a maths...(20PT)_0－香港特色

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a maths...(20PT)

發問:

1. it is given that a= k sin A, b= k sinB and c= k sinC, where ∠A + ∠B + ∠C = π and k > 0. Prove, by the compound angle formulae, that (b^2+c^2-a^2)/(2bc)=cos A.(note: In the above result, we have proved the Cosine Law from the Sine Law.)2. it is given that cos(x+β) = Acos(x-β), where A≠ -1.(a) Prove... 顯示更多 1. it is given that a= k sin A, b= k sinB and c= k sinC, where ∠A + ∠B + ∠C = π and k > 0. Prove, by the compound angle formulae, that (b^2+c^2-a^2)/(2bc)=cos A. (note: In the above result, we have proved the Cosine Law from the Sine Law.) 2. it is given that cos(x+β) = Acos(x-β), where A≠ -1. (a) Prove that tanx = (1-A/1+A)(1/tanβ). (b) Hence, or otherwise, solve cos(x+ (π/6)) = (1/2) cos (x- (π/6)) for 0≦ x ≦ 2π.

最佳解答:

p.s.在"@"邊的是註解因技術問題,sin^2 (x) 會寫成 (sinx)^2,敬祈見諒 1. L.H.S.=(b^2+c^2-a^2)/(2bc) = [ (ksinB)^2 + (ksinC)^2 - (ksinA)^2 ] / [2 (ksinB) (ksinC)] = k^2 [ (sinB)^2 + (sinC)^2 - (sinA)^2 } / k^2 ( 2sinBsinC ) = { (sinB)^2 + (sinC)^2 - [sin(π-B-C)]^2 } / 2sinBsinC @代 A= π-B-C = { (sinB)^2 + (sinC)^2 - [sin(B+C)]^2 } / 2sinBsinC @因為 sin(π-x)=sinx = [ (sinB)^2 + (sinC)^2 - (sinBcosC+sinCcosB)^2 ] / 2sinBsinC = [ (sinB)^2 + (sinC)^2 - (sinBcosC)^2 - 2(sinBcosC)(sinCcosB) - (sinCcosB)^2 ] / 2sinBsinC = { [1-(cosC)^2] (sinB)^2 + [1-(cosB)^2] (sinC)^2 - 2sinBcosBsinCcosC } / 2sinBsinC = [ (sinBsinC)^2 + (sinBsinC)^2 - 2sinBcosBsinCcosC ] / 2sinBsinC = sinBsinC- cosBcosC @分子和分母一起消除2sinBsinC = - cos (B+C) = cos ( π-B-C ) @cos(π-x) = - cos x = cosA = R.H.S. From the above, we are given that a/sinA=b/sinB=c/sinC=k, which is known as the Sine Law. Then we proved that (b^2+c^2-a^2)/(2bc)=cos A. So we have proved the Cosine Law from the Sine Law. 2. (a) R.H.S.= [(1-A)/(1+A)]．(1/tanβ) = { 1-[cos(x+β)/cos(x-β)] / 1+[cos(x+β)/cos(x-β)] }．(1/tanβ) @代 A=cos(x+β)/cos(x-β)] = [ cos(x-β)-cos(x+β) / cos(x-β)+cos(x+β) ]．(1/tanβ) @分子和分母乘以cos(x-β) = ( 2sinxsinβ / 2cosxcosβ )．(1/tanβ) @用a.maths公式cosX+cosY 和cosX-cosY = tanxtanβ ?(1/tanβ) = tanx = L.H.S. (b) cos(x+ (π/6)) = (1/2) cos (x- (π/6)) From the above identity, we know that tanx = (1-A/1+A)(1/tanβ) if cos(x+β) = Acos(x-β), Therefore, substitute β=π/6, A=1/2 we can deduce that tanx = [1-(1/2)] / [1+(1/2)] ．[1/tan(π/6)] tanx = 1/√3 @"√" 是開方根 x = nπ 土 (π/6) , where n is an integer @土是+ 或 - 希望幫到你 2007-12-15 01:29:46 補充：如有問題,歡迎e-mail給我。小弟若有冒犯的地方,請閣下多多包涵。 2007-12-15 11:44:45 補充： Sorry, i missed that 0≦ x ≦ 2π. It should be x = π/6 or 7π/6 2007-12-15 12:51:56 補充： If it is general soultion, x should be nπ + (π/6) , where n is an integer