✔ 最佳答案
我知道第二題的答案了。
2009-01-17 00:43:00 補充:
1
1+x+x^2+...+x^2n=0
(x^(2n+1)-1)/(x-1)=0
So (x^(2n+1)-1)=0
x^(2n+1)=cos2kπ+isin2kπ
x=cos2kπ/(2n+1)+isin2kπ/(2n+1) k=1,2,3,...2n
Note that k ≠ 0 is excluded because x≠1
Notice that cos 2(2n+1-k)π/(2n+1)+isin 2(2n+1-k)π/(2n+1)=cos 2kπ/(2n+1) - isin 2kπ/(2n+1)
So
1+x+x^2+...+x^2n
= Π[x-(cos 2kπ/(2n+1) + isin 2kπ/(2n+1)]
= Π[x-(cos 2kπ/(2n+1) + isin 2kπ/(2n+1)][x-(cos 2kπ/(2n+1) - isin 2kπ/(2n+1)]
=Π[x^2-2x(cos 2kπ/(2n+1)+1]
Put x=1, we have
2n+1=Π[2(1-(cos 2kπ/(2n+1))]=Π[4sin^2 kπ/(n+1))]=2^(2n)Π[sin^2 kπ/(n+1))]
Hence (sinθ)^2 * (sin 2θ)^2 *...* (sin nθ)^2 = (2n+1)/4^n
2
(cosθ+isinθ)^n
=cos^nθ+ncosθ(isinθ)^(n-1)+...+(isinθ)^n
Comparing real and imaginary part
sin nθ=nC1cos^(n-1)θsinθ-nC3cos^(n-3)θsin^3θ+nC5cos^(n-5)θsin^5θ+...
sin nθ=(sin^n θ)(nC1cot^(n-1)θ-nC3cot^(n-3)θ+nC5cot^(n-5)θ+...)
If n is odd and sin nθ= 0 => sin (2m+1)θ= 0 (Let n=2m+1)
θ= kπ/(2m+1) (k=0,1,2,...2m)
Also
sin (2m+1)θ=(sin^(2m+1) θ)(nC1cot^(2m)θ-nC3cot^(2m-2)θ+nC5cot^(2m-4)θ+...)
We notice that cot^2 kπ/(2m+1)= cot^2 (2m+1-k)π/(2m+1) (k=1,2,3,...m)
Let θ= cot^2 θ
Then the root of (nC1θ^m -nC3θ^(m-1)+nC5θ^(m-2)+...= 0)
are cot^2 π/(2m+1), cot^2 2π/(2m+1),... cot^2 mπ/(2m+1)
Sum of roots
=cot^2 π/(2m+1)+ cot^2 2π/(2m+1)+... +cot^2 mπ/(2m+1)
=nC3/nC1
=m(2m-1)/3
So (cotθ)^2 + (cot 2θ)^2 +...+ (cot nθ)^2 = n(2n-1) / 3
2009-01-17 16:50:19 補充:
結果用了3個方法﹐sinθ那條的方法是 cosθ那條的特例