Using these identities
$\sinh(x+1) – \sinh (x) = (-1+\cosh(1))\sinh(x) + \sinh(1)\cosh(x)$
$\cosh (x+1) – \cosh (x) = (-1+\cosh(1))\cosh(x) + \sinh(1)\sinh(x)$
Express the series $C = \cosh 0 + \cosh 1 + \cosh 2 +\dots+ \cosh n$
$ S = \sinh 0 + \sinh 1 + \sinh 2 + \dots+ \sinh n $
In terms of $\cosh(n+1) , \sinh(n+1), \cosh (1)$ and numbers such as $1,2$ etc.
This is what I have so far done:
$x = -1, \cosh(0) = \cosh(1)\cosh(-1) + \sinh(1)\sinh(-1)$
$=> \cosh(0) = \cosh^2 (1) – \sinh^2 (1) = 1$
$x=0, \cosh(1) = \cosh (1)$
$x=1, \cosh(2) = \cosh^2 (1) + \sinh^2 (1)$
$\cosh (2) – \cosh^2 (1) = \cosh ^2 (1) + \sinh^2 (1) – \cosh^(2) (1)$
$=> \cosh(2) = 2\cosh^2 (1) -1 $
$x = n,\cosh(n+1) = \cosh(1)\cosh(n) + \sinh(n)\sinh(1)$
Also, $\sinh(n+1) = \cosh(1)\sinh(n) + \sinh(1)\cosh(n)$
$\cosh(n+1) + \sinh(n+1) = \cosh(1)[\cosh(n)+\sinh(n)] + \sinh(1)[\sinh(n)+\cosh(n)]$
Trying to solve for $\sinh(1)$.I don't really know where to go from here.
Best Answer
You can sum up the left hand side of the prescripted formula $$\sum_{k=0}^n \sinh(k+1)-\sinh(k)=\sinh(n+1)-\sinh(0)$$ on the one hand, and doing the same with the right-hand side $$\sum_{k=0}^n \sinh(k+1)-\sinh(k)=\sum_{k=0}^n(\cosh(1)-1)\sinh(k)+\sinh(1)\cosh(k)\\=(\cosh(1)-1)S+\sinh(1)C$$ on the other hand. Do the same with the other formula. You get a system of two equations. $$\left\{\begin{array}{ccccl} (\cosh(1)-1)S&+&\sinh(1)C&=&\sinh(n+1)\\ \sinh(1)S&+&(\cosh(1)-1)C&=&\cosh(n+1)-1 \end{array}\right.$$ You should be able to conclude.