ⎢cos
2⎣
2
2 ⎥⎦
C
A
B
4 sin sin .sin + 4 (1)
2
2
2
cos A + cos B + cos C + 3 = 2 cos
=
=
=
=
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A+B
A−B
cos
+ sin C
2
2
C
A−B
C
C
= 2 cos cos
+ 2sin cos
2
2
2
2
C⎡
A−B
A + B⎤
= 2 cos ⎢ cos
+ cos
2⎣
2
2 ⎥⎦
C
A
B
= 4 cos cos cos (2)
2
2
2
sin A + sin B + sin C = 2sin
Töø (1) vaø (2) ta coù :
A
B
C
A
B
C
sin
sin
sin
sin sin sin + 1
2 +
2 +
2 =
2
2
2
(*) ⇔
A
B
C
A
B
C
cos
cos
cos
cos cos cos
2
2
2
2
2
2
A⎡
B
C⎤
B⎡
A
C⎤
C⎡
A
B⎤
⇔ sin ⎢cos cos ⎥ + sin ⎢cos cos ⎥ + sin ⎢cos cos ⎥
2⎣
2
2⎦
2⎣
2
2⎦
2⎣
2
2⎦
A
B
C
= sin sin sin + 1
2
2
2
A⎡
B
C
B
C⎤
A⎡
B
C
C
B⎤
⇔ sin ⎢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+C
+ cos sin
=1
⇔ sin .cos
2
2
2
2
⎡A + B + C⎤
⇔ sin ⎢
⎥⎦ = 1
2
⎣
π
⇔ sin = 1 ( hieån nhieân ñuùng)
2
A
B
C
sin
sin
sin
2
2
2
+
+
=2
Baøi 25 : Cho ΔABC . Chöùng minh:
B
C
C
A
A
B
cos cos
cos cos
cos cos
2
2
2
2
2
2
Caùch 1 :
A
B
A
A
B
B
sin
sin
sin cos + sin cos
2
2
2
2
2
2
+
=
Ta coù :
B
C
C
A
A
B
C
cos cos
cos cos
cos cos cos
2
2
2
2
2
2
2
A+B
A−B
sin
cos
1 sin A + sin B
2
2
=
=
A
B
C
2 cos A cos B cos C
cos cos cos
2
2
2
2
2
2
⎛ A − B⎞
C
A−B
cos ⎜
cos .cos
⎟
2 ⎠
⎝
2
2
=
=
A
B
C
A
B
cos .cos .cos
cos cos
2
2
2
2
2
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⎛ A − B⎞
C
A−B
A+B
cos ⎜
sin
cos
+ cos
⎟
⎝ 2 ⎠+
2
2
2
=
Do ñoù : Veá traùi =
A
B
A
B
A
B
cos cos
cos cos
cos cos
2
2
2
2
2
2
A
B
2 cos cos
2
2 =2
=
A
B
cos cos
2
2
Caùch 2 :
B+C
A+C
A+B
cos
cos
2
2
2
+
+
Ta coù veá traùi =
B
C
C
A
A
B
cos cos
cos cos
cos cos
2
2
2
2
2
2
B
C
B
C
A
C
A
C
cos cos − sin sin
cos cos − sin sin
2
2
2
2
2 +
2
2
2
=
B
C
C
A
cos cos
cos cos
2
2
2
2
A
B
A
B
cos cos − sin sin
2
2
2
2
+
A
B
cos cos
2
2
cos
Maø :
Do ñoù :
A C
A B⎤
⎡ B C
= 3 − ⎢ tg tg + tg tg + tg tg ⎥
2
2
2
2
2⎦
⎣ 2
A B
B C
A B
tg tg + tg tg + tg tg = 1
2
2
2
2
2
2
(ñaõ chöùng minh taïi baøi 10 )
Veá traùi = 3 – 1 = 2
A
B
C
, cot g , cot g theo töù töï taïo caáp soá coäng.
2
2
2
A
C
Chöùng minh cot g .cot g = 3
2
2
A
B
C
Ta coù : cot g , cot g , cot g laø caáp soá coäng
2
2
2
A
C
B
⇔ cot g + cot g = 2 cot g
2
2
2
A+C
B
sin
2 cos
2
2
=
⇔
A
C
B
sin sin
sin
2
2
2
Baøi 26 : Cho ΔABC . Coù cot g
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B
2
=
⇔
A
C
B
sin sin
sin
2
2
2
B
1
2
⇔
(do 0 0 )
=
A
C
A+C
2
sin sin
cos
2
2
2
A
C
A
C
cos cos − sin sin
2
2
2
2 = 2 ⇔ cot g A cot g C = 3
⇔
A
C
2
2
sin .sin
2
2
cos
B
2
2 cos
Baøi 27 : Cho ΔABC . Chöùng minh :
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
o
o
cos15 + sin15
= 3
b/
cos15o − sin15o
2π
4π
6π
1
+ cos
+ cos
=−
c/ cos
7
7
7
2
3
3
d/ sin 2x sin 6x + cos 2x.cos 6x = cos3 4x
e/ tg20o.tg40o.tg60o.tg80o = 3
π
2π
5π
π 8 3
π
+ tg
+ tg
+ tg =
cos
6
9
18
3
3
9
π
2π
3π
4π
5π
6π
7π 1
.cos .cos
.cos .cos .cos
=
g/ cos .cos
15
15
15
15
15
15
15 27
f/ tg
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