✔ 最佳答案
1
Consider the equation x^(2n)-2(x^n)(a^n)cos nθ+(a^2n)
treating it as quadratic in x^n, its too roots are
a^n cos nθ √[(a^2n)cos^2 nθ- (a^2n)]
=a^n cos nθ (cos nθ isin nθ)
and hence the 2n roots of x are
a(cos nθ isin nθ)^(1/n)
or a[cos (θ+ kπ/n) (isin n + kπ/n)], k=0,1,2,...n-1
Therefore
x^(2n)-2(x^n)(a^n)cos nθ+(a^2n) = Π{x^2- 2xa[cos (θ+ 2kπ/n) + a^2] }
Put a=1 and x=-1 and n is odd n=2m+1
2(1+ cos (2m+1)θ)=Π 2{1+ [cos (θ+ 2kπ/(2m+1)) ] }
4cos^2 (2m+1)θ/2= 4^(2m+1) Πcos^2 (θ/2 + kπ/(2m+1))
Put β=θ/2
cos^2 (2m+1)β= 4^(2m) Πcos^2 (β + kπ/(2m+1))
Put θ=β
cos^2 (2m+1)θ= 4^(2m) Πcos^2 (θ + kπ/(2m+1))
Take θ= 0
4^(2m) Πcos^2 ( kπ/(2m+1)) = 1
We notice that cos^2 kπ/(2m+1)= cos^2 (2m+1-k)π/(2m+1)
4^(m) Πcos^2 ( kπ/(2m+1)) = 1
(cosθ)^2 *(cos 2θ)^2 *...*(cos nθ)^2 = 1/ 4^n
2
cos(θ1+θ2+...+θn)+isin(θ1+θ2+...+θn)
=(cosθ1+isinθ1)(cosθ2+isinθ2)...(cosθn+isinθn)
=cosθ1cosθ2...cosθn(1+ itanθ1)(1+ itanθ2)...(1+ itanθn)
Equating real and imaginary part
cos(θ1+θ2+...+θn)=cosθ1cosθ2...cosθn(1-Σ2+Σ4-...)
sin(θ1+θ2+...+θn)=cosθ1cosθ2...cosθn(Σ1-Σ3+Σ5-...)
where Σr denotes the sum of the product of t1,t2...tn taken r at a time
So tan(θ1+θ2+...+θn)=(Σ1-Σ3+Σ5-...)/(1-Σ2+Σ4-...)
Put θ1=θ2=...=θn=θ, then Σr=nCr tan^rθ
tan nθ=(nC1 tanθ- nC3 tan^3θ+ nC5 tan^5θ-...)/(1- nC2 tan^2θ- nC4 tan^4θ+...)
Put n=2m+1
tan (2m+1)θ=(nC1 tanθ- nC3 tan^3θ+ nC5 tan^5θ-...)/(1- nC2 tan^2θ- nC4 tan^4θ+...)
Consider the equation tan (2m+1)θ= 0 , we have θ= kπ/(2m+1)
Hence the roots of nC1 - nC3 tan^2θ+ nC5 tan^4θ-... are
θ= kπ/(2m+1)
We notice that tan^2 kπ/(2m+1)= tan^2 (2m+1-k)π/(2m+1)
Take β= tan^2θ, then the roots of nC1 - nC3 β + nC5 β^2 -... are
tan^2 kπ/(2m+1)
Sum of roots
=[tan^2 π/(2m+1)+ tan^2 2π/(2m+1)+... + tan^2 mπ/(2m+1)]
=(nCn-1)
=m(2m+1)
(tanθ)^2 + (tan 2θ)^2 +...+(tan nθ)^2 = n(2n+1)
2009-01-17 16:44:14 補充：
煩惱即是菩提
我最初那個回答長到超過字數限制呀
2009-01-17 16:46:02 補充：
Note：Using the method of (cotθ)^2 + (cot 2θ)^2 +...+ (cot nθ)^2 = n(2n-1) /3 cannot derive the previous result﹐as illustrated in the following example。
Prove tan^2 π/(4n)+ tan^2 3π/4n+... tan^2 (2n-1)π/4n = n(2n-1)
(cosθ+isinθ)^n
=cos^nθ+ncosθ(isinθ)^(n-1)+...+(isinθ)^n
2009-01-17 16:46:29 補充：
Comparing real and imaginary part
cos nθ=cos^nθ- nC2cos^(n-2)θsin^2 θ+ nC4cos^(n-4)θsin^4θ+...
cos nθ=(cos^n θ)(1- nC2tan^2θ-nC3tan^4θ+... )
If n is even and cos nθ= 0 => cos 2mθ= 0 (Let n=2m)
2mθ= (k+1/2)π(k=0,1,2,...2m)
θ= (k+1/2)π/(2m)
2009-01-17 16:46:44 補充：
Also
cos 2mθ=(cos^(2m) θ)(1- nC2tan^2θ-nC3tan^4θ+...)
We notice that tan^2 (k+1/2)π/(2m)= tan^2 (2m- k- 1/2)π/(2m) (k=1,2,3,...m)
Let θ= tan^2 θ
Then the root of (1- nC1θ^2 -nC3θ^4+...= 0)
are tan^2 (1/2)π/(2m), tan^2 (3/2)π/(2m),... tan^2 (m-1/2)π/(2m)
2009-01-17 16:46:49 補充：
Sum of roots
=tan^2 (1/2)π/(2m)+ tan^2 (3/2)π/(2m)+... tan^2 (m-1/2)π/(2m)
=nC2
=m(2m-1)
tan^2 (1/2)π/(2n)+ tan^2 (3/2)π/(2n)+... tan^2 (n+1/2)π/(2n) = n(2n+1)
or
tan^2 π/(4n)+ tan^2 3π/4n+... tan^2 (2n-1)π/4n = n(2n-1)
