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發問:
Please answer the questions as possible. thank you so muchQ1, Rewrite 5 sin(x) + 4 cos(x) as A sin(x + phi)A = ?phi=?Q2,Let f(x) = 3.2 sin(x) + 3.6 cos(x). What is the maximum and minimum value of this function?maximum=?minimum=?Q3.Solve 8cos^2(w)-2cos(w)-1 = 0 for all solutions 0 <= w <... 顯示更多 Please answer the questions as possible. thank you so much Q1, Rewrite 5 sin(x) + 4 cos(x) as A sin(x + phi) A = ? phi=? Q2,Let f(x) = 3.2 sin(x) + 3.6 cos(x). What is the maximum and minimum value of this function? maximum=? minimum=? Q3.Solve 8cos^2(w)-2cos(w)-1 = 0 for all solutions 0 <= w < 2pi w =? Q4,Solve 4cos^2(t)-23cos(t)+15 = 0 for all solutions 0 <= t < 2pi t=? Q.5.Solve 12sin^2(t)+5cos(t)-10 = 0 for all solutions 0 <= t < 2pi t=?
最佳解答:
1. √(52 + 42) =√41 cos(φ) = 5/√41 and sin(φ) = 4/√41 tan(φ) = sin(φ)/cos(φ) = 4/5 φ = tan?1(4/5)≈ 38.66° 5 sinx + 4 cosx = √41 [(5/√41) sin(x) + (4/√41) cos(x)] = √41 [cos38.66° sin(x) + sin38.66° cos(x)] = √41 sin(x + 38.66°) A = √41 φ = tan?1(4/5) ≈ 38.66° ==== 2. √(3.22 + 3.62) =√23.2 cos(φ) = 3.2/√23.2 and sin(φ) = 3.6/√23.2 f(x) = 3.2 sin(x) + 3.6 cos(x) = √23.2 [(3.2/√23.2) sin(x) + (3.6/√23.2) cos(x)] = √23.2 [cos(φ) sin(x) + sin(φ) cos(x)] = √23.2 sin(x + φ) -1 ≤ sin(x + φ) ≤ 1 Hence, -√23.2 ≤ √23.2 sin(x + φ) ≤ √23.2 Maximum of f(x) = √23.2 Minimum of f(x) = -√23.2 ==== 3.8 cos2(w) - 2 cos(w) - 1 =0 [2 cos(w) - 1] [4 cos(w) + 1] = 0 cos(w) = 1/2 or cos(w) = -1/4 w = π/3 (rad), 5π/3 (rad) orw = 1.823 (rad), 4.460 (rad) ==== 4. 4 cos2(t) - 23 cos(t) + 15 = 0 [4 cos(t) - 3] [cos(t) - 5] = 0 cos(t) = 3/4 or cos(t) = 5 (rejected) t = 0.732 (rad), 5.560 (rad) ==== 5. 12 sin2(t) + 5 cos(t) - 10= 0 12 [1- cos2(t)] + 5 cos(t) - 10 = 0 12 - 12cos2(t) + 5 cos(t) - 10 = 0 12 cos2(t) - 5 cos(t) -2 = 0 [3 cos(t) - 2] [4 cos(t) + 1] = 0 cos(t) = 2/3 or cos(t) = -1/4 t = 0.841 (rad), 5.442 (rad)or t = 1.823 (rad) or 4.460 (rad)
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Math (sin. cos, tan and other)發問:
Please answer the questions as possible. thank you so muchQ1, Rewrite 5 sin(x) + 4 cos(x) as A sin(x + phi)A = ?phi=?Q2,Let f(x) = 3.2 sin(x) + 3.6 cos(x). What is the maximum and minimum value of this function?maximum=?minimum=?Q3.Solve 8cos^2(w)-2cos(w)-1 = 0 for all solutions 0 <= w <... 顯示更多 Please answer the questions as possible. thank you so much Q1, Rewrite 5 sin(x) + 4 cos(x) as A sin(x + phi) A = ? phi=? Q2,Let f(x) = 3.2 sin(x) + 3.6 cos(x). What is the maximum and minimum value of this function? maximum=? minimum=? Q3.Solve 8cos^2(w)-2cos(w)-1 = 0 for all solutions 0 <= w < 2pi w =? Q4,Solve 4cos^2(t)-23cos(t)+15 = 0 for all solutions 0 <= t < 2pi t=? Q.5.Solve 12sin^2(t)+5cos(t)-10 = 0 for all solutions 0 <= t < 2pi t=?
最佳解答:
1. √(52 + 42) =√41 cos(φ) = 5/√41 and sin(φ) = 4/√41 tan(φ) = sin(φ)/cos(φ) = 4/5 φ = tan?1(4/5)≈ 38.66° 5 sinx + 4 cosx = √41 [(5/√41) sin(x) + (4/√41) cos(x)] = √41 [cos38.66° sin(x) + sin38.66° cos(x)] = √41 sin(x + 38.66°) A = √41 φ = tan?1(4/5) ≈ 38.66° ==== 2. √(3.22 + 3.62) =√23.2 cos(φ) = 3.2/√23.2 and sin(φ) = 3.6/√23.2 f(x) = 3.2 sin(x) + 3.6 cos(x) = √23.2 [(3.2/√23.2) sin(x) + (3.6/√23.2) cos(x)] = √23.2 [cos(φ) sin(x) + sin(φ) cos(x)] = √23.2 sin(x + φ) -1 ≤ sin(x + φ) ≤ 1 Hence, -√23.2 ≤ √23.2 sin(x + φ) ≤ √23.2 Maximum of f(x) = √23.2 Minimum of f(x) = -√23.2 ==== 3.8 cos2(w) - 2 cos(w) - 1 =0 [2 cos(w) - 1] [4 cos(w) + 1] = 0 cos(w) = 1/2 or cos(w) = -1/4 w = π/3 (rad), 5π/3 (rad) orw = 1.823 (rad), 4.460 (rad) ==== 4. 4 cos2(t) - 23 cos(t) + 15 = 0 [4 cos(t) - 3] [cos(t) - 5] = 0 cos(t) = 3/4 or cos(t) = 5 (rejected) t = 0.732 (rad), 5.560 (rad) ==== 5. 12 sin2(t) + 5 cos(t) - 10= 0 12 [1- cos2(t)] + 5 cos(t) - 10 = 0 12 - 12cos2(t) + 5 cos(t) - 10 = 0 12 cos2(t) - 5 cos(t) -2 = 0 [3 cos(t) - 2] [4 cos(t) + 1] = 0 cos(t) = 2/3 or cos(t) = -1/4 t = 0.841 (rad), 5.442 (rad)or t = 1.823 (rad) or 4.460 (rad)
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