<s>[INST] evaluate trigonometric To all the genius out there , here is a question about expresssing summation of hyperboilc functions :
First of all, I've already proved that:
$$\sinh(x + 1)- \sinh(x) = (-1 + \cosh (1)) \sinh(x) + \sinh (1) \cosh(x)$$
$$\cosh(x + 1)- \cosh(x) = (-1 + \cosh (1)) \cosh(x) + \sinh (1) \sinh(x)$$
The later question asked me to use these results and  express the following sums:
$C_n = \cosh (0) + \cosh (1) + \cosh (2) + \ldots +\cosh (n)$ 
$S_n = \sinh (0) + \sinh (1) + \sinh (2) + \ldots+\sinh (n)$
I have learned a few summation of hyberbolic function by myself, please guide me through this new challenge , thank you .
 [/INST] For simplicity of derivation, let $c = \cosh 1$ and $s = \sinh 1$, we have
$$\begin{align}
\sinh(x+1) - \sinh x &= (c - 1) \sinh x + s \cosh x \tag{*1a}\\
\cosh(x+1) - \cosh x &= (c - 1) \cosh x + s \sinh x \tag{*1b}
\end{align}$$
$s \times (*1a) - (c-1) \times (*1b) \implies$
$$
\begin{align}
  & s \text{LHS}(*1a) - (c-1)\text{LHS}(*1b)\\ 
=\; & s\left( (c - 1) \sinh x + s \cosh x \right)
-(c-1)\left( (c - 1) \cosh x + s \sinh x \right)\\
=\; & (s^2 -(c-1)^2)\cosh x\\
=\; & 2(c-1)\cosh x
\end{align}
$$
This leads to
$$\cosh x = \frac12 \left(\frac{s}{c-1}\text{LHS}(*1a) - \text{LHS}(*1b)\right)
= A(x+1) - A(x)$$
where
$$A(x) = \frac12 \left(\frac{s}{c-1}\sinh x - \cosh x\right)$$
From this, we can conclude
$$\begin{align}
C_n = \sum_{k=0}^n \cosh k 
&= \sum_{k=0}^n ( A(k+1) - A(k) ) = A(n+1) - A(0)\\
&= \frac12 \left(\frac{\sinh(1)}{\cosh(1)-1}\sinh(n+1) - (\cosh(n+1) - 1 )\right)  
\end{align}
$$
The derivation of $S_n$ is similar, I will skip it here.
</s>