Chebyshev polynomial practice: medium sum

I'll post a full solution tomorrow, too tired rn.

For now, note that $\displaystyle \sum_1^{45} \csc^2(2k-1)=\displaystyle \sum_1^{45} \sec^2(2k-1)$

And that when you reverse the coefficients of $f(x)$ that the roots become inverted (eg: if the roots of $f(x)$ are $a,b,c$ then when the coefficients are reversed, the new polynomial has roots $\frac{1}{a},~\frac{1}{b},~\frac{1}{c}$ )

Also, we are once again working with $T_{90}(\sqrt x)$

Let $a$ be a positive integer. There is a fairly well-known identity, which we will prove below: $\sum_{k = 1}^{a - 1} \csc^2 \frac {k\pi}a = \frac {a^2 - 1}3. \tag{1}$ In words, this identity gives the sum the square cosecants of all the integer multiples of $\pi/a$ in the first two quadrants. We are interested in the sum of the square cosecants of the odd integer multiples of $1^\circ = \pi/180$ in the first quadrant, so we can find the sum of all the integer multiples in the first two quadrants, subtract the sum of the even integer multiples, and divide by 2: $\sum_{k = 1}^{45} \csc^2 \left( \frac {(2k - 1)\pi}{180} \right) = \frac 12 \left( \sum_{k = 1}^{179} \csc^2 \frac {k\pi}{180} - \sum_{k = 1}^{89} \csc^2 \frac {2k\pi}{180} \right) = \frac 12 \left( \frac {180^2 - 1}3 - \frac {90^2 - 1}3 \right) = \boxed{4050}.$

Now let us prove (1). For brevity, put $\omega_k = \sin^2 (k\pi/a)$ . Let $P_a$ denote the polynomial defined by $P_a(x) = \prod_{k = 1}^{a - 1}(x - \omega_k),$ and let $r_a$ and $s_a$ denote the constant and linear coefficient of $P_a$ . By Vieta, $\sum_{k = 1}^{a - 1} \csc^2 \frac {k\pi}a = \sum_{k = 1}^{a - 1} \frac 1{\omega_k} = \frac {\sum_k \prod_{j \neq k} \omega_j}{\prod_k \omega_k} = -\frac {s_a}{r_a}.$ We will show a recursive process that generates $P_a$ , and then we will use that process to show inductively that $r_a = a^2 \quad \text{and} \quad s_a = -\frac {(a^2 - 1)a^2}3, \tag{2}$ from which (1) follows.

Let $S_a$ denote the $a$ th spread polynomial, which is defined by $S_a(x) = \frac 12 [1 - T_a(1 - 2x)],$ where $T_a$ is the $a$ th Chebyshev polynomial. One property of $S_a$ is that $S_a(\sin^2 \theta) = \sin^2 a\theta$ . Another is that $S_a$ can be generated recursively by starting with $S_1 = x$ and $S_2 = 4x - 4x^3$ and applying the rule $S_{a + 1}(x) = 2(1 - 2x)S_a(x) - S_{a - 1}(x) + 2x. \tag{3}$ Notice that $S_a(x) = 0$ for $x = 0, \omega_1, \omega_2, \dots, \omega_{a - 1}$ . This means that $0, \omega_1, \omega_2, \dots, \omega_{a - 1}$ are exactly the roots of $S_a$ , or equivalently, $\omega_1, \omega_2, \dots, \omega_{a - 1}$ are exactly the roots of $S_a(x)/x$ . Thus we have shown $P_a(x) = \frac {S_a(x)}x.$

Now we will show by induction that (2) holds for all $a \geq 1$ . When we generate $P_a$ for $a = 1$ and $a = 2$ , we see $P_1(x) = 1 + 0x \quad \text{and} \quad P_2(x) = 4 - 4x,$ and we can verify directly that (2) holds. Fix $b \geq 2$ and find $P_b$ and $P_{b - 1}$ . Extract the coefficients $r_b$ , $s_b$ , $r_{b - 1}$ , and $s_{b - 1}$ and suppose that (2) holds when $a = b$ and $a = b - 1$ . By (3), $P_{b + 1}(x) = \frac {S_{b + 1}(x)}x = 2(1 - 2x) \frac {S_b(x)}x - \frac {S_{b - 1}(x)}x + 2 = 2(1 - 2x) P_b(x) - P_{b - 1}(x) + 2.$ From this equation, we find the constant and linear coefficient of $P_{b + 1}$ are $r_{b + 1} = 2r_b - r_{b - 1} + 2 \quad \text{and} \quad s_{b + 1} = - s_{b - 1} - 2(2rb - s_b) .$ Applying (2) and simplifying, we obtain $r_{b + 1} = (b + 1)^2 \quad \text{and} \quad s_{b + 1} = -\frac{[(b + 1)^2 - 1](b + 1)^2}3.$ Thus we have shown that (2) holds when $a = b + 1$ . By induction, (2) holds for all $a$ , and this is the result that confirms the identity in (1).

Note: I found this proof of (1) (along with several others) at StackExchage .