alimov-9-gdz (Алгебра - 9 класс - Алимов), страница 8

π < α <1т.к. 1 + tg2α =2cos α1, то tgα =2cos α−1 =922 5−1 ==.555112π3) Т.к. 0 < α < , то sinα > 0, ctgα =;==tgα 2 2421 + ctg2α =1sin α4) Т.к. π < α <1 + tg2α =cosα = –1sinα =221 + ctg α1=1+18=8=92 2.33π, то cosα < 0,21,cos 2 αtgα =121 + ctg α=111+212==−2;22 − 6.=33340.1) 5sin2α + tgα ⋅ cosα + 5cos2α == 5 (sin2α + cos2α) +sin α⋅ cosα = 5 + sinα;cosα2) ctgα ⋅ sinα – 2cos2α – 2sin2α ==sin α22⋅ sin α – 2 (sin α + cos α) = cosα – 2;cosα1113)4)321 + tg α51 + ctg 2α22= 3 cos α . Т.к.

cos α == 5 sin 2 α . Т.к. sin2α =11 + tg 2α;11 + ctg 2α.341.ππ1) 2sin( – α) ⋅ cos  − α  – 2cos( – α) ⋅ sin  − α  =222= – 2sinα ⋅ sinα – 2cosα ⋅ cosα = – 2sin α – 2cos2α == – 2(sin2α + cos2α) = – 2;ππ− α  + 3sin2  − α  =222) 3sin(π – α)cos = 3sinα ⋅ sinα + 3cos2α = 3(sin2α + cos2α) = 3;3) (1 – tg( – α)) ⋅ (1 – tg(π + α))cos2α = (1 + tgα)(1 – tgα) ⋅ cos2α == (1 – tg2α) ⋅ cos2α = cos2α – sin2α = cos2α;1 = (1 + tg2α) ⋅=2+−α1ctg()1ctgα+4) (1 + tg2( – α))⋅ =(1 + tg 2α ) ⋅ tg 2α1 + tg 2α122= tg α .342. 3π  3π+ α  = cos α − cos α = −2 cos α .- α  + sin  2 21) sin 11, то значение выражения равно – .42ππ32) cos + α  + cos − α  = − sin α − sin α = −2 sin α .2 211Т.к. sinα = , то значение выражения равно – .63Т.к.

cosα =343.1) 2sin75° ⋅ cos75° = sin150° = sin (180° – 150°) = sin30° =2) cos275° – sin275° = cos150° = – cos (180° – 150°) == – cos30° = –1123;21;22 ( 3 − 1)2 32 16− 2;⋅−⋅ ==2 22 2442⋅ 322 ( 3 + 1)6+ 2+==.4) sin75° = sin(45° + 30°) =2⋅22⋅2443) sin15°=sin(45° – 30°) =344.π1) cos2(π – α) – cos2  − α  = cos2α – sin2α = cos2α;2ππ2) 2sin  − α  cos  − α  = 2 ⋅ cosα ⋅ sinα = sin2α;22cos 2 ( 2π + α ) − sin 2 ( 2π + α ) cos 2 α − sin 2 α cos 2α=== ctg 2α ;2 cosα sin αsin 2απ2 cos(2π + α ) cos − α 2π2 sin(π − α ) sin  − α 2 cosα sin αsin 2α24)=== tg 2α .22πcos2α22cossinα−αsin  α −  − sin (α − π )23)345.47π1π π π= sin 8π −  = sin −  = sin −  = − ;662 6 6ππ25π2) tg= tg 6π +  = tg = 1 ;1) sin4443) ctg27ππ π π= ctg 7π −  = ctg −  = ctg −  = −1 ;44 4 44) cos21π2πππ.= cos 5π +  = cos π +  = − cos = −44442346.1)cosπ23π15π π π π= cos 6π −  - sin  4π  = cos  −  − sin  −  =- sin44444 4ππ22+ sin =+= 2;442225π10πππππ2) sin− tg= sin  8π +  − tg  3π +  = sin − tg =333333= cos=33;− 3=−221133) 3cos3660° + sin( – 1560°) = 3cos(10 ⋅ 360° + 60°) ++ sin( – 120° – 4 ⋅ 360°) = 3⋅cos60° – sin120° = 3⋅=1– sin60° =233 3− 3−=;2224) cos( – 945°) + tg1035° = cos( – 3 ⋅ 360° + 135°) ++ tg(2,5 ⋅ 360° + 135°) = cos135° + tg135° = – cos45° – tg45° ==–22+ 2−1 = −.22347.1) sin3 > cos4,т.к.

sin3 > 0, cos4 < 0.2) cos0 > sin5,т.к. sin5 < 0, cos0 = 1.348.1) sin 3,5 ⋅ tg3,5 =sin 2 3,5< 0 , т.к. sin23,5>0, cos3,5<0;cos 3,52) cos5,01 ⋅ sin0,73 > 0, т.к. cos5,01>0, sin0,73>0;3)tg13< 0 , т.к. tg13>0, cos15<0;cos154) sin1 ⋅ cos2 ⋅ tg3 >0, т.к. sin1>0, cos2 и tg3<0.349.1) sin3π3ππππ π 3π ⋅ cos+ sin⋅ cos = sin  + = sin = 1 ;88888 282) sin165° = sin (120° + 45°) = sin120°⋅ cos45° + cos120° ⋅ sin45° =2 3 −13 2 1 26− 2;=⋅− ⋅==2 2 2 2443) sin105° = sin (60° + 45°) = sin60°⋅ cos45° + cos60°⋅ sin45° =2 3 +13 2 1 26+ 2;=⋅+ ⋅==2 2 2 2444) sin=()()ππππππ π = sin  −  = sin ⋅ cos − cos ⋅ sin =1234343 4()2 3 −13 2 1 26− 2⋅− ⋅==;2 2 2 2445) 1 – sin2195° = cos2195° – sin2195° = cos390° = cos(360° + 30°) == cos30° =1143;26) 2 cos 2= cos3π3π3π3π3π3π- 1 = 2 cos 2- cos 2- sin 2= cos 2- sin 2=8888883π2=−;42350.1) (1 + tg( – α)) ⋅ (1 – ctg( – α) –sin( −α )= (1 – tgα) ⋅ (1 + ctgα) +cos(−α )+ tgα = 1 + ctgα – tgα – 1 + tgα = ctgα;ctgα + tg (−α ) tg (−α )ctgα − tgα12)+=−=cosα + sin(−α ) sinαcosα − sinα cosα=cos2 α − sin2 α1cosα + sinα1−=−=cosα ⋅ sinα (cosα − sinα ) cosα cosα ⋅ sinα cosα=cosα1=.cosα ⋅ sinα sinα351.Т.к.π< α < π , то cosα < 0, тогда cosα = –2tgα =1 − sin 2 α = − 1 −52=− ;935sin α3 = − 5 ; ctg α = 1 = − 2 ;=tgαcosα2−2535  24 5⋅−  = −;3  394 51cos2 α = cos2α – sin2 α = − = − ;9 99sin 2α = 2sin α cosα = 2 ⋅352.1) cos3α ⋅ sinα – sin3α ⋅ cosα = cosα ⋅ sinα(cos2α – sin2α) ==11sin2α ⋅ cos2α = sin4α;242)sin α + sin 2αsin α (1 + 2 cosα ) sin α (1 + 2 cosα )=== tgα .1 + cosα + cos 2αcosα (1 + 2 cosα )2 cos 2 α + cosα353.sin 2α − sin 2α ⋅ cos 2α sin 2α(1 − cos 2α) 2 sin α cos α ⋅ 2 sin 2 α1)=== sin3 α ;4 cos α4 cos α4 cos α1152 cos2 2α2 cos2 2α==sin 4α ⋅ cos 4α + sin 4α sin 4α (cos 4α + 1)11 ;2 cos2 2α===22 sin 2α cos 2α (2 cos 2α ) 2 sin 2α cos 2α sin 4αcos 2α + sin 2α ⋅ cos 2α cos 2α (1 + sin 2α )3)==2 sin 2 α − 1sin 2 α − cos 2 αcos 2α (1 + sin 2α )== −(1 + sin 2α ) ;- cos 2α(cos α − sin α ) 21 − 2 cos α sin α==4)sin 2α ⋅ cos 2α − cos 2α cos 2α (sin 2α − 1)−(sin 2α − 1)−1.==cos 2α (sin 2α − 1) cos 2α2)354.cos 2 xcos 2 x − sin x(1 − sin x) 1 − sin x− sin(π − x) ==1)=1;1 − sin x(1 − sin x)1 − sin x2)cos 2 xcos 2 x + sin x(1 + sin x) 1 + sin x+ cos(1,5π + x) === 1;1 + sin x1 + sin x1 + sin x3)sin 2 xsin 2 x + cos x(1 − cos x) 1 + cos x− sin(1,5π + x) === 1;1 + cos x1 + cos x1 + cos x4)sin 2 xsin 2 x − cos x(1 − cos x) 1 − cos x+ cos(3π − x) ===1.1 − cos x1 − cos x1 − cos x355.1tg 2α + 1 (sin 2 α + cos 2 α ) cosα===tgαtgαcos 2 α ⋅ sin απ112, т.к.

α = − , то===1 sin 2α sin 2α12cosα sin α21) tgα + ctgα = tgα + π 612= −4 ;и значение выражения равно2-122) ctgα − tgα =cosα sin α cos 2 α − sin 2 αcos 2α−=== 2ctg 2α .1 sin αsin α cosαcosα ⋅ sin α2sin2 α = sin  −  = –Т.к. α = −116ππ, то 2ctg2 α = 2ctg (- ) = 2 ;84cosαsin α+=cosα + sin α cosα − sin α3)=cos 2 α − cos α ⋅ sin α + cos α ⋅ sin α + sin 2 α22cos α − sin α=1.cos 2α1π=Т.к. α = − , тоcos2α611== 2; π 1cos − 2 3sin αcosα−=cosα + sin α cosα − sin α4)=sin α ⋅ cos α − sin 2 α − cos 2 α − cos α ⋅ sin α22cos α − sin απ1=Т.к.

α = , то3cos 2α=−1.cos 2α−1- 1==2.2π-1cos23356.π2 sin  + α  sin(π − α ) + cos 2α − 12cos 2α + sin α ⋅ cosα − cos 2 α==2 cosα sin α + cos 2 α − sin 2 α − cos 2 α − sin 2 α222cos α − sin α + sin α ⋅ cosα − cos α=2 sin α (cosα − sin α )=2sin α (cosα − sin α )357.3π 1) sin(2 x + 3π ) sin  x + − sin 3x cos 2 x = −1 ;2 – sin2x ⋅ ( – cos3x) – sin3x cos2x = – 1; sin (3x – 2x) = 1, т.е. sinx=1.Тогда x =π+ 2πn ,2n∈Z.3π) ⋅ cos(2 x + 4π ) − sin(5 x + π ) sin 2 x = 0 ;2πТогда 3 x = + πn, n ∈ Zcos 5 x ⋅ cos 2 x + sin 5 x sin 2 x = 0;2cos(5 x − 2 x) = 1;cos 3x = 0.π πnи x= +, n ∈ Z.6 32) sin(5 x −117358.1) tg (α + β) =3tgα + tgβ, т.к.

tg α = – , tgβ = 2,4 , то41 − tgα tgβ3 + 2,41,65 165 334;===31+2,4 2,8 280 5641432) ctg (α + β) =. Т.к. ctg α = , то tg α = ,tg (α + β)34tg (α + β) =т.к. ctg β = – 1, то tg β = – 1; tg (α + β) ==tgα + tgβ=1 − tgα ⋅ tgβ3 −1−144 = − 1 , поэтому ctg (α + β) = – 7.=3371 − ⋅ (−1) 144359. π ππ ππ1) 2 sin  + 2α  sin  − 2α  = 2 sin  + 2α  sin  −  + 2α   =44424  π ππ= 2 sin  + 2α  cos + 2α  = sin  + 4α  = cos 4α ;2 44 π ππππ2) 2 cos + 2α  ⋅ cos − 2α  = 2 cos + 2α  cos −  + 2α   = 2 4444π ππ= 2 sin  + 2α  cos + 2α  = sin  + 4α  = cos 4α ;4 42πππ ππ− α ) − cos 2 ( + α ) = cos 2 ( − ( + α )) − cos 2 ( + α ) =44244πππ= sin 2 ( + α ) − cos 2 ( + α ) = − cos 2 ( + 2α ) = sin 2α ;442π ππππ4) sin 2  + α  − sin 2  − α  = sin 2  −  − α   − sin 2  − α  =3) cos 2 (4424πππ= cos 2  − α  − sin 2  − α  = cos − 2α  = sin 2α .442360.1) 1 + cos2 x = 2cos x;2cos2x – 2cosx = 0;1182) 1 – cos2x = 2sin x;2sin2x – 2sin x = 0;42cosx (cos x – 1) = 0; cos x = 0 ;cos = 1π x = + πn, n ∈ Z ;2 x = 2πk , k ∈ Z2sin x (sin x – 1) = 0; sin x = 0 ;sin x = 1 x = πn, n ∈ Z.π=+π∈x2k,kZ2Глава V.

Прогрессия361.1) a3 = 9; a6 = 36, an = n2;2) аk = 4, если k = 2; аk = 25, если k = 5;аk = n2, если k = n; аk = (n + 1)2, если k = n + 1.362.1) Пусть an = 2n + 3;a1 = 2 ⋅ 1 + 3 = 5;a2 = 2 ⋅ 2 + 3 = 7;a3 = 2 ⋅ 3 + 3 = 9.2) Пусть an = 1 + 3n;a1 = 1 + 3 ⋅ 1 = 4;a2 = 1 + 3 ⋅ 2 = 7;a3 = 1 + 3 ⋅ 3 = 10.3) Пусть an = 100 – 10n2;4) Пусть a n =a1 = 100 – 10 ⋅ 1 = 100 – 10 = 90;a1 =a2 = 100 – 10 ⋅ 4 = 100 – 40 = 60;a3 = 100 – 10 ⋅ 9 = 100 – 90 = 10.5) Пусть a n =a1 = 1; a2 =1;n11; a3 = .23n−2;31− 21=− ;332−2a2 ==0;33− 2 1a3 == .336) Пусть a n = − n 3 ;a1 = – 1; a2 = – 8; a3 = – 27.363.xn = n2если xn = 100, то n = 10; если xn = 144, то n = 12если xn = 225, то n = 1549, 169 – члены последовательности xn = n2, т.к.

49 = 72, 169 = 13248 – не члены последовательности xn = n2.119364.1) пусть an = – 3, тогда – 3 = n2 – 2n – 6;n2 – 2n – 3 = 0. Решим: n1 = 3; n2 = – 1 – не подходит, т.к. n∈N;a3 = – 3 – член an;2) пусть an = 2, тогда 2 = n2 – 2n – 6;n2 – 2n – 8 = 0. Решим: n1 = 4; n2 = – 2 – не подходит, т.к. n∈N;а4 = 2 – член an;3) пусть an = 3, тогда 3 = n2 – 2n – 6;n2 – 2n – 9 = 0. Решим:n 1, 2 =D= 1 + 9 = 10;41 ± 10– не подходят, т.к. n∈N;1an = n2 – 2n – 6 an = – 3 – не член an;4) пусть an = 9, тогда 9 = n2 – 2n – 6;n2 – 2n – 15 = 0.

Решим: n1 = 5; n2 = – 3 – не подходит, т.к. n∈N;а5 = 9 – член an.365.1) a2 = 3а1 + 1 = 3 ⋅ 2 + 1 = 6 + 1 = 7;a3 = 3а2 + 1 = 3 ⋅ 7 + 1 = 21 + 1 = 22;a4 = 3а3 + 1 = 3 ⋅ 22 + 1 = 66 + 1 = 67;2) a2 = 5 – 2a1 = 5 – 2 ⋅ 2 = 5 – 4 = 1;a3 = 5 – 2a2 = 5 – 2 ⋅ 1 = 5 – 2 = 3;a4 = 5 – 2a3 = 5 – 2 ⋅ 3 = 5 – 6 = – 1.366.1) Если an = 150, то150 = (n – 1)(n + 4);150 = n2 + 3n – 4;n2 + 3n – 154 = 0. Решим:D = 9 + 616 = 625 > 0,n 1, 2 =−3 ± 25;2n1 = 11, n2 = – 14 ∉ N ;не подходит, т.к. n∈N.Ответ: n = 11.2) Если an = 104, то104 = (n – 1)(n + 4);104 = n2 + 3n – 4;n2 + 3n – 108 = 0. Решим:D = 9 + 432 = 441 > 0,n 1, 2 =−3 ± 21;2n1 = 9, n2 = – 12 ∉ N ;не подходит, т.к.

n∈N.Ответ: n = 9.367.а2 = а1 = 256 = 162 = 16 ; а3 = а2 = 16 = 42 = 4 ;а4 = а3 = 4 = 22 = 2 .120368.1) а 2 = sin ππ⋅ a 1  = sin = 1 ;22ππа 4 = sin  ⋅ a 3  = sin = 1 ;22ππа 6 = sin  ⋅ a 5  = sin = 1 ;22ππа 3 = sin  ⋅ a 2  = sin = 1 ;22ππа 5 = sin  ⋅ a 4  = sin = 1 ;222) а2 = cosπ = – 1;a4 = cosπ = – 1;а6 = cosπ = – 1.а3 = cos( – π) = – 1;а5 = cos( – π) = – 1;369.а3 = а12 – а2 = 22 – 3 = 1; а4 = а22 – а3 = 32 – 1 = 8;а5 = а32 – а4 = 12 – 8 = – 7.370.1) Пусть an = – 5n + 4;an + 1 = – 5(n + 1) + 4 = – 5n – 5 + 4;an + 1 = – 5n – 1;an – 1 = – 5(n – 1) + 4 = – 5n + 5 + 4;an – 1 = – 5n + 9;an + 5 = – 5(n + 5) + 4 = – 5n – 25 + 4;an + 5 = – 5n – 21.2) Пусть an = 2(n – 10).Тогда an + 1 = 2(n + 1 – 10) = 2n + 2 – 20;an + 1 = 2n – 18;an – 1 = 2(n – 1 – 10) = 2n – 2 – 20;an – 1 = 2n – 22;an + 5 = 2(n + 5 – 10) = 2n + 10 – 20;an + 5 = 2n – 10.3) Пусть an = 2 ⋅ 3n + 1.

Тогда an + 1 = 2 ⋅ 3n + 2;an – 1 = 2 ⋅ 3n; an + 5 = 2 ⋅ 3n + 6.14) Пусть an = 7⋅  n+221Тогда an + 1 = 7⋅  21an – 1 = 7⋅  2n +1n +3.;1; an + 5 = 7⋅  2n +7.121372.1) Т.к. an = a1 + (n – 1)d, тоa2 = 2 + 5 = 7;a3 = 7 + 5 = 12;a4 = 12 + 5 = 17;a5 = 17 + 5 = 22;2) Т.к. a2 = a1 + d, тоa2 = – 3 + 2 = – 1;a3 = – 1 + 2 = 1;a4 = 1 + 2 = 3;a5 = 3 + 2 = 5.373.1) an + 1 = 3 – 4(n + 1);an + 1 – an = 3 – 4(n + 1) – 3 + 4n = 3/ − 4/ n/ − 4 − 3/ + 4/ n/ = −4 ,т.к. разность an + 1 – an не зависит от n, то это – арифметическая прогрессия.2) an + 1 = – 5 + 2(n + 1);an + 1 – an = – 5 + 2(n + 1) + 5 – 2n = – 5/ + 2/ n/ + 2 + 5/ − 2/ n/ = 2 ,т.к. an + 1 – an не зависит от n, то это – арифметическая прогрессия.3) an + 1 = 3(n + 2);an + 1 – an = 3(n + 2) – 3(n + 1) = 3/ n/ + 6 - 3/ n/ - 3 = 3,т.к.