Proof of Hyperbolic Functions _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)?

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!

$\sinh(1 + 1) = \sinh(1) + \left(−1 + \cosh(1)\right) \sinh(1) + \sinh(1) \cosh(1)$

$\sinh(2) = 2\cosh(1)\sinh(1)$