# Trig Manipulations I'm Not Getting

• I
• thatboi
thatboi
Hi all,
I am starting with the following equation: ##2\cot\left(\frac{\theta}{2}\right) = \cot\left(\frac{k_{1}}{2}\right) - \cot\left(\frac{k_{2}}{2}\right)##
with the following definitions: ##k_{1} = \frac{K}{2} + ik, k_{2} = \frac{K}{2}-ik, \theta = \pi(I_{2}-I_{1}) + iNk##, where ##k,K,N\in\mathbb{R}## and ##I_{2},I_{1}\in\mathbb{Z}##. I wish to plug these definitions into the above equation and get the new equation: ##\cos\left(\frac{K}{2}\right)\sinh(Nk)=\sinh[(N-1)k]+\cos[\pi(I_{2}-I_{1})]\sinh(k)##. I have done the following (first use the identity ##\cot\frac{\theta}{2} = \frac{1+\cos\theta}{\sin\theta}##):

\begin{split}
&2\frac{1+\cos\theta}{\sin\theta} = \frac{1+\cos k_{1}}{\sin k_{1}}-\frac{1+\cos k_{2}}{\sin k_{2}} \\
&\rightarrow 2\frac{1+\cos(\pi(I_{2}-I_{1}))\cos(iNk)-\sin(\pi(I_{2}-I_{1}))\sin(iNk)}{\sin(\pi(I_{2}-I_{1}))\cos(iNk)+\sin(iNk)\cos(\pi(I_{2}-I_{1}))} = \frac{1+\cos\frac{K}{2}\cos(ik)-\sin\frac{K}{2}\sin(ik)}{\sin\frac{K}{2}\cos(ik)+\cos\frac{K}{2}\sin(ik)}-\frac{1+\cos\frac{K}{2}\cos(ik)+\sin\frac{K}{2}\sin(ik)}{\sin\frac{K}{2}\cos(ik)-\cos\frac{K}{2}\sin(ik)}\\
&\rightarrow -2i\frac{1+\cos(\pi(I_{2}-I_{1}))\cosh(Nk)}{\sinh(Nk)\cos(\pi(I_{2}-I_{1}))} = \frac{2i\sinh(k)}{\cos\left(\frac{K}{2}\right)-\cosh(k)}
\end{split}

where I dropped terms with ##\sin(\pi(I_{2}-I_{1}))## since they would evaluate to 0. While this is a lot more simplified I still don't know how to get to the final form of equation I want or if I made an error somewhere. Any advice appreciated!

The formula that you start with in your first line looks very unrealistic, when following the definitions you give for ## \theta ## and the various k's. It looks to me like the original problem statement may contain an error or two.

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