2題A. Maths Compound Angles問題－快遞百科

標題:

2題A. Maths Compound Angles問題

發問:

1.It is given that a=ksinA, b=ksinB, c=ksinC, where ∠A+∠B+∠C = 180° and k is larger than 0. Prove, by the compound angle formulae, that (b2+c2-a2)/2bc = cosA.2. It is given that cos(x+β) = Acos(x-β), where A≠-1.(a)Prove that tan x = [(1-A)/(1+A)]*(1/tanβ).(b)Hence, or otherwise, solve cos(x+30°) =... 顯示更多 1. It is given that a=ksinA, b=ksinB, c=ksinC, where ∠A+∠B+∠C = 180° and k is larger than 0. Prove, by the compound angle formulae, that (b2+c2-a2)/2bc = cosA. 2. It is given that cos(x+β) = Acos(x-β), where A≠-1. (a) Prove that tan x = [(1-A)/(1+A)]*(1/tanβ). (b) Hence, or otherwise, solve cos(x+30°) = [cos(x-30°)]/2 for 0°≦x≦360°.

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最佳解答:

1) As A+B+C = 180°, A = 180° - (B+C) sinA = sin (B+C) = sinBcosC + cosBsinC sin2A = sin2Bcos2C + cos2Bsin2C + 2sinBcosBsinCcosC and cosA = -cos(B+C) = sinBsinC - cosBcosC LHS = (k2sin2B + k2sin2C - k2sin2A)/2k2sinBsinC = (sin2B + sin2C - sin2A)/2sinBsinC = (sin2B+sin2C-sin2Bcos2C-cos2Bsin2C-2sinBcosBsinCcosC)/2sinBsinC = [sin2B(1-cos2C)+sin2C(1-cos2B)-2sinBcosBsinCcosC]/2sinBsinC = (sin2Bsin2C+sin2Csin2B-2sinBcosBsinCcosC)/2sinBsinC = (2sin2Bsin2C-2sinBcosBsinCcosC)/2sinBsinC = sinBsinC(sinBsinC-cosBcosC)/sinBsinC = sinBsinC-cosBcosC = cosA ---------------------- 2a) cos(x+β) = Acos(x-β) cos(x-β) + cos(x+β) = cos(x-β) + Acos(x-β) = [1+A]cos(x-β) 2cosxcosβ = [1+A]cos(x-β) ---------- (1) cos(x+β) = Acos(x-β) cos(x-β) - cos(x+β) = cos(x-β) - Acos(x-β) = [1-A]cos(x-β) 2sinxsinβ = [1-A]cos(x-β) ------------- (2) (2) / (1) sinxsinβ/cosxcosβ = (1-A)/(1+A) tanxtanβ = (1-A)/(1+A) tan x = [(1-A)/(1+A)]*(1/tanβ) 2b) Consider β = 30° and A = 1/2 By a, tan x = [(1-1/2)/(1+1/2)]*(1/tan 30°) = 1/sqrt 3 x = 30° or 210°

其他解答:

1. (b2+c2-a2) - 2bc cosA = k2sin2B + k2sin2C - k2 + k2cos2A - 2k2sinBsinC cosA = k2sin2B + k2sin2C - k2 + k2cos2(B+C) - 2k2sinBsinC cos(B+C) = k2sin2B + k2sin2C - k2 + k2(cos(B)cos(C) - sin(B)sin(C))2 - 2k2sinBsinC(cos(B)cos(C) - sin(B)sin(C)) = k2sin2B + k2sin2C - k2 + k2cos2(B)cos2(C) - k2sin2(B)sin2(C) = k2sin2B - k2sin2(B)sin2(C) + k2sin2C - k2 + k2cos2(B)cos2(C) = k2sin2Bcos2C - k2cos2C + k2cos2(B)cos2(C) = k2sin2Bcos2C + k2cos2(B)cos2(C) - k2cos2C = k2cos2C - k2cos2C = 0 2. (a) (1-A)/(1+A) = (cos(x-β) - cos(x+β)) / (cos(x-β) + cos(x+β)) = 2 sin x sin β/ 2 cos x cos β = tan x tan β 2. (b) cos(x+30°) = [cos(x-30°)]/2 From (a), tan x = 1/3 . 1/tan 30° = 1/3^(0.5) so x = 30° or 210°