Math (sin. cos, tan and other)

2015-02-25 6:52 pm
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=?

回答 (1)

2015-02-25 10:29 pm
✔ 最佳答案
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°


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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


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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)


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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)


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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)


收錄日期: 2021-04-15 18:19:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150225000051KK00028

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