1hint : try expanding \tan n\theta as a polynomial on \tan\theta
many good questions are directly solved by experts......let this one be here....
and lets see if an aspirant is able to sove it
well the question is quite basic and trivial :
\sum_{k=1}^{n-1}{\tan^2\frac{k\pi}{2n}}=\frac{(n-1)(2n-1)}{3}
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8 Answers
11Method to kafi simple ai.
solve he sreies of sin2∂ n cos2∂ differently after converting into double angle formulae thru C+iS method.
Will post soon................
1first u must post the proof before quoting it simple
Prove that if the polynomial P(x)=a_0x^n+a_1x^{n-1}+\cdots\cdots+a_{n-1}x+a_nwith integral
coefficients has odd value for x=0 and x=1 , then the equation p(x)=0 can't have integral roots.
1one more
the coefficient of x2 in
\prod_{k=0}^{n}{\left(1+2^kx \right)}
23i kno u wont accept this answer [3]
p(x)=a_{n}+a_{n-1}x+a_{n-2}x^{2}+a_{n-3}x^{3}+...+a_{o}x^{n}
p(0)=a_{n}=odd
p(2x)=a_{n}+a_{n-1}(2x)+a_{n-2}(2x)^{2}+a_{n-3}(2x)^{3}+...+a_{o}(2x)^{n}
thus p(2x) = an + even no. = odd + even no.
and odd + even can never be zero
hence no even solutions
now
p(2x+1)=a_{n}+a_{n-1}(2x+1)+a_{n-2}(2x+1)^{2}+a_{n-3}(2x+1)^{3}+...+a_{o}(2x+1)^{n}
now consider the term (2x+1)k
(2x+1)^{k}= \sum_{r=0}^{k}{^{k}C_{r}}(2x)^{r}
(2x+1)^{k}= \sum_{r=0}^{k}{^{k}C_{r}}(2x)^{r}= ( 1 + ^{k}C_{2}(2x) +.....)
in this expansion only the first term i.e 1 is odd, rest all are even nos,
hence p(2x+1)=a_{n}+a_{n-1}(2x+1)+a_{n-2}(2x+1)^{2}+a_{n-3}(2x+1)^{3}+...+a_{o}(2x+1)^{n}
now take out 1 from each bracket of the type (2x+1)k
\Rightarrow p(2x+1)=(a_{n}+a_{n-1}+a_{n-2}+...+a_{o})+ (a_{n-1}(2x) +a_{n-2}(4x^{2}+4x)+....)
=p(1)+ even \; no.
= odd + even
again wich can never be zero,
hence P(x) can never be zero for odd and even nos, i.e for integers
211)
from (cos∂+ i sin∂)^2n
we get sin 2n A= cos^2n A ( 2nC1 tanA - 2nC3tan^3 A +......)
putting A = ∩/2n we see tan ∩/2n,tan 2∩/2n.....tan (2n-1)∩/2n are roots of
( 2nC1 tanA - 2nC3tan^3 A +......)=0 (highest degree of tanA is 2n-1)
so the equation 2nC1- 2nC3tan^2 ..... =0 has tan ∩/2n,tan 2∩/2n.....tan (2n-2)∩/2n as roots
(highest degree is 2n-2)
if we make this equation as a polynomial in tan^2 A we will see that the it has tan2∩/2n , tan22∩/2n.... tan2(n-1)∩/2n is its roots
so sum of roots is given by 2nC3/2nC1=(2n-1)(n-1)/3