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Could someone give me some hint of how to do this question please.

I've been stuck for more than $3$ hours on this question.

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1 Answer

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Replacing $x$ with $k$ in the given relations (and rearranging somewhat), we have

$$\begin{cases} &\sinh{(1)}\cosh{(k)}+\left(\cosh{(1)}-1\right)\sinh{(k)}=\sinh{(k+1)}-\sinh{(k)},\\ &\left(\cosh{(1)}-1\right)\cosh{(k)}+\sinh{(1)}\sinh{(k)}=\cosh{(k+1)}-\cosh{(k)}.\\ \end{cases}$$

Summing both equations from $k=0$ to $k=n$, we find

$$\begin{cases} &\sinh{(1)}C_{n}+\left(\cosh{(1)}-1\right)S_{n}=\sinh{(n+1)},\\ &\left(\cosh{(1)}-1\right)C_{n}+\sinh{(1)}S_{n}=\cosh{(n+1)}-1.\\ \end{cases}$$

This yields a system of two linear equations in two unknowns, $C_{n}$ and $S_{n}$, which can be solved by standard methods. The result is

$$\begin{cases} &C_{n}=\frac{\sinh{(1)}\sinh{(n+1)}-\left(\cosh{(1)}-1\right)\left(\cosh{(n+1)}-1\right)}{\sinh^2{(1)}-\left(\cosh{(1)}-1\right)^2},\\ &S_{n}=\frac{\sinh{(1)}\left(\cosh{(n+1)}-1\right)-\left(\cosh{(1)}-1\right)\sinh{(n+1)}}{\sinh^2{(1)}-\left(\cosh{(1)}-1\right)^2}.\\ \end{cases}$$

Using the hyperbolic Pythagorean identity, $\sinh^2{(x)}-\cosh^2{(x)}=-1$, the denominators can be simplified as

$$\begin{align} \sinh^2{(1)}-\left(\cosh{(1)}-1\right)^2 &=\sinh^2{(1)}-\left(\cosh^2{(1)}-2\cosh{(1)}+1\right)\\ &=\sinh^2{(1)}-\cosh^2{(1)}+2\cosh{(1)}-1\\ &=2\cosh{(1)}-2\\ &=2\left(\cosh{(1)}-1\right).\\ \end{align}$$

If one desires to eliminate $\sinh{(1)}$ from the numerator as well, one can simply replace it with $\sinh{(1)}=\sqrt{\cosh^2{(1)}-1}$, again using the Pythagorean relation.

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