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?
Best Answer
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?