Tiêu đề Trigonometry & Gymnosperm- Daily-8 MCQ (Home Practice)-With Solve.pdf ✅

2 Ges c`msL ̈v we‡Rvo n‡j, †hvMdj = ga ̈c` + c`msL ̈v – 1 2 7. hw` P  Q Ges sinP + cosP = sinQ + cosQ n‡j, P + Q Gi gvb KZ? 3 2   2 0 DËi:  2 e ̈vL ̈v: sinP + cosP = sinQ + cosQ  sinP – sinQ = cosQ – cosP  2cos    P + Q 2 sin    P – Q 2 = 2sin    P + Q 2 sin    P – Q 2  cos    P + Q 2 sin    P + Q 2 = 1  cot    P + Q 2 = 1  P + Q 2 =  4  P + Q =  2 8. sin9sin27sin45sec63sec81 = ? 1 2 1 0 1 2 DËi: 1 2 e ̈vL ̈v: sin9sec(90 – 9)sec(90 – 27)sin27sin45 = sin9cosec9sin27cosec27sin45 = sin9 1 sin9 sin27 1 sin27 sin45 = 1 2 9. hw` tan + tan = x, cot + cot = y Ges  +  =  nq, Z‡e cot = ? xy x – y x – y xy y – x xy xy y – x DËi: y – x xy e ̈vL ̈v: cot + cot = y  1 tan + 1 tan = y  tan + tan tan  tan = y  tantan = x y  +  =   tan = tan( + ) = tan + tan 1 – tantan = x 1 – x y = xy y – x  cot = y – x xy 10. hw` n Gi gvb †Rvo nq, Z‡e cos + cos( + ) + cos(2 + ) + ...... + cos(n + ) = ? cos ncos 0 cos(n + 0) DËi: cos e ̈vL ̈v: cos + cos( + ) + cos(2 + ) + cos(3 + ) + cos(4 + ) + ...... n = 2 wb‡q cvB, cos – cos + cos = cos 11. tan  12 + tan  6 + tan  12 tan  6 Gi gvb KZ? 3 1 1 3 0 DËi: 1 e ̈vL ̈v: tan     12 +  6 = tan  12 + tan  6 1 – tan  12tan  6  tan     4 = tan  12 + tan  6 1 – tan  12tan  6  tan  12 + tan  6 + tan  12 tan  6 = 1 12. ABC G A = 45, B = 75 n‡j, a + c 2 = KZ? 3b 2b b 5b DËi: 2b e ̈vL ̈v: C = 180 – A – B = 60 b = ccosA + acosC  2b = 2(ccos45 + acos60) = 2    c 2 + a 2  2b = 2c + a 13. tan = nsincos 1 – nsin2 n‡j, tan( – ) = ? ntan (1 + n)tan (1 – n)tan (2 + n)tan DËi: (1 – n)tan e ̈vL ̈v: awi,  =  = 45 tan( – ) = 0

4 20. C B D A AB = 4, BC = 3, DCB = 60 n‡j, BD = ? 1.5 2.25 0.75 5.19 DËi: 5.19 e ̈vL ̈v: BDC-G tan60 = BD BC  3 = BD 3  BD = 3 3 = 3  1.73 = 5.19 21. 15 B 9 D C A 30 ABC Gi cwimxgv KZ? 48 + 12 2 48 48 – 12 2 48 + 12 3 DËi: 48 + 12 3 e ̈vL ̈v: AD = 152 – 9 2 = 12; DC = AD cot30 = 12 3 AC = DC sec30 = 24  AB + BC + AC = 48 + 12 3 22. sin  16 Gi gvb †KvbwU? 2 + 2 + 2 2 – 2 + 2 1 2 2 – 2 + 2 1 2 2 + 2 + 2 DËi: 1 2 2 – 2 + 2 e ̈vL ̈v: 1 2 2sin  16 = 1 2 2 . 2sin2  16 = 1 2 2     1 – cos  8 = 1 2 2 – 2cos  8 = 1 2 2 – 2 . 2cos2  8 = 1 2 2 – 2     1 + cos  4 = 1 2 2 – 2 + 2 Shortcut: 2sin  2 n = 2 – 2 + 2 + ... (n – 1) msL ̈K 23. †h †Kv‡bv wÎfz‡Ri evû a, b, c Ges Gi †ÿÎdj  n‡j, sinA = KZ? 2 ca 2 bc 2 ab 2 abc DËi: 2 bc e ̈vL ̈v:  = 1 2  bc sinA  sinA = 2 bc A B C b a c 24. 3sinx + 5cosx = 7 n‡j, sinx + cosx = ? 7 9 9 7 7 8 7 DËi: 8 7 e ̈vL ̈v: Shortcut: asinx + bcosx = c n‡j, sinx + cosx = a + b c  sinx + cosx = 3 + 5 7 = 8 7 25. ABC wÎfz‡R 1 a + c + 1 b + c = 3 a + b + c n‡j, C Gi gvb KZ? 30 45 90 60 DËi: 60 e ̈vL ̈v: 1 a + c + 1 b + c = 3 a + b + c  b + c + a + c ab + ac + bc + c2 = 3 a + b + c  a 2 + ab + ac + ab + b2 + bc + 2ac + 2bc + 2c2 – 3ab – 3ac – 3bc – 3c2 = 0  a 2 + b2 – c 2 = ab  a 2 + b2 – c 2 2ab = 1 2  cosC = cos60  C = 60 26. cosxsin    x –  6 Gi gvb e„nËg n‡j, x Gi gvb KZ?  6  3  4  2 DËi:  3 e ̈vL ̈v: P = cosxsin    x –  6 P = 1 2 2cosxsin   x –   6 [⸪ sin(A + B) – sin(A – B) = 2cosAsinB]