標題:
發問:
1. It is given that sinθ/m = cosθ/n = k and sin^2 θ/n^2 + cos^2 θ/m^2 = 10/[3(m^2+n^2)], where k,m,n are constants.(a) Prove that m^2+n^2 = 1/k^2 and also 3m^4 - 10m^2n^2 + 3n^4 = 0.(b) find the general solution of θ.2. In triangle ABC, it is given that sinA cosB - sinB = sinC - sinA cosC.(a) Prove that sinA =... 顯示更多 1. It is given that sinθ/m = cosθ/n = k and sin^2 θ/n^2 + cos^2 θ/m^2 = 10/[3(m^2+n^2)], where k,m,n are constants. (a) Prove that m^2+n^2 = 1/k^2 and also 3m^4 - 10m^2n^2 + 3n^4 = 0. (b) find the general solution of θ. 2. In triangle ABC, it is given that sinA cosB - sinB = sinC - sinA cosC. (a) Prove that sinA = tan [(B+C)/2] and also prove that tan[(B+C)/2] = cot(A/2). (b) From the above results, prove that triangle ABC is right-angled.
最佳解答:
(1a) sin2 θ/m2 = cos2 θ/n2 = k2 sin2 θ = m2k2 cos2 θ = n2k2 1 = k2(m2 + n2) m2 + n2 = 1/k2 sin2 θ/n2 + cos2 θ/m2 = 10/[3(m2 + n2)] m2sin2 θ + n2cos2 θ = 10m2n2/[3(m2 + n2)] m4k2 + n4k2 = 10m2n2/[3(m2 + n2)] (m4 + n4)/(m2 + n2) = 10m2n2/[3(m2 + n2)] m4 + n4 = 10m2n2/3 3m4 + 3n4 = 10m2n2 3m4 - 10m2n2 + 3n4 = 0 (b) 3m4 - 10m2n2 + 3n4 = 0 (3m2 - n2)(m2 - 3n2) = 0 3m2 = n2 or m2 = 3n2 m/n = ±1/√3 or ±√3 tan θ = ±1/√3 or ±√3 θ = 180n° ± 30° or 180n° ± 60° (2a) sin A cos B - sin B = sin C - sin A cos C sin A (cos B + cos C) = sin B + sin C 2 sin A cos [(B + C)/2] cos [(B - C)/2] = 2 sin [(B + C)/2] cos [(B - C)/2] sin A cos [(B + C)/2] = sin [(B + C)/2] sin A = tan [(B + C)/2] (2b) sin A = tan [(B + C)/2] sin A = tan [(180° - A)/2] sin 2(A/2) = tan (90° - A/2) 2 sin (A/2) cos (A/2) = cos (A/2)/sin (A/2) 2 sin2 (A/2) = 1 sin (A/2) = 1/√2 A/2 = 45° A = 90° Hence ABC is right-angled. 2008-02-14 00:11:22 補充： More on 2a:tan [(B 十 C)/2] = tan (90° - A/2)= cot (A/2)
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A-MTH Trigonometry (Quick)發問:
1. It is given that sinθ/m = cosθ/n = k and sin^2 θ/n^2 + cos^2 θ/m^2 = 10/[3(m^2+n^2)], where k,m,n are constants.(a) Prove that m^2+n^2 = 1/k^2 and also 3m^4 - 10m^2n^2 + 3n^4 = 0.(b) find the general solution of θ.2. In triangle ABC, it is given that sinA cosB - sinB = sinC - sinA cosC.(a) Prove that sinA =... 顯示更多 1. It is given that sinθ/m = cosθ/n = k and sin^2 θ/n^2 + cos^2 θ/m^2 = 10/[3(m^2+n^2)], where k,m,n are constants. (a) Prove that m^2+n^2 = 1/k^2 and also 3m^4 - 10m^2n^2 + 3n^4 = 0. (b) find the general solution of θ. 2. In triangle ABC, it is given that sinA cosB - sinB = sinC - sinA cosC. (a) Prove that sinA = tan [(B+C)/2] and also prove that tan[(B+C)/2] = cot(A/2). (b) From the above results, prove that triangle ABC is right-angled.
最佳解答:
(1a) sin2 θ/m2 = cos2 θ/n2 = k2 sin2 θ = m2k2 cos2 θ = n2k2 1 = k2(m2 + n2) m2 + n2 = 1/k2 sin2 θ/n2 + cos2 θ/m2 = 10/[3(m2 + n2)] m2sin2 θ + n2cos2 θ = 10m2n2/[3(m2 + n2)] m4k2 + n4k2 = 10m2n2/[3(m2 + n2)] (m4 + n4)/(m2 + n2) = 10m2n2/[3(m2 + n2)] m4 + n4 = 10m2n2/3 3m4 + 3n4 = 10m2n2 3m4 - 10m2n2 + 3n4 = 0 (b) 3m4 - 10m2n2 + 3n4 = 0 (3m2 - n2)(m2 - 3n2) = 0 3m2 = n2 or m2 = 3n2 m/n = ±1/√3 or ±√3 tan θ = ±1/√3 or ±√3 θ = 180n° ± 30° or 180n° ± 60° (2a) sin A cos B - sin B = sin C - sin A cos C sin A (cos B + cos C) = sin B + sin C 2 sin A cos [(B + C)/2] cos [(B - C)/2] = 2 sin [(B + C)/2] cos [(B - C)/2] sin A cos [(B + C)/2] = sin [(B + C)/2] sin A = tan [(B + C)/2] (2b) sin A = tan [(B + C)/2] sin A = tan [(180° - A)/2] sin 2(A/2) = tan (90° - A/2) 2 sin (A/2) cos (A/2) = cos (A/2)/sin (A/2) 2 sin2 (A/2) = 1 sin (A/2) = 1/√2 A/2 = 45° A = 90° Hence ABC is right-angled. 2008-02-14 00:11:22 補充： More on 2a:tan [(B 十 C)/2] = tan (90° - A/2)= cot (A/2)
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