226 bài tập Lượng giác lớp 10 (Có đáp án)

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Mời các bạn học sinh cùng tham khảo tài liệu dưới đây với 226 bài tập Lượng giác lớp 10 có lời giải nhằm giúp các em học sinh tự học, tự ôn thi hiệu quả, chuẩn bị tốt nhất cho kỳ thi sắp tới. Đồng thời đây cũng là tài liệu tham khảo hữu ích dành cho các thầy cô giáo trong việc biên soạn đề thi và giáo án. Mời các bạn cùng tham khảo.

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Nội dung Text: 226 bài tập Lượng giác lớp 10 (Có đáp án)

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 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 ( ) 30o π ( ) 45o π ( ) 60o π ( ) 90o 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 + tg2α = vôùi α ≠ + kπ ( k ∈ Z ) cos α 2 2 1 t + cot g2 = 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 ) www.MATHVN.com
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 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 ⎠ 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 tga ± tgb tg ( a ± b ) = 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 sin 2 a = (1 − cos2a ) 2 1 cos2 a = (1 + cos2a ) 2 1 − cos2a tg2a = 1 + cos2a IX. 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 − 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 + 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 ) ⎤ 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 2 x + 1 − 2 cos x + cos2 x ⎞ 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 x 3 sin x sin x 3 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 + 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 − 1 Baøi 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 < 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 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 < 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 www.MATHVN.com
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 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 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 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 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 π Baøi 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) 4 Baøi :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 sin16x 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 a/ sin4 x + cos4 x = ( 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/ Ta coù: sin 4 x + cos4 x = ( sin 2 x + cos2 x ) − 2 sin 2 x cos2 x 2 2 =1− sin2 2x 4 1 = 1 − (1 − cos 4 x ) 4 3 1 = + cos 4x 4 4 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 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 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 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 =− 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 8 3 Baøi 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 3 Baøi 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[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 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 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 : 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 2 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 : = 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 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 ⎛ 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 C Baøi 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 sin Baøi 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 C Baøi 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 2 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 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