Tiêu đề Trigonometriy Engineering Practice Sheet Solution.pdf ✅

g ̈vwUa· I wbY©vqK  Engineering Question Bank Solution 1 07 mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ Trigonometric Ratios of Associated Angles WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. hw` xcos + ysin = z = xsec + ycosec nq Ges x  y  z  0 nq, Zvn‡j sin2 †K x, y Ges z Gi gva ̈‡g cÖKvk Ki| [BUET 23-24] mgvavb: xcos + ysin = z .................. (i) xsec + ycosec = z  x cos + y sin = z ..................... (ii) (i)  (ii) K‡i cvB, x 2 + xytan + xycot + y2 = z 2  tan + cot = z 2 – x 2 – y 2 xy  sin cos + cos sin = z 2 – x 2 – y 2 xy  sin2  + cos2  2sincos = z 2 – x 2 – y 2 2xy     1 2 Øviv ̧Y K‡i  1 sin2 = z 2 – x 2 – y 2 2xy  sin2 = 2xy z 2 – x 2 – y 2 (Ans.) 2. hw` x 2 – 9x + 8 = 0, mgxKiYwU exp{(sin2 x + sin4 x + sin6 x + .... )ln2 Øviv wm× nq, Zvn‡j cosx sinx + cosx Gi gvb wbY©q Ki, †hLv‡b 0 < x <  2 [BUET 22-23] mgvavb: x 2 – 9x + 8 = 0  (x – 1)(x – 8) = 0  x = 1, 8 GLb, exp(sin2 x + sin4 x + sin6 x + .... )ln2 = (eln2) sin2x + sin4x + sin6x + ......  = 2 sin2 x 1 – sin2 x = 2tan2x kZ©g‡Z, 2 tan2x = 1, 8 ⸪ elnx = x Amxg avivi †hvMdj, a 1 – r = sin2 x 1 – sin2 x GLb, 2 tan2x = 1 = 20  tan2 x = 0  x = 0     hv AMÖnY‡hvM ̈ 0 < x <  2 Avevi, 2 tan2x = 8  2 tan2x = 23  tan2 x = 3  tanx =  3  tanx = 3     0 < x <  2  x =  3  cosx sinx + cosx = cos  3 cos  3 + sin 3 = – 1 + 3 2 (Ans.) 3. cosx + cosy = a Ges sinx + siny = b n‡j, cos(x + y) Gi gvb KZ? [BUET 19-20; MIST 19-20] mgvavb: cosx + cosy = a  2cos x + y 2 cos x – y 2 = a ........ (i) sinx + siny = b  2sin x + y 2 cos x – y 2 = b ........ (ii) (i)2  (ii)2 K‡i, cos2 x + y 2 sin2 x + y 2 = a 2 b 2  cos2 x + y 2 – sin2 x + y 2 cos2 x + y 2 + sin2 x + y 2 = a 2 – b 2 a 2 + b2 [we‡qvRb-†hvRb K‡i]  cos(x + y) = a 2 – b 2 a 2 + b2 (Ans.) 4. tan + tan     3 +  + tan    2 3 +  †K tan3 Gi gva ̈‡g cÖKvk Ki| [BUET 16-17] mgvavb: tan + tan     3 +  + tan    2 3 +  = tan + tan  3 + tan 1 – tan  3 tan + tan 2 3 + tan 1 – tan 2 3 tan
2  Higher Math 1st Paper Chapter-7 = tan + 3 + tan 1 – 3 tan + – 3 + tan 1 + 3 tan = tan + 3 + 3tan + tan + 3 tan2  – 3 + 3tan + tan – 3 tan2  (1 – 3 tan)(1 + 3 tan) = tan + 8 tan 1 – 3tan2  = tan – 3tan3  + 8tan 1 – 3tan2  = 3(3tan – tan3 ) 1 – 3tan2  = 3tan3 (Ans.) 5. hw` sinx + siny = a Ges cosx + cosy = b nq Z‡e †`LvI †h, sin1 2 (x – y) =  1 2 4 – a 2 – b 2 . [BUET 16-17] mgvavb: sinx + siny = a ........ (i) cosx + cosy = b ...... (ii) mgxKiY (i)2 + (ii)2 K‡i cvB, 1 + 1 + 2sinxsiny + 2cosxcosy = a2 + b2  2 + 2(cosxcosy + sinxsiny) = a2 + b2  2 + 2cos(x – y) = a2 + b2  2 + 2      1 – 2sin2 1 2 (x – y) = a2 + b2  4 – 4sin2 1 2 (x – y) = a2 + b2  sin2 1 2 (x – y) = 4 – a 2 – b 2 4  sin1 2 (x – y) =  1 2 4 – a 2 – b 2 (Showed) 6. cÖgvY Ki †h, 16cos2 15 cos 4 15 cos 8 15 cos 14 15 = 1 [BUET 15-16, 00-01] mgvavb: awi, 2 15 =  L.H.S = 16 cos cos2 cos4 cos7 = 8 sin  (2sin cos)cos2 cos4 cos7 = 4 sin  (2sin2 cos2)cos4 cos7 = 2 sin  (2sin4 cos4)cos7 = 1 sin  (2sin8 cos7) = 1 sin  (sin15 + sin) = 1 sin  (sin2 + sin) = 1 = R.H.S (Proved) 7. hw` tan  2 = 1 – p 1 + p tan  2 nq Zvn‡j †`LvI †h, cos = cos – p 1 – pcos [BUET 14-15] mgvavb: tan  2 = 1 – p 1 + p tan  2  1 tan  2 = 1 – p ( 1 + p) tan  2  1 – tan2  2 1 + tan2  2 = (1 – p) – (1 + p) tan2  2 (1 – p) + (1 + p) tan2  2 [eM© K‡i we‡qvRb-†hvRb] GLb, cos    2   2 = 1 – tan2  2 – p     1 + tan2  2 1 + tan2  2 + p    tan2  2 – 1 = 1 – tan2  2 1 + tan2  2 – p 1 + tan2  2 1 + tan2  2 1 + tan2  2 1 + tan2  2 – p 1 – tan2  2 1 + tan2  2  cos = cos – p 1 – pcos (Showed) 8. hw`  =  36 nq, Z‡e sin2 3 + sin2 4 + sin2 5 + ........ + sin2 15 Gi gvb wbY©q Ki| [BUET 13-14] mgvavb: sin2 3 + sin2 4 + sin2 5 + ........ + sin2 15 = sin2 15 + sin2 20 + sin2 25 + ...... + sin2 75 = (sin2 15 + cos2 15) + (sin2 20 + cos2 20) + (sin2 25 + cos2 25) + (sin2 30 + cos2 30) + (sin2 35 + cos2 35) + (sin2 40 + cos2 40) + sin2 45 = 6 + 1 2 = 13 2 (Ans.)
mshy3 †Kv‡Yi w·KvYwgwZK AbycvZ  Engineering Practice Sheet Solution 3 9. hw`  =  20 nq, Z‡e cotcot3cot5 ........ cot19 Gi gvb wbY©q Ki| [BUET 11-12] mgvavb: cot  cot3  cot5  cot7  cot9  cot11  cot13  cot15  cot17  cot19 = cot  cot3  1  cot7  cot9  cot     2 +   cot     2 + 3  (– 1)  cot     2 + 7  cot     2 + 9    ∵   =  20 = – cot  (– tan)  cot3  (– tan3)  cot7  (– tan 7)  cot9 (– tan9) = – (cot  tan)  (cot 3  tan3)  (cot7  tan7)  (cot9  tan9) = – 1 (Ans.) 10. hw` A + B + C =  nq Z‡e cÖgvY Ki †h, sin2A + sin2B + sin2C = 4sinA sinBsinC. [BUET 09-10] mgvavb: L.H.S = sin2A + sin2B + sin2C = 2sin(A + B) cos(A – B) + sin2C = 2sin( – C) cos(A – B) + sin2C = 2sinC cos(A – B) + 2sinC cosC = 2sinC [cos(A – B) – cos(A + B)] = 2sinC  2sinA sinB = 4sinA sinB sinC = R.H.S (Proved) 11. ABC GKwU ̄’~j‡KvYx wÎfzR| cÖgvY Ki †h, cotAcotB + cotBcotC + cotCcotA = 1 [BUET 05-06] mgvavb: †h‡Kv‡bv wÎfzR ABC Gi Rb ̈, A + B + C =   A + B =  – C  cot(A + B) = cot( – c)  cotAcotB – 1 cotB + cotA = – cotC  cotAcotB + cotBcotC + cotCcotA = 1 (Proved) 12. cÖgvY Ki †h,  †ÿÎdjwewkó ABC wÎfz‡Ri Rb ̈ 1 a sinA + 1 b sinB + 1 c sinC = 6 abc [BUET 05-06] mgvavb:  = 1 2 bc sinA = 1 2 ca sinB = 1 2 ab sinC  sinA = 2 bc ; sinB = 2 ca ; sinC = 2 ab  L.H.S = 1 a  2 bc + 1 b  2 ca + 1 c  2 ab = 6 abc = R.H.S (Proved) 13. ABC wÎfz‡R cosA = sinB – cosC n‡j †`LvI †h, wÎfzRwU mg‡KvYx| [BUET 05-06, 04-05] mgvavb: cosA = sinB – cosC  cosA + cosC = sinB  cosA – cos(A + B) = sinB  2sin     2A + B 2 sin B 2 = 2sinB 2 cos B 2  sin    2A + B 2 = cos B 2  sin   A +  B 2 = sin     2 – B 2  A + B 2 =      2 – B 2  A + B =  2  ABC mg‡KvYx| (Showed) 14. †`LvI †h, cot3A – 3cotA 3cot2A – 1 = cot3A [BUET 04-05] mgvavb: R.H.S = cot3A = cot(2A + A) = cotAcot2A – 1 cotA + cot2A = cotA    cot2A – 1 2cotA – 1 cotA + cot2A – 1 2cotA = cot3A – cotA – 2cotA 2cotA 2cot2A + cot2A – 1 2cotA = cot3A – 3cotA 3cot2A – 1 = L.H.S (Showed) 15. cÖgvY Ki †h, tan20 tan40 tan80 = 3 [BUET 04-05; BUTex 11-12] mgvavb: L.H.S = tan20 tan40 tan80 = tan20 tan(60 – 20) tan(60 + 20) = tan20  tan60 – tan20 1 + tan60 tan20  tan60 + tan20 1 – tan60 tan20 = tan20  3 – tan20 1 + 3 tan20  3 + tan20 1 – 3 tan20 = tan20 . 3 – tan2 20 1 – 3tan2 20 = 3tan20 – tan3 20 1 – 3tan2 20 = tan(3  20) = tan60 = 3 = R.H.S (Proved)
4  Higher Math 1st Paper Chapter-7 16. hw` a = 2b Ges A = 3B nq, Z‡e wÎfy‡Ri †KvY ̧‡jv wbY©q Ki| [BUET 03-04] mgvavb: Avgiv Rvwb, ABC-G, a sinA = b sinB  2b sin3B = b sinB  2sinB = sin3B  2sinB = 3sinB – 4sin3B  4sin2B = 1  sinB = 1 2  B = 30 (Ans.)  A = 3B = 90 (Ans.)  C = 180 – (90 + 30) = 60 (Ans.) 17. hw` a = 2, b = 1 + 3 , C = 60 nq, Z‡e wÎfzRwU mgvavb K‡iv| [BUET 02-03] mgvavb: cosC = a 2 + b2 – c 2 2ab  cos60 = 4 + 1 + 3 + 2 3 – c 2 2  2  (1 + 3)  1 2 = 4 + 1 + 3 + 2 3 – c 2 2  2  (1 + 3)  c = 6  cosA = b 2 + c2 – a 2 2bc  cosA = 1 + 3 + 2 3 + 6 – 4 2(1 + 3) 6  cosA = 1 2  A = 45 Avevi, A + B + C = 180  B = 180 – (45 + 60)  B = 75  wb‡Y©q mgvavb: c = 6, A = 45, B = 75 (Ans.) 18. hw` sinx + siny = 1 Ges cosx + cosy = 0 nq Z‡e cÖgvY Ki †h, x + y =  [BUET 02-03; RUET 18-19; MIST 17-18] mgvavb: sinx + siny = 1  2sin    x + y 2 cos    x – y 2 = 1 ...... (i) cosx + cosy = 0  2cos    x + y 2 cos    x – y 2 = 0 ...... (ii) (ii)  (i) K‡i, cos    x + y 2 sin    x + y 2 = 0  cot    x + y 2 = 0  x + y 2 =  2    ⸪ cot   2 = 0  x + y =  (Ans.) 19. hw`  I  abvZ¥K I m~2‡KvY nq Ges cos2 = 3cos2 – 1 3 – cos2 nq, Z‡e †`LvI †h, tan = 2tan [BUET 01-02; CUET 09-10] mgvavb: cos2 = 3cos2 – 1 3 – cos2  1 – cos2 1 + cos2 = 4(1 – cos2) 2(1 + cos2)  2sin2 2cos2 = 2 . sin2  cos2   tan2tan2  tan = 2tan (Showed) [⸪ ,  abvZ¥K m~ÿ¥‡KvY, ZvB FYvZ¥K gvb MÖnY‡hvM ̈ bq] 20. †h‡Kv‡bv wÎfzR ABC Gi Rb ̈ †`LvI †h, 1 a cos2 A 2 + 1 b cos2 B 2 + 1 c cos2 C 2 = s 2 abc [BUET 00-01] mgvavb: L.H.S = 1 a cos2A 2 + 1 b cos2B 2 + 1 c cos2C 2 = 1 a  s(s – a) bc + 1 b  s(s – b) ca + 1 c  s(s – c) ab = s(3s – a – b – c) abc = s(3s – 2s) abc = s 2 abc = R.H.S (Showed) 21. cÖgvY Ki †h, cos15 + sin15 cos15 – sin15 = 3 = 1 + tan15 1 – tan15 [BUET 98-99] mgvavb: L.H.S = cos15 + sin15 cos15 – sin15 = 1 + tan15 1 – tan15 = tan45 + tan15 1 – tan45 tan15 = tan (45 + 15) = tan60 = 3 = R.H.S (Proved)