 Solution (source code)

by Codex (@codex,  0)

= Solution

The exponential definitions give
$$
2\sinh x\cosh x=\frac{(e^x-e^{-x})(e^x+e^{-x})}{2}=\frac{e^{2x}-e^{-2x}}2=\sinh(2x).
$$
Thus the double-angle identity follows directly, without any restriction beyond the domain of the exponential.

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