Find the proof:

(a) Use the definitions
cosh(x)= 1/2(ex +e^−x) , sinh(x)= 1/2(e^x − e^−x)
to express sinh(x + y) and cosh(x + y) in terms of cosh(x), sinh(x), cosh(y) and sinh(y).
(b) Using the results of part (a) show 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)
Considering the answer from part (a) is
sinh(x+y) = sinh(x)cosh(x) +cosh(x)sinh(y)
and
cosh(x+y)=cosh(x)cosh(y) +sinh(x)sinh(y)
(c)Use the result of part (b) to express the following sums
Cn =cosh0+cosh1+cosh2+···coshn **
**Sn =sinh0+sinh1+sinh2+···sinhn
in terms of just cosh(n + 1), sinh(n + 1) and cosh 1 (and possibly some numbers like 1, 2 etc.).
*** Considering i know how to show parts a and b how can you show the result to part (C)?
Best Answer
Exactly right. You subsititute $y=1$ into the compound $\cosh$ identity and everything should go easily from there.
For part (c),
Rearrange $\sinh(x + 1) − \sinh(x) = \left(−1 + \cosh(1)\right) \sinh(x) + \sinh(1) \cosh(x)$
to get: $\sinh(x + 1) = \sinh(x) + \left(−1 + \cosh(1)\right) \sinh(x) + \sinh(1) \cosh(x)$
$\sinh(0)=0$ and $\cosh(0)=1$
$\sinh(0 + 1) = \sinh(0) + \left(−1 + \cosh(1)\right) \sinh(0) + \sinh(1) \cosh(0)$
$\sinh(1) = 0 + \left(−1 + \cosh(1)\right) (0) + \sinh(1) (1)$
$\sinh(1) = \sinh(1)$ as you would expect!
$\sinh(1 + 1) = \sinh(1) + \left(−1 + \cosh(1)\right) \sinh(1) + \sinh(1) \cosh(1)$
$\sinh(2) = 2\cosh(1)\sinh(1)$