[proofplan]
We derive the derivative formulas for $T_{2n}'$ and $T_{2n+1}'$ from the representation $T_m'(x) = m\sin(m\theta)/\sin\theta$ (obtained by differentiating $T_m(\cos\theta) = \cos(m\theta)$ via the chain rule). For the even case $m = 2n$, we show $\sin(2n\theta)/\sin\theta = 2\sum_{k=1}^{n}\cos((2k-1)\theta)$ by multiplying both sides by $\sin\theta$ and verifying that the right-hand side telescopes to $\sin(2n\theta)$ via the product-to-sum identity. The odd case $m = 2n+1$ is handled analogously with a similar telescoping argument.
[/proofplan]
[step:Derive the formula $T_m'(x) = m\sin(m\theta)/\sin\theta$ from the chain rule]
Let $x = \cos\theta$ with $\theta \in (0, \pi)$ (so that $\sin\theta > 0$). By definition, $T_m(x) = \cos(m\theta)$. Differentiating with respect to $x$ using the chain rule:
\begin{align*}
T_m'(x) = \frac{d}{dx}\cos(m\theta) = -m\sin(m\theta) \cdot \frac{d\theta}{dx}.
\end{align*}
Since $x = \cos\theta$, we have $\frac{dx}{d\theta} = -\sin\theta$, so $\frac{d\theta}{dx} = \frac{-1}{\sin\theta}$. Substituting:
\begin{align*}
T_m'(x) = -m\sin(m\theta) \cdot \frac{-1}{\sin\theta} = \frac{m\sin(m\theta)}{\sin\theta}.
\end{align*}
[/step]
[step:Prove the even case: $T_{2n}'(x) = 4n\sum_{k=1}^{n} T_{2k-1}(x)$ via telescoping]
For $m = 2n$, the derivative formula gives $T_{2n}'(x) = \frac{2n\sin(2n\theta)}{\sin\theta}$. The claimed identity becomes
\begin{align*}
\frac{2n\sin(2n\theta)}{\sin\theta} = (2n) \cdot 2\sum_{k=1}^{n} T_{2k-1}(\cos\theta) = 4n\sum_{k=1}^{n}\cos((2k-1)\theta).
\end{align*}
Dividing both sides by $2n$ (which is nonzero for $n \geq 1$), it suffices to show
\begin{align*}
\frac{\sin(2n\theta)}{\sin\theta} = 2\sum_{k=1}^{n}\cos((2k-1)\theta).
\end{align*}
Multiply both sides by $\sin\theta$. The right-hand side becomes
\begin{align*}
2\sum_{k=1}^{n}\cos((2k-1)\theta)\sin\theta.
\end{align*}
Applying the product-to-sum identity $2\cos\alpha\sin\beta = \sin(\alpha + \beta) - \sin(\alpha - \beta)$ with $\alpha = (2k-1)\theta$ and $\beta = \theta$:
\begin{align*}
2\cos((2k-1)\theta)\sin\theta = \sin(2k\theta) - \sin((2k-2)\theta).
\end{align*}
Summing from $k = 1$ to $k = n$:
\begin{align*}
\sum_{k=1}^{n}\bigl[\sin(2k\theta) - \sin(2(k-1)\theta)\bigr] = \sin(2n\theta) - \sin(0) = \sin(2n\theta),
\end{align*}
where the sum telescopes (the $\sin(2k\theta)$ term from index $k$ cancels with the $-\sin(2(k-1+1)\theta)$ term from index $k+1$). This confirms $\sin(2n\theta) = 2\sin\theta\sum_{k=1}^{n}\cos((2k-1)\theta)$, and dividing by $\sin\theta > 0$ yields the result.
[guided]
The telescoping mechanism is: each summand $\sin(2k\theta) - \sin(2(k-1)\theta)$ jumps from one value to the next in an arithmetic progression of arguments $0, 2\theta, 4\theta, \ldots, 2n\theta$. Writing out the first few terms:
\begin{align*}
k=1: &\quad \sin(2\theta) - \sin(0), \\
k=2: &\quad \sin(4\theta) - \sin(2\theta), \\
k=3: &\quad \sin(6\theta) - \sin(4\theta), \\
&\quad \vdots \\
k=n: &\quad \sin(2n\theta) - \sin((2n-2)\theta).
\end{align*}
Adding these, every intermediate term cancels in pairs, leaving only $\sin(2n\theta) - \sin(0) = \sin(2n\theta)$.
Note the parity structure: $T_{2n}'$ is a sum of odd-degree Chebyshev polynomials $T_{2k-1}$. This is expected because $T_{2n}$ is an even polynomial (in $x$), so its derivative must be an odd polynomial, which is a linear combination of odd-degree Chebyshev polynomials.
[/guided]
[/step]
[step:Prove the odd case: $T_{2n+1}'(x) = (2n+1)[T_0(x) + 2\sum_{k=1}^{n} T_{2k}(x)]$ via telescoping]
For $m = 2n+1$, the derivative formula gives $T_{2n+1}'(x) = \frac{(2n+1)\sin((2n+1)\theta)}{\sin\theta}$. The claimed identity becomes
\begin{align*}
\frac{(2n+1)\sin((2n+1)\theta)}{\sin\theta} = (2n+1)\left[1 + 2\sum_{k=1}^{n}\cos(2k\theta)\right],
\end{align*}
where $T_0(\cos\theta) = 1$ and $T_{2k}(\cos\theta) = \cos(2k\theta)$. Dividing by $(2n+1)$, it suffices to show
\begin{align*}
\frac{\sin((2n+1)\theta)}{\sin\theta} = 1 + 2\sum_{k=1}^{n}\cos(2k\theta).
\end{align*}
Multiply both sides by $\sin\theta$. The right-hand side becomes
\begin{align*}
\sin\theta + 2\sum_{k=1}^{n}\cos(2k\theta)\sin\theta.
\end{align*}
Applying the product-to-sum identity $2\cos\alpha\sin\beta = \sin(\alpha + \beta) - \sin(\alpha - \beta)$ with $\alpha = 2k\theta$ and $\beta = \theta$:
\begin{align*}
2\cos(2k\theta)\sin\theta = \sin((2k+1)\theta) - \sin((2k-1)\theta).
\end{align*}
Summing from $k = 1$ to $k = n$:
\begin{align*}
\sum_{k=1}^{n}\bigl[\sin((2k+1)\theta) - \sin((2k-1)\theta)\bigr] = \sin((2n+1)\theta) - \sin(\theta),
\end{align*}
by the same telescoping argument as in the even case (intermediate terms cancel in pairs). Adding the initial $\sin\theta$ term:
\begin{align*}
\sin\theta + \sin((2n+1)\theta) - \sin\theta = \sin((2n+1)\theta).
\end{align*}
This confirms $\sin((2n+1)\theta) = \sin\theta\bigl[1 + 2\sum_{k=1}^{n}\cos(2k\theta)\bigr]$, and dividing by $\sin\theta > 0$ yields the result.
[guided]
The telescoping structure is identical to the even case, but now the arithmetic progression of arguments is $\theta, 3\theta, 5\theta, \ldots, (2n+1)\theta$ (odd multiples). Writing out the terms:
\begin{align*}
k=1: &\quad \sin(3\theta) - \sin(\theta), \\
k=2: &\quad \sin(5\theta) - \sin(3\theta), \\
&\quad \vdots \\
k=n: &\quad \sin((2n+1)\theta) - \sin((2n-1)\theta).
\end{align*}
The sum telescopes to $\sin((2n+1)\theta) - \sin(\theta)$. Adding the standalone $\sin\theta$ from the $T_0$ term recovers $\sin((2n+1)\theta)$.
Note the parity structure: $T_{2n+1}'$ is a sum of even-degree Chebyshev polynomials $T_0, T_2, \ldots, T_{2n}$. This is because $T_{2n+1}$ is an odd polynomial (in $x$), so its derivative is even, hence a combination of even-degree Chebyshev polynomials. The constant term $T_0 = 1$ appears with coefficient $1$ (not $2$), which is consistent with the standard Chebyshev expansion convention where the $T_0$ term has a different normalisation.
[/guided]
[/step]
