Question

Write equations whose roots are equal to numbers
$si{n}^2\frac{\pi }{2n+1},si{n}^2\frac{2\pi }{2n+1},si{n}^2\frac{3\pi }{2n+1},.....,si{n}^2\frac{n\pi }{2n+1}$

Answer 1

From Demoivre's theorem we know that
$sin(2n+1)\alpha {=}^{2n+1}{C}_1(1-si{n}^2\alpha {)}^{n}sin\alpha {-}^{2n+1}{C}_3(1-si{n}^2\alpha {)}^{n-1}si{n}^3\alpha +......+(-1{)}^{n}si{n}^{2n+1}\alpha$.
It follows that the numbers
$sin\frac{\pi }{2n+1},sin\frac{2\pi }{2n+1},........,sin\frac{n\pi }{2n+1}$,
$sin\left(\frac{-\pi }{2n+1}\right)=-sin\left(\frac{\pi }{2n+1}\right),sub\left(\frac{-2\pi }{2n+1}\right)=-sin\left(\frac{2\pi }{2n+1}\right),........,sin\left(\frac{-n\pi }{2n+1}\right)=-sin\frac{n\pi }{2n+1}$
where the roots of the equation.
${}^{2n+1}{C}_1(1-x^2{)}^{n}x{-}^{2n+1}{C}_3(1-x^2{)}^{n-1}{x}^3+.....+(-1{)}^n{x}^{2n+1}=0$
of the (2n + 1)th degree
Consequently, the numbers $si{n}^2\frac{\pi }{2n+1},si{n}^2\frac{2\pi }{2n+1},.........,si{n}^2\frac{n\pi }{2n+1}$ are the roots of the equation
${}^{2n+1}{C}_1(1-x{)}^n{-}^{2n+1}{C}_3(1-x{)}^{n-1}x+........+(-1{)}^n{x}^n=0$ of $n^{th}$ degree.