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MATHVN.COM CHÖÔNG 1: COÂNG THÖÙC LÖÔÏNG GIAÙCI. 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 αII. Baûng giaù trò löôïng giaùc cuûa moät soá cung (hay goùc) ñaëc bieät Goùc α ( ) 0 0o π ( ) 30o π ( ) 45o π ( ) 60o π ( ) 90oGiaù trò 6 4 3 2sin α 0 1 2 3 1 2 2 2cos α 1 3 2 1 0 2 2 2tgα 0 3 1 3 || 3cot gα || 3 1 3 0 3III. Heä thöùc cô baûn sin 2 α + cos2 α = 1 1 π 1 + tg2α = vôùi α ≠ + kπ ( k ∈ Z ) cos α 2 2 1 t + cot g2 = vôùi α ≠ kπ ( k ∈ Z ) sin 2 αIV. 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 MATHVN.COMb. Buø nhau: α vaø π − αsin ( π − α ) = sin αcos ( π − α ) = − cos αtg ( π − α ) = − tgαcot g ( π − α ) = − cot gαc. Sai nhau π : α vaø π + αsin ( π + α ) = − sin αcos ( π + α ) = −cosαtg ( π + α ) = t gαcot g ( π + α ) = cot gα πd. 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 πe.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 ⎠ 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 gxV. Coâng thöùc coäng sin ( a ± b ) = sin a cos b ± sin b cosa cos ( a ± b ) = cosa cos b ∓ sin asin b tga ± tgb tg ( a ± b ) = 1 ∓ tgatgbVI. 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 gaVII. Coâng thöùc nhaân ba: sin3a = 3sin a − 4sin3 a cos3a = 4 cos3 a − 3cosaVIII. Coâng thöùc haï baäc: 1 sin 2 a = (1 − cos2a ) 2 1 cos2 a = (1 + cos2a ) 2 1 − cos2a tg2a = 1 + cos2aIX. Coâng thöùc chia ñoâi a Ñaët t = tg (vôùi a ≠ π + k2 π ) 2 www.MATHVN.com MATHVN.COM 2t sin a = 1 + t2 1 − t2 cosa = 1 + t2 2t tga = 1 − t2X. 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 bXI. 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 2Baøi 1: Chöùng minh = 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 1 + cos x ⎡ (1 − cos x ) ⎤ 2Baøi 2: Ruùt goïn bieåu thöùc A = = ⎢1 + ⎥ sin x ⎢⎣ sin 2 x ⎥⎦ 1 πTính giaù trò A neáu cos x = − vaø 0 2 3 Vaäy sin x = 2 2 4 4 3 Do ñoù A = = = sin x 3 3Baøi 3: Chöùng minh caùc bieåu thöùc sau ñaây khoâng phuï thuoäc x: a.

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A = 2 cos4 x − sin 4 x + sin2 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 + sin2 x cos2 x + 3sin2 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.cosx ≠ 0,tgx ≠ 1 2 cot gx + 1 Ta coù: B = + tgx − 1 cot gx − 1 www.MATHVN.com MATHVN.COM 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 − 1Baøi 4: Chöùng minh 1 + cosa ⎡ (1 − cosa ) ⎤ cos2 b − sin 2 c 2 ⎢1 − 2 ⎥+ 2 2 − cot g2 b cot g2 c = cot ga − 1 2sin a ⎢ sin a ⎥ sin bsin c ⎣ ⎦ Ta coù: cos2 b − sin 2 c * − cot g2 b.cot g2 c sin b.sin c 2 2 cotg2 b 1 = − 2 − cot g2 b cot g2 c sin c sin b 2 ( ) ( ) = cot g2 b 1 + cot g2 c − 1 + cot g2 b − cot g 2 b cot g2 c = −1 (1) 1 + cosa ⎡ (1 − cosa ) ⎤ 2 * ⎢1 − ⎥ 2sin a ⎢ sin 2 a ⎥ ⎣ ⎦ 1 + cosa ⎡ (1 − cosa ) ⎤ 2 = ⎢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= ⎪⎩ 0 y ” = − (1 − t ) + 4t 3 2 Ta coù : y ” = 0 Ù (1 − t ) = 8t 3 3 ⇔ 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 MATHVN.COMBaøi 7: Cho haøm soá y = sin4 x + cos4 x − 2m sin x cos xTìm 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 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 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 2Baøi 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 ABaøi 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 www.MATHVN.com MATHVN.COM A C B C A B ⇔ tg tg + tg tg + tg tg = 1 2 2 2 2 2 2 π π π πBaøi 11 : Chöùng minh : 8 + 4tg + 2tg + tg = cot g ( *) 8 16 32 32 π π π π Ta coù : (*) ⇔ 8 = cot g − tg − 2tg − 4tg 32 32 16 8 cos a sin a cos a − sin a 2 2 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) 4Baøi :12 : Chöùng minh : ⎛ 2π ⎞ ⎛ 2π ⎞ 3 a/ cos2 x + cos2 ⎜ + x ⎟ + cos2 ⎜ − x⎟ = ⎝ 3 ⎠ ⎝ 3 ⎠ 2 1 1 1 1 b/ + + + = cot gx − cot g16x 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 + ⎟ ⎥ + ⎢1 + cos ⎜ − 2x ⎟ ⎥ 2 2⎣ ⎝ 3 ⎠⎦ 2 ⎣ ⎝ 3 ⎠⎦ 3 1⎡ ⎛ 4π ⎞ ⎛ 4π ⎞⎤ = + ⎢ cos 2x + cos ⎜ 2x + ⎟ + cos ⎜ − 2x ⎟ ⎥ 2 2⎣ ⎝ 3 ⎠ ⎝ 3 ⎠⎦ 3 1⎡ 4π ⎤ = + ⎢ cos 2x + 2 cos 2x cos ⎥ 2 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 Do ñoù : cot gx − cot g2x = = (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 sin16xBaøi 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 MATHVN.COM 1 = ( sin4 x + cos4 x ) − sin2 2x 4 ⎛3 1 ⎞ 1 = ⎜ + cos 4x ⎟ − (1 − cos 4x ) ( do keát quaû caâu a ) ⎝4 4 ⎠ 8 3 5 = cos 4x + 8 8 c/ Ta coù : sin 8 x + cos8 x = ( sin 4 x + cos4 x ) − 2 sin 4 x cos4 x 2 1 2 ( 3 + cos 4x ) − sin4 2x 2 = 16 16 2 1 1 ⎡1 ⎤ = 16 ( 9 + 6 cos 4x + cos 4x ) − 8 ⎢⎣ 2 (1 − cos 4x )⎥⎦ 2 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 64Baøi 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 +1Baøi 16 : Chöùng minh : cos12o + cos18o − 4 cos15o.cos 21o cos 24 o = − 2 Ta coù : cos12 + cos18 − 4 cos15 ( cos 21 cos 24 ) o o o o o = 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 =− 2Baøi 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 8 3Baøi 18 : Chöùng minh : tg30o + tg40o + tg50o + tg60o = cos 20o 3 sin ( a + b ) AÙp duïng : tga + tgb = cos a cos b Ta coù : ( tg50 + tg40 ) + ( tg30o + tg60o ) o o sin 90o sin 90o = + cos 50o cos 40o cos 30o cos 60o 1 1 = + sin 40 cos 40 o o 1 cos 30o 2 2 2 = + sin 80o cos 30o ⎛ 1 1 ⎞ = 2⎜ + ⎟ ⎝ cos10 cos 30o ⎠ o www.MATHVN.com MATHVN.COM ⎛ cos 30o + cos10o ⎞ = 2⎜ o ⎟ ⎝ cos10 cos 30 ⎠ o cos 20p cos10o =4 cos10o cos 30o 8 3 = cos 20o 3Baøi 19 : Cho ΔABC, Chöùng minh : A B C a/ sin A + sin B + sin C = 4 cos 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+B A−B a/ Ta coù : sin A + sin B + sin C = 2sin cos + sin ( A + B ) 2 2 A + B⎛ A−B A + B⎞ = 2 sin ⎜ 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 b/ Ta coù : cos A + cos B + cos C = 2 cos cos − cos ( A + B ) 2 2 A+B A−B ⎛ A+B ⎞ = 2 cos cos − ⎜ 2 cos2 − 1⎟ 2 2 ⎝ 2 ⎠ A+B⎡ A−B A + B⎤ = 2 cos ⎢ cos − cos +1 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 C cos(A − B) + 2 sin C cos C = 2sin C = −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 g/ Ta coù : tg = cot g 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 2Baøi 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 = – 4cosAcosBcosC Do ñoù : cos2A + cos2B + cos2C + 1 + 4cosAcosBcosC = 0 www.MATHVN.com MATHVN.COMBaøi 21 : Cho ΔABC. Chöùng minh : 3A 3B 3C cos3A + cos3B + cos3C = 1 – 4 sin 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 3C 3 ( A − B) 3C = −2 sin cos − 2sin 2 +1 2 2 2 3C ⎡ 3 ( A − B) 3C ⎤ = −2 sin ⎢cos + sin ⎥ +1 2 ⎣ 2 2 ⎦ 3C ⎡ 3 ( A − B) 3 ⎤ = −2 sin ⎢cos − cos ( A + B ) ⎥ + 1 2 ⎣ 2 2 ⎦ 3C 3A −3B = 4 sin sin sin( ) +1 2 2 2 3C 3A 3B = −4 sin sin sin +1 2 2 2Baøi 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 : = cos A + cos B − cos C + 1 A+B A−B 2 C 2 cos cos + 2 sin 2 2 2 C⎡ A−B C⎤ A−B A+B 2 cos ⎢cos − sin ⎥ cos − cos 2⎣ 2 2⎦ C 2 2 = = cot g. C⎡ A−B C⎤ 2 cos A − B + cos A + B 2 sin ⎢cos + sin ⎥ 2⎣ 2 2⎦ 2 2 A ⎛ B⎞ −2 sin .sin ⎜ − ⎟ C 2 ⎝ 2⎠ = cot g. 2 A B 2 cos .cos 2 2 www.MATHVN.com MATHVN.COM C A B = cot g .tg .tg 2 2 2Baøi 23 : Cho ΔABC. Chöùng minh : A B C B C A C A Bsin 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 ⎛ A B⎞ C Ta coù : = − 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 ⎡ 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 (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 ⇔ sin cos + cos sin =1 2 2 2 2 A+B+C π ⇔ sin = 1 ⇔ sin = 1 ( hieån nhieân ñuùng) 2 2 A B C 3 + cos A + cos B + cos CBaøi 24 : Chöùng minh : tg + tg + tg = ( *) 2 2 2 sin A + sin B + sin C Ta coù : A+B A−B ⎡ C⎤ cos A + cos B + cos C + 3 = 2 cos cos + ⎢1 − 2 sin 2 ⎥ + 3 2 2 ⎣ 2⎦ C A−B C = 2sin cos + 4 − 2sin2 2 2 2 C⎡ A−B C⎤ = 2 sin ⎢cos − sin ⎥ + 4 2⎣ 2 2⎦ C⎡ A−B A + B⎤ = 2 sin ⎢cos − cos +4 2⎣ 2 2 ⎥⎦ C A B = 4 sin sin .sin + 4 (1) 2 2 2 www.MATHVN.com MATHVN.COM A+B A−B sin A + sin B + sin C = 2sin 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 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 ⇔ sin .cos + cos sin =1 2 2 2 2 ⎡A + B + C⎤ ⇔ sin ⎢ ⎥⎦ = 1 ⎣ 2 π ⇔ sin = 1 ( hieån nhieân ñuùng) 2 A B C sin sin sinBaøi 25 : Cho ΔABC. Chöùng minh: 2 + 2 + 2 =2 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 Ta coù : 2 + 2 = 2 2 2 2 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 = = 2 cos A cos B cos C A cos cos cos B C 2 2 2 2 2 2 C A−B ⎛ 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 Do ñoù : Veá traùi = ⎝ 2 ⎠+ 2 = 2 2 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 cos Ta coù veá traùi = 2 + 2 + 2 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 ⎡ B C A C A B⎤ = 3 − ⎢ tg tg + tg tg + tg tg ⎥ ⎣ 2 2 2 2 2 2⎦ A B B C A B Maø : tg tg + tg tg + tg tg = 1 2 2 2 2 2 2 (ñaõ chöùng minh taïi baøi 10 ) Do ñoù : Veá traùi = 3 – 1 = 2 A B CBaøi 26 : Cho ΔABC. Coù cot g, 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 www.MATHVN.com MATHVN.COM B B cos 2 cos ⇔ 2 = 2 A C B sin sin sin 2 2 2 1 2 B ⇔ = (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 2Baøi 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 TAÄP1. 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 www.MATHVN.com

Chuyên mục: Tổng hợp

Bạn đang xem: Bài tập lượng giác lớp 10 cơ bản có đáp án MATHVN.COM CHÖÔNG 1: COÂNG THÖÙC LÖÔÏNG GIAÙCI. 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 αII. Baûng giaù trò löôïng giaùc cuûa moät soá cung (hay goùc) ñaëc bieät Goùc α ( ) 0 0o π ( ) 30o π ( ) 45o π ( ) 60o π ( ) 90oGiaù trò 6 4 3 2sin α 0 1 2 3 1 2 2 2cos α 1 3 2 1 0 2 2 2tgα 0 3 1 3 || 3cot gα || 3 1 3 0 3III. Heä thöùc cô baûn sin 2 α + cos2 α = 1 1 π 1 + tg2α = vôùi α ≠ + kπ ( k ∈ Z ) cos α 2 2 1 t + cot g2 = vôùi α ≠ kπ ( k ∈ Z ) sin 2 αIV. 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 MATHVN.COMb. Buø nhau: α vaø π − αsin ( π − α ) = sin αcos ( π − α ) = − cos αtg ( π − α ) = − tgαcot g ( π − α ) = − cot gαc. Sai nhau π : α vaø π + αsin ( π + α ) = − sin αcos ( π + α ) = −cosαtg ( π + α ) = t gαcot g ( π + α ) = cot gα πd. 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 πe.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 ⎠ 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 gxV. Coâng thöùc coäng sin ( a ± b ) = sin a cos b ± sin b cosa cos ( a ± b ) = cosa cos b ∓ sin asin b tga ± tgb tg ( a ± b ) = 1 ∓ tgatgbVI. 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 gaVII. Coâng thöùc nhaân ba: sin3a = 3sin a − 4sin3 a cos3a = 4 cos3 a − 3cosaVIII. Coâng thöùc haï baäc: 1 sin 2 a = (1 − cos2a ) 2 1 cos2 a = (1 + cos2a ) 2 1 − cos2a tg2a = 1 + cos2aIX. Coâng thöùc chia ñoâi a Ñaët t = tg (vôùi a ≠ π + k2 π ) 2 www.MATHVN.com MATHVN.COM 2t sin a = 1 + t2 1 − t2 cosa = 1 + t2 2t tga = 1 − t2X. 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 bXI. 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 2Baøi 1: Chöùng minh = 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 1 + cos x ⎡ (1 − cos x ) ⎤ 2Baøi 2: Ruùt goïn bieåu thöùc A = = ⎢1 + ⎥ sin x ⎢⎣ sin 2 x ⎥⎦ 1 πTính giaù trò A neáu cos x = − vaø 0 2 3 Vaäy sin x = 2 2 4 4 3 Do ñoù A = = = sin x 3 3Baøi 3: Chöùng minh caùc bieåu thöùc sau ñaây khoâng phuï thuoäc x: a.Xem thêm: Điểm Chuẩn Học Viện Kỹ Thuật Quân Sự 2019, Điểm Chuẩn Học Viện Kỹ Thuật Quân Sự A = 2 cos4 x − sin 4 x + sin2 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 + sin2 x cos2 x + 3sin2 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.cosx ≠ 0,tgx ≠ 1 2 cot gx + 1 Ta coù: B = + tgx − 1 cot gx − 1 www.MATHVN.com MATHVN.COM 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 − 1Baøi 4: Chöùng minh 1 + cosa ⎡ (1 − cosa ) ⎤ cos2 b − sin 2 c 2 ⎢1 − 2 ⎥+ 2 2 − cot g2 b cot g2 c = cot ga − 1 2sin a ⎢ sin a ⎥ sin bsin c ⎣ ⎦ Ta coù: cos2 b − sin 2 c * − cot g2 b.cot g2 c sin b.sin c 2 2 cotg2 b 1 = − 2 − cot g2 b cot g2 c sin c sin b 2 ( ) ( ) = cot g2 b 1 + cot g2 c − 1 + cot g2 b − cot g 2 b cot g2 c = −1 (1) 1 + cosa ⎡ (1 − cosa ) ⎤ 2 * ⎢1 − ⎥ 2sin a ⎢ sin 2 a ⎥ ⎣ ⎦ 1 + cosa ⎡ (1 − cosa ) ⎤ 2 = ⎢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= ⎪⎩ 0 y ” = − (1 − t ) + 4t 3 2 Ta coù : y ” = 0 Ù (1 − t ) = 8t 3 3 ⇔ 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 MATHVN.COMBaøi 7: Cho haøm soá y = sin4 x + cos4 x − 2m sin x cos xTìm 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 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 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 2Baøi 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 ABaøi 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 www.MATHVN.com MATHVN.COM A C B C A B ⇔ tg tg + tg tg + tg tg = 1 2 2 2 2 2 2 π π π πBaøi 11 : Chöùng minh : 8 + 4tg + 2tg + tg = cot g ( *) 8 16 32 32 π π π π Ta coù : (*) ⇔ 8 = cot g − tg − 2tg − 4tg 32 32 16 8 cos a sin a cos a − sin a 2 2 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) 4Baøi :12 : Chöùng minh : ⎛ 2π ⎞ ⎛ 2π ⎞ 3 a/ cos2 x + cos2 ⎜ + x ⎟ + cos2 ⎜ − x⎟ = ⎝ 3 ⎠ ⎝ 3 ⎠ 2 1 1 1 1 b/ + + + = cot gx − cot g16x 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 + ⎟ ⎥ + ⎢1 + cos ⎜ − 2x ⎟ ⎥ 2 2⎣ ⎝ 3 ⎠⎦ 2 ⎣ ⎝ 3 ⎠⎦ 3 1⎡ ⎛ 4π ⎞ ⎛ 4π ⎞⎤ = + ⎢ cos 2x + cos ⎜ 2x + ⎟ + cos ⎜ − 2x ⎟ ⎥ 2 2⎣ ⎝ 3 ⎠ ⎝ 3 ⎠⎦ 3 1⎡ 4π ⎤ = + ⎢ cos 2x + 2 cos 2x cos ⎥ 2 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 Do ñoù : cot gx − cot g2x = = (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 sin16xBaøi 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 MATHVN.COM 1 = ( sin4 x + cos4 x ) − sin2 2x 4 ⎛3 1 ⎞ 1 = ⎜ + cos 4x ⎟ − (1 − cos 4x ) ( do keát quaû caâu a ) ⎝4 4 ⎠ 8 3 5 = cos 4x + 8 8 c/ Ta coù : sin 8 x + cos8 x = ( sin 4 x + cos4 x ) − 2 sin 4 x cos4 x 2 1 2 ( 3 + cos 4x ) − sin4 2x 2 = 16 16 2 1 1 ⎡1 ⎤ = 16 ( 9 + 6 cos 4x + cos 4x ) − 8 ⎢⎣ 2 (1 − cos 4x )⎥⎦ 2 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 64Baøi 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 +1Baøi 16 : Chöùng minh : cos12o + cos18o − 4 cos15o.cos 21o cos 24 o = − 2 Ta coù : cos12 + cos18 − 4 cos15 ( cos 21 cos 24 ) o o o o o = 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 =− 2Baøi 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 8 3Baøi 18 : Chöùng minh : tg30o + tg40o + tg50o + tg60o = cos 20o 3 sin ( a + b ) AÙp duïng : tga + tgb = cos a cos b Ta coù : ( tg50 + tg40 ) + ( tg30o + tg60o ) o o sin 90o sin 90o = + cos 50o cos 40o cos 30o cos 60o 1 1 = + sin 40 cos 40 o o 1 cos 30o 2 2 2 = + sin 80o cos 30o ⎛ 1 1 ⎞ = 2⎜ + ⎟ ⎝ cos10 cos 30o ⎠ o www.MATHVN.com MATHVN.COM ⎛ cos 30o + cos10o ⎞ = 2⎜ o ⎟ ⎝ cos10 cos 30 ⎠ o cos 20p cos10o =4 cos10o cos 30o 8 3 = cos 20o 3Baøi 19 : Cho ΔABC, Chöùng minh : A B C a/ sin A + sin B + sin C = 4 cos 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+B A−B a/ Ta coù : sin A + sin B + sin C = 2sin cos + sin ( A + B ) 2 2 A + B⎛ A−B A + B⎞ = 2 sin ⎜ 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 b/ Ta coù : cos A + cos B + cos C = 2 cos cos − cos ( A + B ) 2 2 A+B A−B ⎛ A+B ⎞ = 2 cos cos − ⎜ 2 cos2 − 1⎟ 2 2 ⎝ 2 ⎠ A+B⎡ A−B A + B⎤ = 2 cos ⎢ cos − cos +1 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 C cos(A − B) + 2 sin C cos C = 2sin C

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