[proofplan]
We verify directly that $u(\xi,\tau) = 4\arctan(e^\eta)$ with $\eta = a\xi + a^{-1}\tau$ satisfies the sine-Gordon equation $u_{\xi\tau} = \sin u$. The proof proceeds in three steps: first compute the mixed partial derivative $u_{\xi\tau}$ using the chain rule and hyperbolic identities; then evaluate $\sin u = \sin(4\arctan(e^\eta))$ using the double-angle formula and inverse trigonometric identities; finally confirm that both expressions equal $-2\operatorname{sech}(\eta)\tanh(\eta)$.
[/proofplan]
[step:Compute $u_\xi$, $u_\tau$, and $u_{\xi\tau}$]
Set $\eta = a\xi + a^{-1}\tau$, so $u = 4\arctan(e^\eta)$. Since $\partial \eta / \partial \xi = a$ and $\partial \eta / \partial \tau = a^{-1}$, the chain rule gives
\begin{align*}
u_\xi = \frac{4e^\eta}{1 + e^{2\eta}} \cdot a = \frac{4a \, e^\eta}{1 + e^{2\eta}}.
\end{align*}
To simplify, divide numerator and denominator by $e^\eta$:
\begin{align*}
u_\xi = \frac{4a}{e^\eta + e^{-\eta}} \cdot \frac{e^\eta}{e^\eta} \cdot \frac{1}{1} = \frac{4a}{e^\eta + e^{-\eta}} = \frac{4a}{2\cosh(\eta)} = 2a\operatorname{sech}(\eta).
\end{align*}
By an identical computation with $\partial \eta / \partial \tau = a^{-1}$:
\begin{align*}
u_\tau = 2a^{-1}\operatorname{sech}(\eta).
\end{align*}
Now differentiate $u_\xi = 2a\operatorname{sech}(\eta)$ with respect to $\tau$. Using $\frac{d}{d\eta}\operatorname{sech}(\eta) = -\operatorname{sech}(\eta)\tanh(\eta)$ and $\partial \eta / \partial \tau = a^{-1}$:
\begin{align*}
u_{\xi\tau} = 2a \cdot \bigl(-\operatorname{sech}(\eta)\tanh(\eta)\bigr) \cdot a^{-1} = -2\operatorname{sech}(\eta)\tanh(\eta).
\end{align*}
[/step]
[step:Evaluate $\sin(4\arctan(e^\eta))$ using double-angle identities]
We need to show that $\sin u = \sin(4\arctan(e^\eta)) = -2\operatorname{sech}(\eta)\tanh(\eta)$.
Set $\theta = \arctan(e^\eta)$, so $u = 4\theta$ and $\tan\theta = e^\eta$. The standard identities for inverse tangent give
\begin{align*}
\sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta} = \frac{2e^\eta}{1 + e^{2\eta}}, \qquad \cos(2\theta) = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1 - e^{2\eta}}{1 + e^{2\eta}}.
\end{align*}
Applying the double-angle formula $\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$:
\begin{align*}
\sin(4\arctan(e^\eta)) = 2 \cdot \frac{2e^\eta}{1 + e^{2\eta}} \cdot \frac{1 - e^{2\eta}}{1 + e^{2\eta}} = \frac{4e^\eta(1 - e^{2\eta})}{(1 + e^{2\eta})^2}.
\end{align*}
[/step]
[step:Convert to hyperbolic functions and confirm equality with $u_{\xi\tau}$]
Dividing numerator and denominator of $\frac{4e^\eta(1 - e^{2\eta})}{(1 + e^{2\eta})^2}$ by $e^{2\eta}$:
\begin{align*}
\frac{4e^\eta(1 - e^{2\eta})}{(1 + e^{2\eta})^2} = \frac{4(e^{-\eta} - e^{\eta})}{(e^{-\eta} + e^{\eta})^2} = \frac{-4 \cdot 2\sinh(\eta)}{(2\cosh(\eta))^2} = \frac{-2\sinh(\eta)}{\cosh^2(\eta)} = -2\operatorname{sech}(\eta)\tanh(\eta),
\end{align*}
where we used $e^\eta - e^{-\eta} = 2\sinh(\eta)$ and $e^\eta + e^{-\eta} = 2\cosh(\eta)$.
Since both $u_{\xi\tau}$ and $\sin u$ equal $-2\operatorname{sech}(\eta)\tanh(\eta)$, the sine-Gordon equation $u_{\xi\tau} = \sin u$ is satisfied identically for all $(\xi, \tau) \in \mathbb{R}^2$.
[/step]
