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
Best Answer
I'm not going to spoon-feed you, assuming you're doing your ENG1091 assignment. But here's a hint. You are on the right track, and for the last part you might want to consider the following:
cosh^2(x) - sinh^2(x) = 1
You have sinh(1), rearrange the above identity and then you'll be able to express your answer in terms of cosh(1).