Hyperbolic function identity proof? On a question i am working thru it says:
Obtain the formula:$$ \sinh 2x - \sinh 2y = 2\cosh(x+y)\sinh(x-y)
$$and prove that
$$\coshθ + \cosh2θ +...+\cosh nθ =\cosh(0.5(n+1)θ)\sinh(0.5nθ)\text{csch}(0.5θ).$$
I am fine with the first part but am really stuck on the second. I have tried using geometric series but this does not seem to work. I presume i have to use the prevous part of the question but can not see how. Could some one give me some ideas/hints thanks?  
 A: You are correct in thinking that geometric progression is the way to go. I do not think it is necessary to use the first part of the question.
Expressing the $\cosh$ function in terms of its constituent exponentials, we have two geometric series to consider
$$\sum_{k=1}^n\cosh k\theta=\frac{1}{2}\left(\sum_{k=1}^ne^{k\theta}+\sum_{k=1}^ne^{-k\theta}\right)\\=\frac{1}{2}\left(e^\theta\frac{1-e^{n\theta}}{1-e^\theta}+e^{-\theta}\frac{1-e^{-n\theta}}{1-e^{-\theta}}\right)\\=\frac{1}{2}\left(e^\theta\frac{e^{n\theta/2}\color{blue}{(e^{-n\theta/2}-e^{n\theta/2})}}{e^{\theta/2}\color{blue}{(e^{-\theta/2}-e^{\theta/2})}}+e^{-\theta}\frac{e^{-n\theta/2}\color{blue}{(e^{n\theta/2}-e^{-n\theta/2})}}{e^{\theta/2}\color{blue}{(e^{\theta/2}-e^{-\theta/2})}}\right)$$
The fraction highlighted in blue is simply $\sinh(n\theta/2)\text{cosech}(\theta/2)$ which we can factorise out, leading to
$$\sum_{k=1}^n\cosh k\theta=\sinh(n\theta/2)\text{cosech}(\theta/2)\frac{1}{2}(e^{(n+1)\theta/2}+e^{-(n+1)\theta/2})$$ 
Can you take it from here?
A: If you want to use the first part of the problem to prove the second, you can do so by using telescoping series.
Let
$$S=\cosh \theta+\cosh 2\theta+\cdots +\cosh n\theta$$
Multiply both the sides with $2\sinh\theta$ and then use the following:
$2\cosh \theta \sinh \theta=\sinh(2\theta)$
$2\cosh 2\theta \sinh \theta=\sinh 3\theta-\sinh \theta$
$2\cosh 3\theta \sinh \theta=\sinh 4\theta-\sinh 2\theta$
$2\cosh 4\theta \sinh \theta=\sinh 5\theta-\sinh 3\theta$
$\vdots$
$2\cosh (n-1)\theta \sinh \theta=\sinh n\theta-\sinh (n-2)\theta$
$2\cosh n\theta \sinh \theta=\sinh (n+1)\theta-\sinh (n-1)\theta$
Hence,
$$2S\sinh\theta=\sinh(n+1)\theta+\sinh(n\theta)-\sinh\theta$$
$$=2\cosh\left(\frac{n+1}{2}\theta\right)\sinh\left(\frac{n+1}{2}\theta\right)+2\cosh\left(\frac{n+1}{2}\theta\right)\sinh\left(\frac{n-1}{2}\theta\right)$$
$$=2\cosh\left(\frac{n+1}{2}\theta\right)\left(\sinh\left(\frac{n+1}{2}\theta\right)+\sinh\left(\frac{n-1}{2}\theta\right)\right)$$
$$=4\cosh\left(\frac{n+1}{2}\theta\right)\sinh\left(\frac{n\theta}{2}\right)\cosh\left(\frac{\theta}{2}\right)$$
$$\Rightarrow S\sinh\left(\frac{\theta}{2}\right)=\cosh\left(\frac{n+1}{2}\theta\right)\sinh\left(\frac{n\theta}{2}\right)$$
$$\Rightarrow \boxed{S=\cosh\left(\dfrac{n+1}{2}\theta\right)\sinh\left(\dfrac{n\theta}{2}\right)\text{csch}\left(\dfrac{\theta}{2}\right)}$$
