a maths...(20PT)

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