Starting from the definition of cosh in terms of exponentials, prove that
$$2 \cosh ^ { 2 } A = \cosh 2 A + 1$$
The curve \(C\) has parametric equations
$$\mathrm { x } = 2 \cosh 2 \mathrm { t } + 3 \mathrm { t } , \quad \mathrm { y } = \frac { 3 } { 2 } \cosh 2 \mathrm { t } - 4 \mathrm { t } , \quad \text { for } - \frac { 1 } { 2 } \leqslant t \leqslant \frac { 1 } { 2 }$$
The area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(y\)-axis is denoted by \(A\).
Show that \(A = 10 \pi \int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } ( 2 \cosh 2 t + 3 t ) \cosh 2 t d t\).
Hence find \(A\) in terms of \(\pi\) and e.
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