Answer: (a) \(2\cosh^2 A=\cosh 2A+1\).
(b)(i) \(A=10\pi\displaystyle \int_{-\frac12}^{\frac12}(2\cosh 2t+3t)\cosh 2t\,dt\).
(b)(ii) \(A=10\pi\left(\frac14(e^2-e^{-2})+1\right)\).
(a) Starting from the definition,
\(\cosh A=\frac12(e^A+e^{-A})\).
Then
\(2\cosh^2 A=2\left(\frac12(e^A+e^{-A})\right)^2\)
\(\phantom{2\cosh^2 A}=\frac12(e^{2A}+2+e^{-2A})\)
\(\phantom{2\cosh^2 A}=\frac12(e^{2A}+e^{-2A})+1\).
Since \(\cosh 2A=\frac12(e^{2A}+e^{-2A})\), it follows that
\(2\cosh^2 A=\cosh 2A+1\).
(b)(i) For rotation about the \(y\)-axis, the surface area is
\(A=2\pi\int x\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).
Here
\(x=2\cosh 2t+3t,\qquad y=\frac32\cosh 2t-4t\).
Differentiate:
\(\frac{dx}{dt}=4\sinh 2t+3,\qquad \frac{dy}{dt}=3\sinh 2t-4\).
So
\(\left(4\sinh 2t+3\right)^2+\left(3\sinh 2t-4\right)^2\)
\(=16\sinh^2 2t+24\sinh 2t+9+9\sinh^2 2t-24\sinh 2t+16\)
\(=25\sinh^2 2t+25\)
\(=25(\sinh^2 2t+1)=25\cosh^2 2t\).
Hence
\(\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=5\cosh 2t\),
since \(\cosh 2t\ge 0\).
Therefore
\(A=2\pi\int_{-\frac12}^{\frac12}(2\cosh 2t+3t)(5\cosh 2t)\,dt\)
\(\phantom{A}=10\pi\int_{-\frac12}^{\frac12}(2\cosh 2t+3t)\cosh 2t\,dt\).
(b)(ii) Using part (i),
\(A=10\pi\int_{-\frac12}^{\frac12}(2\cosh 2t+3t)\cosh 2t\,dt\)
\(=10\pi\left(2\int_{-\frac12}^{\frac12}\cosh^2 2t\,dt+3\int_{-\frac12}^{\frac12}t\cosh 2t\,dt\right)\).
Now \(t\cosh 2t\) is odd, so
\(\int_{-\frac12}^{\frac12}t\cosh 2t\,dt=0\).
Thus
\(A=20\pi\int_{-\frac12}^{\frac12}\cosh^2 2t\,dt\).
From part (a), with \(A=2t\),
\(2\cosh^2 2t=\cosh 4t+1\),
so
\(A=10\pi\int_{-\frac12}^{\frac12}(\cosh 4t+1)\,dt\).
Now
\(\int (\cosh 4t+1)\,dt=\frac14\sinh 4t+t\).
Hence
\(A=10\pi\left[\frac14\sinh 4t+t\right]_{-\frac12}^{\frac12}\)
\(=10\pi\left(\frac14\sinh 2+\frac12-\left(-\frac14\sinh 2-\frac12\right)\right)\)
\(=10\pi\left(\frac12\sinh 2+1\right)\).
Using \(\sinh 2=\frac12(e^2-e^{-2})\),
\(A=10\pi\left(\frac14(e^2-e^{-2})+1\right)\).
