Inthe above formula, let us choose the 3rd formula. cos 2A = 2cos2A - 1. = 2 (15/17)2 - 1. = 2 (225/289) - 1. = (450 - 289)/289. cos 2A = 161/289. (ii) sin A = 4/5. Solution : cos 2A = 1 - 2sin2A. B Find the exact value of each expression (if possible) without using a calculator. Circle your final answer. 1. sin arccosž 5 CSC cos 4 cos sin 5. 8. 11. 14. cot arctan 13 cos arccos(O.7 arcsin (sin 5m) o cos (arcsin (1 O O 6. 12. sin arcsin tan (arctan (20 arccos (co z) -z 10. 13. sin (arcsin arccos cos 15. arctan (tan 8m) ( ILO) Name 1 Since both A & B are in (0,π/4), both (A + B) & (A - B) will be only in (0, π/2); hence tan(A+B) & tan(A-B) both will be positive. 2) As cos(A+B) = 4/5, tan (A+B) = 3/4. [Applying right triangle rule, adj = 4; hyp = 5; ==> opp = 3] Thefollowing relationship is known to be true for two angles A and B: cos (A)cos (B)-sin (A)sin (B)=0.957269 Express A in terms of the angle B. Work in degrees and report numeric values accurate to 2 decimal places. So I'm pretty lost on how to even begin. Diketahui∆ABC dengan sin A = 5/13 dan sin B = 4/5. Jika B sudut tumpul, nilai tan (A-B) = rebbose. Tuesday, 6 October 2020 Bank soal Edit. Sin B = 4/5 dan sudut B tumpul maka diperoleh bahwa cos B = - 3/5. Sin A = 5/13 dan sudut A lancip maka diperoleh cos A = 12/13. Sehingga : can you drink tap water in bali. Given as sin A = 4/5 and cos B = 5/13 As we know that cos A = √1 – sin2 A and sin B = √1 – cos2 B, where 0 < A, B < π/2 Therefore let us find the value of sin A and cos B cos A = √1 – sin2 A = √1 – 4/52 = √1 – 16/25 = √25 – 16/25 = √9/25 = 3/5 sin B = √1 – cos2 B = √1 – 5/132 = √1 – 25/169 = √169 – 25/169 = √144/169 = 12/13 i sin A + B As we know that sin A +B = sin A cos B + cos A sin B Therefore, sin A + B = sin A cos B + cos A sin B = 4/5 × 5/13 + 3/5 × 12/13 = 20/65 + 36/65 = 20 + 36/65 = 56/65 ii cos A + B As we know that cos A +B = cos A cos B – sin A sin B Therefore, cos A + B = cos A cos B – sin A sin B = 3/5 × 5/13 – 4/5 × 12/13 = 15/65 – 48/65 = -33/65 iii sin A – B As we know that sin A – B = sin A cos B – cos A sin B Therefore, sin A – B = sin A cos B – cos A sin B = 4/5 × 5/13 – 3/5 × 12/13 = 20/65 – 36/65 = -16/65 iv cos A – B As we know that cos A - B = cos A cos B + sin A sin B Therefore, cos A - B = cos A cos B + sin A sin B = 3/5 × 5/13 + 4/5 × 12/13 = 15/65 + 48/65 = 63/65

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