Find the equation whose roots are: 1^° sin^2dfracπ2n+1…

Find the equation whose roots are: $\ 1^\circ\ \sin^2\dfrac\pi{2n+1},\ \sin^2\dfrac{2\pi}{2n+1},\ \ldots,\ \sin^2\dfrac{n\pi}{2n+1};\quad 2^\circ\ \cot^2\dfrac\pi{2n+1},\ \cot^2\dfrac{2\pi}{2n+1},\ \ldots,\ \cot^2\dfrac{n\pi}{2n+1}$.

Part $1^\circ$ · Polynomial with roots $\sin^2\frac{k\pi}{2n+1}$

Step 1 · Expand $\sin((2n+1)\theta)$ via the binomial theorem

From $(\cos\theta+i\sin\theta)^{2n+1}=\cos((2n+1)\theta)+i\sin((2n+1)\theta)$, taking the imaginary part (terms with an odd power of $i\sin\theta$, index $k=2j+1$, using $i^{2j+1}=i(-1)^j$):

$$ \sin((2n+1)\theta)=\sum_{j=0}^n(-1)^j\binom{2n+1}{2j+1}\cos^{2n-2j}\theta\,\sin^{2j+1}\theta $$

Step 2 · Divide by $\sin\theta$ and substitute $x=\sin^2\theta$

Using $\cos^2\theta=1-x$:

$$ \frac{\sin((2n+1)\theta)}{\sin\theta}=\sum_{j=0}^n(-1)^j\binom{2n+1}{2j+1}(1-x)^{n-j}x^j =: P_n(x) $$

a polynomial of degree $n$ in $x$.

Step 3 · Identify the roots

$P_n(x)=0$ exactly when $\sin((2n+1)\theta)=0$ with $\sin\theta\neq0$, i.e. $\theta=\dfrac{k\pi}{2n+1}$ for $k=1,\ldots,n$ (the values $k=n+1,\ldots,2n$ repeat the same $x=\sin^2\theta$ by symmetry, so exactly $n$ distinct roots — matching the degree):

$$ \boxed{P_n(x)=\sum_{j=0}^n(-1)^j\binom{2n+1}{2j+1}(1-x)^{n-j}x^j=0} $$

Step 4 · The polynomial explicitly, small $n$

(leading coefficient $(-4)^n$, constant term $2n+1$ — visible patterns in every case).

Part $2^\circ$ · Polynomial with roots $\cot^2\frac{k\pi}{2n+1}$

Step 5 · Expand $(\cot\theta+i)^{2n+1}$ directly

Let $y=\cot\theta$. By the binomial theorem, the imaginary part of $(y+i)^{2n+1}$ (again picking out odd powers of $i$) is:

$$ \text{Im}\big[(y+i)^{2n+1}\big]=\sum_{j=0}^n(-1)^j\binom{2n+1}{2j+1}y^{2n-2j} $$

Step 6 · Why this vanishes exactly at $\theta=\frac{k\pi}{2n+1}$

Since $\cos\theta+i\sin\theta=(\sin\theta)(y+i)$ (with $y=\cot\theta$), raising to the power $2n+1$: $\ (y+i)^{2n+1}=\dfrac{\cos((2n+1)\theta)+i\sin((2n+1)\theta)}{\sin^{2n+1}\theta}$. At $\theta=\frac{k\pi}{2n+1}$: $\sin((2n+1)\theta)=\sin(k\pi)=0$, so $(y+i)^{2n+1}$ is real there — its imaginary part vanishes.

Step 7 · Reduce to a degree-$n$ polynomial in $y^2$

The expression from Step 5 involves only even powers of $y$, so writing $Y=y^2=\cot^2\theta$:

$$ \boxed{Q_n(Y)=\sum_{j=0}^n(-1)^j\binom{2n+1}{2j+1}Y^{n-j}=0} $$

Its $n$ roots are exactly $Y=\cot^2\dfrac{k\pi}{2n+1}$ for $k=1,\ldots,n$ (again, $k$ and $2n+1-k$ give the same $Y$ by $\cot(\pi-\varphi)=-\cot\varphi$, collapsing $2n$ candidate values of $y$ down to $n$ distinct values of $Y$).

Step 8 · The polynomial explicitly, small $n$

(leading coefficient $2n+1$, constant term $(-1)^n$ — the coefficients are exactly $\pm\binom{2n+1}{2j+1}$, alternating in sign).

Verification

Both $P_n$ and $Q_n$ were checked by direct numerical root-finding for $n=1,2,3,4$: in every case the computed roots matched $\sin^2\frac{k\pi}{2n+1}$ and $\cot^2\frac{k\pi}{2n+1}$ (respectively) to full floating-point precision.