An identity involving the Chebyshev polynomials

Question

Let $n \in \{0, 1, 2, \dots\}$ and let $T_n$ denote the Chebyshev polynomial of degree $n$: $T_n(x) = \cos\left(n \arccos(x)\right)$. Let $t_0, t_1, \dots, t_n$ be $T_{n + 1}$'s roots: $t_i = \cos\left(\frac{2i + 1}{2n + 2} \pi\right)$. How can I show that for every $i \in \{0, 1, \dots, n\}$, $$ T'_{n + 1}(t_i) T_n(t_i) = n + 1 $$

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My attempt at a solution $$ \begin{align} T'_{n + 1}(t_i) T_n(t_i) &= \frac{\sin\left((n + 1) \arccos(t_i)\right) \cdot (n + 1)}{\sqrt{1 - t_i^2}} \cdot \cos\left(n \arccos(t_i)\right) \\ &= (n + 1) \cdot \frac{\sin\left((n + 1) \arccos\left(\cos\left(\frac{2i + 1}{2n + 2} \pi\right)\right)\right)}{\sqrt{1 - \cos^2\left(\frac{2i + 1}{2n + 2} \pi\right)}} \cdot \cos\left(n \arccos\left(\cos\left(\frac{2i + 1}{2n + 2} \pi\right)\right)\right) \\ &= (n + 1) \cdot \frac{\sin\left(\frac{2i + 1}{2} \pi\right) \cdot \cos\left(\frac{2i + 1}{2n + 2} n \pi \right)}{\sin\left(\frac{2i + 1}{2n + 2} \pi\right)} \\ &=\ ? \end{align} $$

Answer 1

$$T'_{n+1}\left(t_i\right)T_n\left(t_i\right) =\frac{\sin\left(\left(n+1\right)\arccos\left(t_i\right)\right)}{\sqrt{1-t_i^2}}\left(n+1\right)\cos\left(n\arccos\left(t_i\right)\right) $$

$$=\left(n+1\right)\frac{\sin\left(\left(2i+1\right)\frac{\pi}{2}\right)}{\sqrt{1-t_i^2}}\cos\left(n\arccos\left(t_i\right)\right) =\left(n+1\right)\frac{\sin\left(i\pi\right)\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\cos\left(i\pi\right)}{\sqrt{1-t_i^2}}\cos\left(n\arccos\left(t_i\right)\right)$$

$$ =\left(n+1\right)\frac{0+\left(-1\right)^{i}}{\sqrt{1-t_i^2}}\cos\left(n\arccos\left(t_i\right)\right) =\left(n+1\right)\left(-1\right)^{i}\frac{\cos\left(n\arccos\left(t_i\right)\right)}{\sqrt{1-t_i^2}} $$

$$ =\left(n+1\right)\left(-1\right)^{i}\frac{\cos\left(\left(n+1-1\right)\arccos\left(t_i\right)\right)}{\sqrt{1-t_i^2}}$$

$$=\left(n+1\right)\left(-1\right)^{i}$$$$\times\frac{\cos\left(\left(n+1\right)\arccos\left(t_i\right)\right)\cos\left(\arccos\left(t_i\right)\right)+\sin\left(\left(n+1\right)\arccos\left(t_i\right)\right)\sin\left(\arccos\left(t_i\right)\right)}{\sqrt{1-t_i^2}} $$

$$=\left(n+1\right)\left(-1\right)^{i}\frac{T_{n+1}\left(t_i\right)\cos\left(\arccos\left(t_i\right)\right)+\sin\left(\left(n+1\right)\arccos\left(t_i\right)\right)\sqrt{1-t_i^2}}{\sqrt{1-t_i^2}}$$

$$=\left(n+1\right)\left(-1\right)^{i}\sin\left(\left(n+1\right)\arccos\left(t_i\right)\right) =\left(n+1\right)\left(-1\right)^{i}\sin\left(\left(2i+1\right)\frac{\pi}{2}\right)=\left(n+1\right)\left(-1\right)^{i}\left(-1\right)^{i}=n+1.$$

Answer 2

We have $$\dfrac{dT_{n+1}(\cos(\theta))}{dt} = \dfrac{dT_{n+1}(\cos(\theta))}{d\theta}\times \dfrac{d \theta}{dt} = \dfrac{d(\cos((n+1)\theta))}{d\theta} \times \dfrac{-1}{\sin(\theta)} = (n+1) \dfrac{\sin((n+1)\theta)}{\sin(\theta)}$$ Hence, we have $$\dfrac{dT_{n+1}(\cos(\theta))}{dt} \times T_n(\cos(\theta)) = (n+1) \dfrac{\sin((n+1)\theta) \cos(n\theta)}{\sin(\theta)} \,\,\, (\spadesuit)$$ Given $\theta = \dfrac{2i+1}{2n+2} \pi$, we have $\sin((n+1)\theta) = \sin\left(\dfrac{(2i+1)}2\pi \right) \,\,\, (\clubsuit)$ and \begin{align} \cos(n\theta) & = \cos\left(\dfrac{n}{n+1} (2i+1) \dfrac{\pi}2\right) = \cos\left((2i+1)\dfrac{\pi}2 - \dfrac{(2i+1)}{2(n+1)}\pi\right)\\ & = \sin\left((2i+1)\dfrac{\pi}2\right) \sin\left(\dfrac{2i+1}{2(n+1)}\pi\right) = \sin\left((2i+1)\dfrac{\pi}2\right) \sin(\theta) \,\,\, (\heartsuit) \end{align} Hence, using $(\clubsuit)$ and $(\heartsuit)$ in $(\spadesuit)$, we obtain \begin{align} \dfrac{dT_{n+1}(\cos(\theta))}{dt} \times T_n(\cos(\theta)) & = (n+1) \dfrac{\sin\left((2i+1)\dfrac{\pi}2\right) \sin\left((2i+1)\dfrac{\pi}2\right) \sin(\theta)}{\sin(\theta)}\\ & = (n+1)\sin^2\left((2i+1)\dfrac{\pi}2\right) = (n+1) \end{align}