Relation between root and coeff

難度是有﹐不過還是比較好做﹐因為有些更難做的
(a) cot6α= ( cot 6 α-15 cot4α+15cot²α-1)/(6 cot 5α- 20 cot³α+6 cot α)

(b)(i)Put α= ß to (a) then replace cot α by x , we get
x 6 - 6x5 cot 6ß – 15x4 +20 x3 cot 6ß +15x2-6x cot 6ß -1=0

Σ (k=0 to 5)cot (ß+k(pi)/6)= 6 cot 6ß

(ii) Hence, deduce Σ (k=1 to 5) csc² (ß +k(pi)/6)= 36 csc² 6ß -csc²ß
(iii) Find Σ(k=1 to 5)cot ²(4k+3)(pi)/24
SOLUTION
(a) Have been done
(b)
(i)
Put α= ß
cot6ß= ( cot 6 ß-15 cot4ß+15cot²ß-1)/(6 cot 5ß- 20 cot³ß+6 cot ß)
Put
cot α by x
cot6ß= ( x^ 6-15 x^4+15 x^2-1)/(6 x^5- 20 x^3+6 x)
So (cot6ß)(6 x^5- 20 x^3+6 x)= ( x^ 6-15 x^4+15 x^2-1)
Or
x 6 - 6x5 cot 6ß – 15x4 +20 x3 cot 6ß +15x2-6x cot 6ß -1=0
Since x=cotß is a root of this equation. But using fundamental theorem of algebra, there are 6 roots
So we want to find 5 more roots.
We notice that cot 6(ß+k(pi)/6)=cot (6ß+k(pi))=cot 6ß
So cot(ß+k(pi)/6) k=1 to 5 is the other 5 roots
You can just think that you want to find a value such that cot6x=0
if ß is one then the other five is ß+k(pi)/6
(ii)

Σ (k=0 to 5)cot (ß+k(pi)/6)= 6 cot 6ß
[Σ (k=0 to 5)cot (ß+k(pi)/6)]^2= 36 cot^2 6ß
Σ (k=0 to 5) cot² (ß +k(pi)/6)-30= 36 cot² 6ß
Since csc^2x=cot^2x+1
Σ (k=0 to 5) [csc² (ß +k(pi)/6)-1]-30= 36 (csc² 6ß-1)
Σ (k=1 to 5) csc² (ß +k(pi)/6)= 36 csc² 6ß-csc^2ß
(iii)
sub ß=3pi/24
Σ (k=0 to 5) cot² ((4k+3)(pi)/24) -30= 36 cot² (3/4)pi=36
Σ (k=0 to 5) cot² ((4k+3)(pi)/24) =6
Σ(k=1 to 5)cot ²(4k+3)(pi)/24=6-cot^2(pi/8)=6-[1/(√2-1)]^2=66-(3+2√2)=63-2√2

(b) (i)

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(ii)

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(iii)

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jan08/tinhoi7.jpg