Tài liệu Lượng giác toàn tập

MATHVN.COM 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 ) www.MATHVN.com ( ) 1 0 MATHVN.COM 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 ⎠ www.MATHVN.com MATHVN.COM 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 www.MATHVN.com MATHVN.COM 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 www.MATHVN.com MATHVN.COM 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 www.MATHVN.com MATHVN.COM 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 ⇔ www.MATHVN.com MATHVN.COM 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 www.MATHVN.com x∈ D MATHVN.COM 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 www.MATHVN.com MATHVN.COM 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 : www.MATHVN.com MATHVN.COM ⇔ 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 = www.MATHVN.com MATHVN.COM = 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 ) www.MATHVN.com MATHVN.COM = ( 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 www.MATHVN.com MATHVN.COM 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 www.MATHVN.com 8 3 cos 20o 3 MATHVN.COM ⎛ 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 www.MATHVN.com MATHVN.COM = 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 www.MATHVN.com MATHVN.COM 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 www.MATHVN.com MATHVN.COM = 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 = = = = www.MATHVN.com MATHVN.COM 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 www.MATHVN.com MATHVN.COM ⎛ 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 www.MATHVN.com MATHVN.COM 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 www.MATHVN.com