alimov-9-gdz (Алгебра - 9 класс - Алимов), страница 7

0 < α <π, то2cosα > 0, тогдаcos α = 1 − sin 2 α =2;3πππcos + α  = cos ⋅ cos α − sin ⋅ sin α =333=1 23 12− 3⋅−⋅=.2 3232 3π2) Т.к. < α < π , то2sinα > 0, тогда2sin α = 1 − cos α = 1 −1 2 2=;93πππ1 2 2 22 4− 2+⋅=.cos α −  = cos α ⋅ cos + sin α ⋅ sin = − ⋅4443 2326296.1) cos3α ⋅ cosα – sinα ⋅ sin3α = cos(3α + α) = cos4α;2) cos5β ⋅ cos2β + sin5β ⋅ sin2β = cos(5β – 2β) = cos3β;π 5ππ 5π3) cos + α  cos- α  sin  + α  sin -α  = 147 1475πππ= cos + α +- α  = cos = 0 ;1427 7π 2π 7π 2π4) cos+ α  ⋅ cos+ α  + sin + α  ⋅ sin +α  = 5 5 5 52π 7π= cos+α −− α  = cos π = −1 .55297. ππ1) cos(α + β ) + cos − α  cos − β  = cos α ⋅ cos β −22 − sin α ⋅ sin β + sin α ⋅ sin β = cosα ⋅ cos β ;96ππ π π2) sin  - α  sin  - β  - cos(α − β ) =  sin ⋅ cosα − cos ⋅ sin α x2222 π πx  sin ⋅ cos β − cos ⋅ sin β  - (cos α ⋅ cos β + sin α ⋅ sin β ) =22= cosα ⋅ cos β − cosα ⋅ cos β − sin α ⋅ sin β = − sin α ⋅ sin β .298.1) sin73° ⋅ cos17° + cos73° ⋅ sin17° = sin(73° + 17°)=sin90°=1;2) sin73° ⋅ cos13° – cos73° ⋅ sin13° = sin(73° – 13°)=sin60°=3;25π π5π5ππππ+ sin ⋅ cos=sin  +  =sin =1;⋅ cos12121212121227π π 7π7ππππ4) sin– sin ⋅ cos=sin −  =sin =1.⋅ cos121212122 12 12 3) sin299.1) Т.к.

π < α <3π, то2sinα < 0, тогдаsin α = − 1 − cos 2 α = − 1 −94=− ;255πππ4 3 3 1− ⋅ =sin α +  = sin α ⋅ cos + cos α ⋅ sin = − ⋅6665 2 5 2=4 3 +3−4 3 −3=−.10102) Т.к.π< α < π , то2cosα < 0, тогдаcosα = – 1 − sin 2 α = − 1 −27;=93ππ2  7 2 2π⋅−⋅=sin − α  = sin ⋅ cos α − cos ⋅ sin α =442  3  2 34=− 14 − 214 + 2.=−6697300.1) sin(α + β) + sin( – α)cos( – β) = sinα⋅cosβ ++ cosα⋅sinβ – sinα⋅cosβ = cosα ⋅ sinβ;2) cos( – α)sin( – β) – sin(α – β) == – cosα⋅ sinβ – ( sinα⋅cosβ – cosα⋅sinβ) = – cosα⋅sinβ – sinα⋅cosβ ++ cosα⋅sinβ = – sinα⋅cosβ;3)πππ π3) cos − α  sin  − β  − sin(α − β ) =  cos cos α + sin sin α  ×222 2π π×  sin cos β − cos sin β  − sin α cos β + cos α sin β = sin α cos β −22− sin α cos β + cos α sin β = cos α sin β ;π4) sin (α + β ) + sin  − α  sin(− β ) = sin α cos β + cos α sin β −2− cos α sin β = sin α cos β .301.Т.к.3π< α < 2π , то2cosα > 0,тогдаcosα = 1 − sin 2 α = 1 −Т.к.

0 < β <94= .25 5π, то2cosβ > 0,тогдаcos β = 1 − sin 2 α = 1 −6415;=289 17cos(α + β) = cosα⋅cosβ – sinα⋅sinβ =4 153 8 60 24 84= ⋅ −  −  ⋅ = + = ;5 17  5  17 85 85 854 153 8 60 24 36−=.cos(α – β) = ⋅ +  −  ⋅ =5 17  5  17 85 85 8598302.Т.к.π< α < π , то sinα > 0;2sinα = 1 − cos 2 α = 1 − 0,64 = 0,36 = 0,6 .3πТ.к. π < β <,2то cosβ < 0;cosβ = – 1 − sin 2 α = − 1 −1445=− ;16913sin(α – β) = sinα⋅cosβ – cosα⋅sinβ =5 12 −15 48 63− =− . − (−0,8) ⋅  −  = 13 65 65 65 13= 0,6 ⋅  −303.2π2ππ21) cos π − α  + cosα +  = cos⋅ cos α + sin⋅ sin α +3333+ cos1313ππ⋅ cosα − sin ⋅ sin α = − cosα +sin α + cosα −sin α = 0 ;2222332 2π2ππ2) sinα + π  − sin − α  = sinα ⋅ cos + cosα ⋅ sin −333 31331ππ− sin ⋅ cosα + cos ⋅ sinα = − sinα +cosα −cosα + sinα = 0 ;2233222 cos α sin β + sin(α − β )2 cos α sin β + sin α cos β − cos α sin β3)==2 cos α cos β − cos(α − β ) 2 cos α cos β − cos α cos β − sin α sin β=cos α sin β + sin α cos βsin(α + β )== tg (α + β ) ;cos α cos β − sin α sin β cos(α + β )4)cos α cos β − cos(α + β ) cos α cos β − cos α cos β + sin α sin β==cos(α − β ) − sin α sin βcos α cos β + sin α sin β − sin α sin β=sin α sin β= tg α ⋅ tg β .cos α cos β304.1) sin(α – β)⋅cos(α + β) = (sinαcosβ – cosα sinβ)( sinαcosβ ++ cosα sinβ) = sin2α cos2β – cos2α sin2β = sin2α(1 – sin2β) – (1 ––sin2α) sin2β = sin2α – sin2α⋅ sin2β – sin2β + sin2α ⋅ sin2β=sin2α – sin2β;992) sin(α – β)⋅cos(α + β) = (cosαcosβ – sinα sinβ)( cosαcosβ ++ sinα sinβ) = cos2α cos2β – sin2α sin2β = cos2α(1 – sin2β) –– (1 – cos2α) sin2β = cos2α – cos2α⋅ sin2β – sin2β + cos2α ⋅ sin2β == cos2α – sin2β; 22π2 cos α − 2cos α +sin α 2 cos α − 2 cos − α 24 23)==π132 sin  + α  − 3 sin α2 cos α +sin α  − 3 sin α2622 cos α − 2 cos α − 2 sin α=cos α + 3 sin α − 3 sin α=− 2 sin α= − 2tgα ;cos α13πcos α − 2 cos α −sin α cos α − 2 cos + α 223=4)=π 312 sin α −  − 3 sin α 2sin α + cos α  − 3 sin α6 22cos α − cos α + 3 sin α=3 sin α − cos α − 3 sin α=3 sin α= − 3tgα .− cos α305.1) cos6x ⋅ cos5x + sin6x ⋅ sin5x = – 1;cos (6x – 5x) = – 1.

Тогда cos x = – 1;x = π + 2πn, n ∈ ∧;2) sin3x ⋅ cos5x – sin5x ⋅ cos3x = – 1;sin (3x – 5x) = – 1; – sin2x = – 1;sin2x = 1. Значит, 2x =π+ 2πn;2x=π+ 2πn,n ∈ ∧;43)π2 cos + x  − cos x = 1 ;4 222cos x −sin x  − cos x = 1 ; 22cos x – sin x – cos x = 1;sin x = – 1. Поэтому x = –100π+ 2πn, n ∈ ∧;2xπ x 2 sin  −  + sin = 1 ;2 4 24) 22xxx2cos −sin  + sin = 1 ; 22222xxx− sin + sin = 1 ;222xcos = 1 .2xЗначит, = 2πn и x = 4πn, n ∈ ∧.2cos306.1)tg 29° + tg31°= tg ( 29° + 31°) = tg 60° = 3 ;1 − tg 29° ⋅ tg31°7π3π− tg1616 = tg 7π − 3π  = tg π = 1 .2)7π3π4 16 16 ⋅ tg1 + tg1616tg307.sinα cos β cosα sin β+cosα cos β cosα cos β tgα + tgβsin(α + β ) sinα cosβ + cosα sin β===;1)sinα cos β cosα sin βtgα − tgβsin(α − β ) sinα cosβ − cosα sin β−cosα cos β cosα cos βcosα cos β sinα sin β+sinα sin β sinα sin β ctgα ⋅ ctgβ + 1cos(α − β ) cosα cosβ + sinα sin β==.2)=cosα cos β sinα sin β ctgα ⋅ ctgβ − 1cos(α + β )ctgα ⋅ ctgβ −1−sinα sin β sinα sin β308.1) 2sin15°cos15° = sin2 ⋅ 15° = sin30° =1;22) cos215° – sin215° = cos2 ⋅ 15° – cos30° =3;23) (cos75° – sin75°)2 = cos275° – 2sin75°cos75° + sin275° == 1 – sin150° = 1 – sin30° = 1 –11= ;224) (cos15° + sin15°)2 = cos215° + 2sin15°cos15° + sin215° =31= 1 + sin30° = 1 + = .22101309.π81) 2 sin cosπ83) sin cos22πππππ; 2) cos 2 − sin 2 = cos =;= sin =84288422 1π 1 1π 1+ = sin + =+ =8 4 24 4442 +1;424)=ππππ2 2 −  cos + sin  =− 1 + 2 sin cos  =2 882 882 2 2 22π− 1 + sin  =− 1+=−1−= −1 .2 42 2 22310.1) Т.к.π< α < π , то cos α < 0, тогда294=− ;2553 424sin2α = 2 sin αcos α = 2 ⋅  −  = − .5 5253π2) Т.к.

π < α <, то2cosα = − 1 − sin 2 α = − 1 −sin α < 0, тогда sin α = –1−163=− ;2553424sin2α = 2sin αcos α = 2 ⋅  −  ⋅  −  = − . 5  525311.1) sin2α = 1 – cos2α;2)cos2α = 1 – sin2α;sin2α = 1 –cos2α = 1 –169=.25 25916=.25 25Т.к. cos2α = cos2α – sin2α, тоТ.к.

cos2α = cos2α – sin2α, тоcos2α =cos2α =16 97;−=25 25 25312.2 sin α cosα sin 2α;=22sin 2απ2) cosα cos − α  = sin α cosα =;221) sin α cosα =10216 97.−=25 25 253) cos4α + sin22α = cos22α – sin22α + sin22α = cos22α;4) sin2α + (sinα – cosα)2 = 2sinαcosα + sin2α – 2sinαcosα ++ cos2α = 1.313.cos 2α + 1 cos 2 α − sin 2 α + cos 2 α + sin 2 α 2 cos 2 α=== cosα ;2 cosα2 cosα2 cosαsin 2α2 sin α cosα 2 cosα2)=== 2ctgα ;2sin α1 − cos αsin 2 α1)sin 2 α3)=(sin α + cos α )2 − 1=sin 2 αsin 2 α + 2 sin α cos α + cos 2 α − 1=sin 2 α1= tgα ;2 sin α cosα 24)1 + cos 2α cos 2 α + sin 2 α + cos 2 α − sin 2 α 2 cos 2 α2=== ctg α .21 − cos 2α cos 2 α + sin 2 α − cos 2 α + sin 2 α2 sin α314.1) (sinα + cosα)2 – 1 = 1 + 2sinαcosα – 1 = 2sinαcosα = sin2α;2) (sinα – cosα)2 = sin2α – 2sinαcosα + cos2α = 1 – sin2α;3) cos4α – sin4α = (cos2α – sin2α)( cos2α + sin2α) = cos2α;4) 2cos2α – cos2α = 2cos2α – cos2α + sin2α = cos2α + sin2α = 1.315.1) sinα + cosα =1.2Возведем в квадрат.Получим: (sinα + cosα)2 =1;4311; sin2α = – 1 = – .44412) sinα – cosα = – .31 + 2sinαcosα =Возведем в квадрат(sinα – cosα)2 =8111; 1 – 2sinαcosα = ; sin2α = 1 – = .9999316.1) 1 + cos2α = sin2α + cos2α + cos2α – sin2α = 2cos2α;2) 2sin2α = sin2α + cos2α – cos2α + sin2α = 1 – cos2α.103317.1) 2 cos215° – 1 = 2 cos215° – (sin215° + cos215°) == cos215° – sin215° = cos30° =3;22) 1 – sin222,5° = sin222,5° + cos222,5° – 2sin222,5° == cos222,5° – sin222,5° = cos45° =3) 2 cos 2= cosππ ππππ− 1 = 2 cos 2 −  cos 2 + sin 2  = cos 2 − sin 2 =88 88882π;=424) 1 - 2 sin 2= cos2;2ππππππ- 2 sin 2- sin 2= cos 2+ sin 2= cos 2=1212121212123π.=62318.1) 1 – 2sin25α=sin25α + cos25α – 2sin25α=cos25α – sin25α=cos10α;2) 2cos23α – 1 = 2cos23α – (sin23α + cos23α) == cos23α – sin23α = cos6α;1 − cos 2αsin 2 α + cos 2 α − cos 2 α + sin 2 α 4 sin 2 α=== 4 sin α ;αα1sin αsin cossin α222αα2α2α2 cos- 1 2 cos- cos 2 - sin 2cos α12222 =4).==2 sin α ⋅ cos α2 sin α cos α 2 sin αsin 2α3)319.1)cos 2α2sin α cos α + sin αcos α sin α=−= ctg − 1;sin α sin α2)sin 2α − 2 cosα2sin α − sin α==(cos α − sin α )(cos α + sin α ) = cos α − sin αsin α (cos α + sin α )sin α2 cosα (sin α − 1)2 cosα=−= −2ctgα ;sin α (1 − sin α )sin α3) tgα ⋅ (1 + cos 2α ) = tgα ⋅ (cos 2 α + sin 2 α + cos 2 α − sin 2 α ) ==sin α⋅ 2 cos 2 α = 2 sin α cos α = sin 2α ;cos α104=4)1 − cos 2α + sin 2α=1 + cos 2α + sin 2αcos 2 α + sin 2 α − cos 2 α + sin 2 α + 2 sin α cos α cos α⋅=cos 2 α + sin 2 α + cos 2 α − sin 2 α + 2 sin α cos α sin α2 sin α (sin α + cos α ) ⋅ cos α== 1.2 cos α (sin α + cos α ) ⋅ sin α=320.1) sin2x – 2cosx = 0;2cosx ⋅ sinx – 2cosx = 0;2cosx (sinx – 1) = 0.πx = + πncos x = 0;;Тогда 2sin x − 1 = 0 sin x = 1π x = 2 + πn, n ∈ Z. x = π + 2πn, n ∈ Z2π+ πn .22) cos2x + 3sinx = 1;cos2x – sin2x + 3sinx – sin2x – cos2x = 0;3sinx – 2sin2x = 0;sinx ( – 2sinx + 3) = 0;n∈Zsin x = 0 x = πn,.− 2 sin x + 3 = 0; sin x = 1,5− нет решенияОтвет:Ответ: πn; n ∈ ∧.3) 2sinx = sin2x;2sinx – 2sinx ⋅ cosx = 0;2sinx (1 – cosx) = 0;sin x = 0 x = πn, n ∈ Z  x = πn, n ∈ Z; .1 − cos x = 0; cos x = 1 x = 2πnОтвет: πn.4) sin2x = – cos2x;sin2x + cos2x – sin2x = 0;cos2x = 0;cosx = 0.πОтвет: + πn ; n ∈ ∧.2105321.Т.к.

tg 2α =2 tgα21 − tg α, то tg 2α =2 ⋅ 0,61,2 1207===1 .1 − 0,36 0,64 648322.2 tg1)π8π1 − tg82= tg6 tg15°1π= 1 ; 2)= 3 ⋅ tg30° = 3 ⋅= 3.2431 − tg 15°323.1) sin13πππ = sin  6π +  = sin = 1 ;2222) sin17π = sin (18π – π) = – sinπ = 0;3) cos7π = cos (8π – π) = cosπ = – 1;4) cos11πππ = cos 6π −  = cos = 0 ;2225) sin720° = sin (2 ⋅ 360°) = 0;6) cos540° = cos (360° + 180°) = cos 180° = – 1.324.1) cos420° = cos (360° + 60°) = cos60° =2) tg570° = tg (3 ⋅ 180° + 30°) = tg30° =1;2133) sin3630° = sin (10⋅ 360° + 30°) = sin30° =4) ctg960° = ctg (5 ⋅ 180° + 60°) = ctg60° =;1;213;13πππ 1= sin  2π +  = sin = ;666 211π1ππ= tg 2π −  = − tg = −6) tg.66635) sin325.1) cos150° = cos (90° + 60°) = – sin60° = –2) sin135° = sin (90° + 45°) = cos45° =1062;23;23) cos120° = – cos60° = –1;24) sin315° = sin (360° – 45°) = – 45° = –2.2326.5πππ= tg π +  = tg = 1 ;4447π1ππsin= sin  π +  = − sin = − ;66625πππ 1cos= cos 2π −  = cos = ;333 211πππ 1sin  − = − sin  2π −  = sin = ;66 2 6 ππ 1 7π cos − = cos 2π +  = cos = ;33 2 3 ππ 2π tg − = − tg π −  = tg = 3 .33 3 1) tg2)3)4)5)6)327.1) cos630° – sin1470° – ctg1125° = cos(720° – 90°) –– sin(1440° + 30°) – ctg(1080° + 45°) = cos90° – sin30° – ctg45° =31=0– –1=– ;222) tg1800° – sin495° + cos945° = 0 – sin135° + cos225° == –sin(90° + 45°)+cos(180° + 45°)= –cos45° –cos45° = – 2⋅2=–22;31π7π− tg= − sin(6π + π ) −34ππππ− 2 cos10π +  − tg  2π −  = − sin π − 2 cos + tg =34341= 0 − 2 ⋅ + 1 = −1 + 1 = 0 ;2π 21π  49π 4) cos( −9π ) + 2 sin  − = cos π − 2 sin  8π +  + − ctg  −6 6 4 πππ1+ ctg  5π +  = −1 − sin + ctg = −1 − 2 ⋅ + 1 = −1 − 1 + 1 = −1 .46423) sin( −7π ) − 2 cos107328.1) cos2(π – α) + sin2(α – π) = cos2α + sin2α = 1;2) cos(π – α)cos(3π – α) – sin(α – π)sin(α – 3π) == cos(π – α)cos(3π – α) – sin(π – α)sin(3π – α) = cos(π – α + 3π – α) == cos(4π – 2α) = cos2α.329.1) cos723° + sin900° = cos(360°⋅20 + 30°) + sin(360°⋅2 + 180°) == cos30° + sin180° =33;+0 =222) sin300° + tg150° = sin(360° – 60°) + tg(180° – 30°) == – sin60° – tg30° = –33 −5 3;−=2363) 2 sin 6,5π − 3 sin19πππ= 2 sin  6π +  - 3 sin  6π +  =323ππ33 1= 2− = ;- 3 sin = 2 − 3 ⋅2322 2161ππ 1π4)) 2 cos 4,25π −coscos10π +  == 2 cos 4π +  −= 2 sin36436ππ2131 11−cos = 2 ⋅−⋅= 1− = ;4622 233 2π− sin  6π +  − tg (6π + π )sin(−6,5π ) + tg (−7π )25)==cos(−7π ) + ctg (−16,25π )πcos(6π + π ) − ctg 16π + 4π− sin − tgπ−1− 0 12=== ;π−1−1 2cos π − ctg4cos(−540°) + sin 480° cos(720° − 180°) + sin(360° + 120°)6)==tg 405° − ctg 330°tg (360° + 45°) − ctg (360° − 30°)= 2 cos3−1 +( 3 − 2)(1 − 3 ) 5 − 3 3cos 180° + sin 120°3 −22 =.====tg 45° + ctg 30°41+ 32(1 + 3 ) 2(1 + 3 )(1 − 3 )108330.πsin  − α  + sin(π − α )cosα + sin α2== −1 ;1)cos(π − α ) + sin( 2π − α ) − cosα − sin απcos(π − α ) + cos − α 2 − cosα + sin α2)== 1;πsin α − cosαsin(π − α ) − sin  − α 2− sin α ⋅ ( − tgα )sin(α − π ) tg (π − α )⋅== 1;3)tg (α + π )tgα ⋅ sin απcos − α 2πsin 2 (π − α ) + sin  − α 2sin 2 α + cos 2 α4)⋅ tg (π − α ) =⋅ (−tgα ) =sin(π − α )sin α=1  sin α 1⋅−.=−sin α  cosα cosα331.Пусть α , β , γ – углы треугольника,sinγ = sin(180° – (α + β)) = sin180°⋅cos (α + β) –– cos180° ⋅ sin (α + β) = 0⋅cos (α + β) – ( – 1)⋅sin (α + β) = sin (α + β).332.1)πππsin  + α  = sin ⋅ cos α + cos ⋅ sin α =222= 1 ⋅ cos α + 0 ⋅ sin α = cos α ;2)πππcos + α  = cos ⋅ cos α − sin ⋅ sin α =222= 0 ⋅ cos α − 1 ⋅ sin α = − sin α ;3)3π3π 3πcos− α  = cos⋅ cos α + sin⋅ sin α =222= 0 ⋅ cos α + (−1) ⋅ sin α = − sin α ;4)3π3π 3πsin − α  = sin⋅ cos α − cos⋅ sin α =22 2= −1 ⋅ cos α − 0 ⋅ sin α = − cos α .109333.π1) cos − x  = 1 ;2sinx = 1.πТогда x = + 2πn, n ∈ Z .43) cos (x – π) = 0;cosx = 0.Поэтому x =π+ 2πn, n ∈ Z .42) sin (π – x) = 1;sinx = 1.π+ 2πn,4π4) sin  x −  = 1 ;2– cosx = 1;Значит x =n∈Z .cosx = – 1.Тогда x = π + 2πn, n ∈ Z.334.ππ + α  - cos - α  =441) sin 2222cos α +sin α −cos α −sin α = 0 ;2222=π6π32) cos - α  - sin  + α  =3131cos α + sin α −cos α − sin α = 0 .2222=336.1) I четв.;3) III четв.;5) II четв.337.1) sin3π = 0; cos3π = – 1;3) sin3,5π = – 1;2) III четв.;4) IV четв.;2) sin4π = 0; cos4π = 1;π 5π 4) sin   = sin = 1 ;2 2 5π=0;2cos3,5π = 0;cos5) sinπn = 0;n − четное 1,cos πn = ;−1,n−нечетное6) sin ((2n + 1)π) = 0;110cos ((2n + 1)π) = – 1, n ∈ Z.338.3π= 0 – 0 = 0;21) sin3π – cos2) cos0 – cos3π + cos3,5π = 1 – ( – 1) + 0 = 2;3) sinπk + cos2πk = 0 + 1 = 1;4) cos(2k + 1)π(4k + 1) π= 0 - 1 = −1 .- sin22339.1) Т.к.π< α < π , то cosα < 0, тогда2cosα = − 1 − sin 2 α = − 1 −16.=−333π, то tgα < 0,22) Т.к.