Tài liệu ôn thi đại học phần lượng giác - sách hay

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 sñAM=β vôùi 02≤β≤π
Ñaët k2,kZα=β+π∈
Ta ñònh nghóa: sinOKα= cosOHα= sintgcosαα=α vôùi co s0α≠coscotgsinαα=α vôùi sin0α≠
II. Baûng giaù trò löôïng giaùc cuûa moät soá cung (hay goùc) ñaëc bieät
Goùc α
Giaù trò
()o00 ()o306π ()o454π ()o603π ()o902π
sinα
0 12 22 32
1
cosα
1 32 22 12
0
tgα
0 33
1 3
||
cotgα
|| 3
1 33
0
III. Heä thöùc cô baûn
22sincos1α+α= 2211tgcos+α=α vôùi ()kkZ2πα≠+π∈ 221tcotgsin+=α vôùi ()kkZα≠π∈
IV. Cung lieân keát (Caùch nhôù: cos ñoái, sin buø, tang sai π; phuï cheùo)
a. Ñoái nhau: vaø ưα α
()sinsinưα=ưα
()coscosưα=α
()()tgtgưα=ưα
()()cotgcotgưα=ưα
b. Buø nhau: vaø απưα
()()()()sinsincoscostgtgcotgcotgπưα=απưα=ưαπưα=ưαπưα=ưα
c. Sai nhau : vaø π+ παα
()()()()sinsincoscostgtgcotgcotgπ+α=ưαπ+α=ưαπ+α=απ+α=α
d. Phuï nhau: vaø α2πưα sincos2cossin2tgcotg2cotgtg2π⎛⎞ưα=α⎜⎟⎝⎠π⎛⎞ưα=α⎜⎟⎝⎠π⎛⎞ưα=α⎜⎟⎝⎠π⎛⎞ưα=α⎜⎟⎝⎠
e.Sai nhau 2π: α vaø 2π+α sincos2cossin2tgcotg2cotgtg2π⎛⎞+α=α⎜⎟⎝⎠π⎛⎞+α=ưα⎜⎟⎝⎠π⎛⎞+α=ưα⎜⎟⎝⎠π⎛⎞+α=ưα⎜⎟⎝⎠
f.
()()()()()()+π=ư∈+π=ư∈+π=∈+π=kksinxk1sinx,kZcosxk1cosx,kZtgxktgx,kZcotgxkcotgx
V. Coâng thöùc coäng
()()()sinabsinacosbsinbcosacosabcosacosbsinasinbtgatgbtgab1tgatgb±=±±=±±=mm
VI. Coâng thöùc nhaân ñoâi ==ư=ư==ưư=222222sin2a2sinacosacos2acosasina12sina2cosa12tgatg2a1tgacotga1cotg2a2cotga
VII. Coâng thöùc nhaân ba:
33sin3a3sina4sinacos3a4cosa3cosa=ư=ư
VIII. Coâng thöùc haï baäc: ()()2221sina1cos2a21cosa1cos2a21cos2atga1cos2a=ư=+ư=+
IX. Coâng thöùc chia ñoâi
Ñaët attg2= (vôùi ak) 2≠π+π
22222tsina1t1tcosa1t2ttga1t=+ư=+=ư
X. Coâng thöùc bieán ñoåi toång thaønh tích ()()ababcosacosb2coscos22ababcosacosb2sinsin22ababsinasinb2cossin22ababsinasinb2cossin22sinabtgatgbcosacosbsinbacotgacotgbsina.sinb+ư+=+ưư=ư+ư+=+ưư=±±=±±=
XI. Coâng thöùc bieån ñoåi tích thaønh toång ()()()( ()()1cosa.cosbcosabcosab21sina.sinbcosabcosab21sina.cosbsinabsinab2=⎡++ư⎤⎣⎦ư=⎡+ưư⎣⎦=⎡++ư⎤⎣⎦
Baøi 1: Chöùng minh 4466sinacosa12sinacosa13+ư=+ư
Ta coù:
()24422222sinacosa1sinacosa2sinacosa12sinacosa+ư=+ưư=ư
Vaø:
()()()662242244422222222sinacosa1sinacosasinasinacosacosa1sinacosasinacosa112sinacosasinacosa13sinacosa+ư=+ư+=+ưư=ưưư=ư
Do ñoù: 44226622sinacosa12sinacosa2sinacosa13sinacosa3+ưư==+ưư
Baøi 2: Ruùt goïn bieåu thöùc ()221cosx1cosxA1sinxsinx⎡⎤ư+==+⎢⎥⎢⎥⎣⎦
Tính giaù trò A neáu 1cosx2=ư vaø x2π<<π
Ta coù: 2221cosxsinx12cosxcosxAsinxsinx⎛⎞++ư+=⎜⎟⎝⎠ ()221cosx1cosxA.sinxsinxư+⇔=
()223321cosx2sinx2Asinxsinxsinxư⇔=== (vôùi sinx0≠)
Ta coù: 2213sinx1cosx144=ư=ư=
Do: x2π<<π neân sin x0>
Vaäy 3sinx2=
Do ñoù 244Asinx33===
Baøi 3: Chöùng minh caùc bieåu thöùc sau ñaây khoâng phuï thuoäc x:
a. 4422A2cosxsinxsinxcosx3sinx=ư++
b. 2cotgxBtgx1cotgx1+=+ưư
a. Ta coù:
4422A2cosxsinxsinxcosx3sinx=ư++
()()()()24222242424A2cosx1cosx1cosxcosx31cosxA2cosx12cosxcosxcosxcosx33cosx⇔=ưư+ư+ư⇔=ưư++ư+ư
A2⇔= (khoâng phuï thuoäc x)
b. Vôùi ñieàu kieän sinx.cosx0,tgx1≠≠
Ta coù: 2cotgxBtgx1cotgx1 +=+ưư
1122 gxtgxB1tgx1tgx11tgx1tgx++⇔=+=+ưưư
()21tgx1tgxB1tgx1tgx1ưưư⇔===ưưư (khoâng phuï thuoäc vaøo x)
Baøi 4: Chöùng minh ()222222221cosa1cosacosbsinc1cotgbcotgccotga12sinasinasinbsinc⎡⎤ư+ưư+ư=⎢⎥⎢⎥⎣⎦
Ta coù:
* 222222cosbsinccotgb.cotgcsinb.sincưư 22222cotgb1cotgbcotgcsincsinb=ưư
()()22222cotgb1cotgc1cotgbcotgbcotgc=+ư+ư (1)
* ()221cosa1cosa12sinasina⎡⎤ư+ư⎢⎥⎢⎥⎣⎦ ()221cosa1cosa12sina1cosa⎡⎤ư+=ư⎢⎥ư⎢⎥⎣⎦ 1cosa1cosa12sina1cosa+ư⎡⎤=ư⎢⎥+⎣⎦ 1cosa2cosa.c ga2sina1cosa+==+ (2)
Laáy (1) + (2) ta ñöôïc ñieàu phaûi chöùng minh xong.
Baøi 5: Cho tuøy yù vôùi ba goùc ñeàu laø nhoïn. ABCΔ
Tìm giaù trò nhoû nhaát cuûa PtgA.tgB.tgC=
Ta coù: ABC+=πư
Neân: ()tgABtgC+=ư tgAtgBtgC1tgA.tgB+⇔=ư
tgAtgBtgCtgA.tgB.tgC⇔+=ư+
Vaäy: PtgA.tgB.tgCtgAtgBtgC==+
AÙp duïng baát ñaúng thöùc Cauchy cho ba soá döông tgA,tgB,tgC ta ñöôïc 3tgAtgBtgC3tgA.tgB.tgC++≥
3P3P⇔≥ 32P3P33⇔≥⇔≥
Daáu “=” xaûy ra ==⎧π⎪⇔⇔=⎨π<<⎪⎩tgAtgBtgCABC30A,B,C2
Do ñoù: MinP33ABC3π=⇔===
Baøi 6 : Tìm giaù trò lôùn nhaát vaø nhoû nhaát cuûa
a/ 84y2sinxcos2x=+
b/ 4ysinxcos=ư
a/ Ta coù : 441cos2xy2cos2x2ư⎛⎞=+⎜⎟⎝⎠
Ñaët vôùi thì tcos2x=1t1ư≤≤()441y1t8=ư+
=> ()331y'1t4t2=ưư+
Ta coù : () y'0=331t8tư=
⇔ 1t 2tư=
⇔ 1t3=
Ta coù y(1) = 1; y(-1) = 3; 11y32⎛⎞=⎜⎟⎝⎠
Do ñoù : ∈=xy3Max vaø ∈=x1yMin27
b/ Do ñieàu kieän : sin vaø co neân mieàn xaùc ñònh x0≥sx0≥π⎡⎤=π+π⎢⎥⎣⎦Dk2,k22 vôùi ∈k
Ñaët tcos= vôùi thì 0t1≤≤422tcosx1sin==ư
Neân 4sinx1t=ư
Vaäy 84y1t=ưư treân []D'0,1=
Thì ()ư=ư< ∀∈
Neân y giaûm treân [ 0, 1 ]. Vaäy data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEXh5PJm+yKVAAAAAXRSTlMAQObYZgAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=" class="mceSmilieSprite mceSmilie3" alt="" title="Frown ">)∈==xDmaxyy01, ()∈==ưxDminyy11
Baøi 7: Cho haøm soá 44ysinxcosx2msinxcos=+ư
Tìm giaù trò m ñeå y xaùc ñònh vôùi moïi x
Xeùt 44f(x)sinxcosx2msinxcosx=+ư
()()2222fxsinxcosxmsin2x2sinxcosx=+ưư
()21fx1sin2xmsin2x2=ưư
Ñaët : vôùi tsin2x=[]t1,∈ư
y xaùc ñònh ⇔ x∀()fx0xR≥∀∈
⇔ 211tmt0 ∀∈
⇔ ()2gtt2mt20=+ư≤[]t1,∀∈ư
Do neân g(t) coù 2 nghieäm phaân bieät t1, t2 2'm20Δ=+>m∀
Luùc ñoù t t1 t2
g(t) + 0 - 0
Do ñoù : yeâu caàu baøi toaùn ⇔ 12t11≤ư<≤
⇔ ⇔ ()()1g101g10ư≤⎧⎪⎨≤⎪⎩2m102m10ưư≤⎧⎨ư≤⎩
⇔ 1m21m2ư⎧≥⎪⎪⎨⎪≤⎪⎩ ⇔ 11m22ư≤≤
Caùch khaùc :
gt ()2 []t1, ∀∈
{}[,]max()max(),()tgtgg∈ư⇔≤
{}max),)mm⇔ưưư+≤21210⇔ 1m21m2ư⎧≥⎪⎪⎨ m⇔ư≤≤1122
Baøi 8 : Chöùng minh 4444357Asinsinsinsin161616162ππππ=+++
Ta coù : 7sin πππ⎛⎞=ư=⎜⎟⎝⎠55sincoscos1621616
Maët khaùc : ()244222 sin
2212sincos=ưα
211sin22=ưα
Do ñoù : 444473Asinsinsinsin16161616ππππ=+++
444433sincossincos16161616πππ⎛⎞⎛=+++⎜⎟⎜⎝⎠⎝
22111sin1sin2828ππ⎛⎞⎛=ư+ư⎜⎟⎜⎝⎠⎝
22132sinsin288ππ⎛⎞=ư+⎜⎟⎝⎠
2212sincos288 ⎛⎞⎜⎟
13222=ư=
Baøi 9 : Chöùng minh [IMG]data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAAEAAAABAQMAAAAl21bKAAAAA1BMVEXh5PJm+yKVAAAAAXRSTlMAQObYZgAAAApJREFUCNdjYAAAAAIAAeIhvDMAAAAASUVORK5CYII=" class="mceSmilieSprite mceSmilie9" alt=":eek:" title="Eek! :eek:">ooo16sin10.sin30.sin50.sin701=
Ta coù : ooAcos101Acos10cos10== (16sin10ocos10o)sin30o.sin50o.sin70o
⇔ ()ooo11 A8sin20cos40.cos202cos10⎛⎞=⎜⎟⎝⎠
⇔ ()0oo1 A4sin20cos20.cos40cos10=
⇔ ()ooo1A2sin40cos40cos10=
⇔ oooo1cos1 Asin801cos10cos10==
Baøi 10 : Cho ABCΔ. Chöùng minh : ABBCCAtgtgtgtgtgtg1222222++=
Ta coù : ABC22+π =ư
Vaäy : ABCtgcotg22+=
⇔ ABtgtg122ABC1tg.tgtg222+=ư
⇔ ABCAtgtgtg1tgtg2222⎡⎤+=ư⎢⎥⎣⎦
⇔ ACBCABtgtgtgtgtgtg1222222++
Baøi 11 : Chöùng minh : ()ππππ+++=84tg2tgtgcotg*8163232
Ta coù
8cotgtg2tg4tg3232168πππ=ưưư 22cosasinacosasinacotgatgasinacosasinacosaưư=ư= cos2a2cotg2a1sin2a2==
Do ñoù : cotgtg2tg4tg83232168π⎡⎢⇔ 2cotg2tg4tg816168πππ⎡⎤ưư⎢⎥⎣⎦
⇔ 4cotg4tg8⇔ π= 8cotg8
⇔
4
Baøi :12 : Chöùng minh : 22222cosxcosxcosx33ππ⎛⎞⎛⎞+++ư⎜⎟⎜⎟⎝⎠⎝⎠
a/ 1111cotgxcotg16x
sin2 22222cosxcosxcosx33ππ⎛⎞⎛+++ư⎜⎟⎜⎝⎠⎝ ()114141cos2x1cos2x1c
o 3144cos2xcos2x
c 314cos2x2cos2xcos223π⎡⎤=++⎢⎥⎣⎦ 311cos2x2cos2x222⎡⎤⎛⎞=++ư⎜⎟⎢⎥⎝⎠⎣⎦
3= cosacosbsinbcosasinacosbcotga
cot ()sinbasinasinbư
= ()()sin2xx1cotgxcotg2x1sinxsin2xsin2xưư==
()()sin4x2x1cotg2xcotg4x2sin2xsin4xsin4xưư==
()()sin8x4x1cotg4xcotg8x3sin4xsin8xsin8xưư==
()()
sincotg8xcotg16xư=1111cotgxcotg16xsin2xsin4xsin8xsin16xư=+++
: Chöùng minh : 38sin18
+=
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 (do 0 < sin18
Caùch khaùc :
( 1 ) ⇔ 8sin2180
()
a/ 44si+=
b/ ()1sin6 ()881
sin ()244222sinxcosxsinxcosx2sinxcosx+=+ư
a/ Ta coù 221sin24=ư
()111cos44=ưư
31cos4x44=+
()(224224sinxcosxsinxsinxcosxcosx=+ư+
()4421sinxcosxsin2x4=+ư ()311cos4x1cos4x448⎛⎞=+ưư⎜⎟⎝⎠ 35cos4x88=+
()+=+ư288444sinxcosxsinxcosx2sinxcosx
()=+ư24123cos4xsin2x1616 ()()⎡⎤=++ưư⎢⎥⎣⎦2211196cos4xcos4x1cos4x1682 ()()29311cos4x1cos8x12cos4xcos4x1683232=+++ưư+ ()