Use the definition \(\cosh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)\) to show that \(\cosh 2 x = 2 \cosh ^ { 2 } x - 1\).
(2 marks)
The arc of the curve \(y = \cosh x\) between \(x = 0\) and \(x = \ln a\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Show that \(S\), the surface area generated, is given by
$$S = 2 \pi \int _ { 0 } ^ { \ln a } \cosh ^ { 2 } x \mathrm {~d} x$$
Hence show that
$$S = \pi \left( \ln a + \frac { a ^ { 4 } - 1 } { 4 a ^ { 2 } } \right)$$