# Partials sums of cosh(x) and sinh(x)

Ok, i asked this question yesterday but then hit a snag again.

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 $C = \cosh 0 + \cosh 1 + \cosh 2 +\dots+ \cosh n$ and

$S = \sinh 0 + \sinh 1 + \sinh 2 + \dots+ \sinh n$

In terms of $\cosh(n+1) , \sinh(n+1), \cosh (1)$ and $1,2$ etc.

This is where i'm at

sum of sinh(x+1)-sinh(x), from 0 to n, = sinh(n+1)

sum of cosh(x+1)-cosh(x), from 0 to n, = cosh(n+1)-1

C = cosh(0) + cosh(1)... +cosh(n)

=> C = sum of cosh(x) from 0 to n

=cosh(n)

S = sinh(0) + sinh(1) ... + sinh(n)

=> S = sum of sinh(x) from 0 to n

= sinh(n)

From here i was unsure what to do, i could sum the RHS of the given formula, and equate with the LHS, but that would require me to sub in S and C. Is it even possible to substitue entire partial sums in equations?

I tried it out and got

sinh(n+1)

= -sinh(n)+cosh(1)sinh(n)+sinh(1)cosh(n)

= -S+cosh(1)S + sinh(1)C

sinh(n+1) = (cosh(1) -1)S + sinh(1)C

______________________________-

cosh(n+1) - 1

= (-1+cosh(1))cosh(n) + sinh(1)sinh(n))

= (-1+cosh(1))C + sinh(1)S

cosh(n+1) - 1 = sinh(1)S + (cosh(1) -1)C

which i then tried to solve for S and C, but gave me answers in terms of sinh(1) which i can't have