If $\alpha=\ln\sqrt{\cot\frac{\pi}{12}}$ and $$\displaystyle\frac{\sum_{k=0}^{\infty}e^{-2k\alpha}}{\sum_{k=0}^{\infty}(-1)^ke^{-2k\alpha}}=T$$ and $$S=\frac{\binom{n}{1}\sin\frac{2\pi}{3n}+\binom{n}{2}\sin\frac{4\pi}{3n}+\cdots+\binom{n}{n}\sin\frac{2\pi}{3}}{1+\binom{n}{1}\cos\frac{2\pi}{3n}+\binom{n}{2}\cos\frac{4\pi}{3n}+\cdots+\binom{n}{n}\cos\frac{2\pi}{3}}(n\in\mathbb{N})$$ then find the value of $T+2\sqrt3S$
I was able to calculate the value of $T$ with much effort. It may seem that it is just a simple infinite Geometric Progression but the manipulation was tough. Anyways I got the value of $T$ as $\sqrt3$ but for $S$ I have no idea. The binomial coefficients make the calculation tough. What I think that the expressions are expansion of a function but I can't figure out the exact function. Any help is greatly appreciated.
Best Answer
Here's my try for evaluating S; $$S=\frac{\binom{n}{1}\sin\frac{2\pi}{3n}+\binom{n}{2}\sin\frac{4\pi}{3n}+\cdots+\binom{n}{n}\sin\frac{2\pi}{3}}{1+\binom{n}{1}\cos\frac{2\pi}{3n}+\binom{n}{2}\cos\frac{4\pi}{3n}+\cdots+\binom{n}{n}\cos\frac{2\pi}{3}}(n\in\mathbb{N})$$ Rewriting numerator and denominator in terms of a general term; $$S=\frac{\sum_{r=0}^{n}\binom{n}{r}\sin\frac{2\pi r}{3n}}{\sum_{r=0}^{n}\binom{n}{r}\cos\frac{2\pi r}{3n}}$$ Using the Euler's Identity to generate $\sin\Phi$ and $cos\Phi$; $$Z=e^{i \Phi} =\cos\Phi+i\sin\Phi$$ $$\operatorname{Re}(z)=\cos\Phi=\frac{e^{i \Phi}+e^{-i \Phi}}{2}$$ $$\operatorname{Im}(z)=\sin\Phi=\frac{e^{i \Phi}+e^{-i \Phi}}{2i}$$ Substituting $\Phi=\frac{2\pi r}{3n}$; $$e^{i \frac{2\pi r}{3n}} =\cos\frac{2\pi r}{3n}+i\sin\frac{2\pi r}{3n}$$ $$S=\frac{\sum_{r=0}^{n}\binom{n}{r}\operatorname{Re}(e^{i \frac{2\pi r}{3n}})}{\sum_{r=0}^{n}\binom{n}{r}\operatorname{Im}(e^{i \frac{2\pi r}{3n}})}$$ $$S=\frac{\frac{\sqrt 3}{2} \left(e^{\frac{i\pi}{3n}}+e^{\frac{-i\pi}{3n}}\right)^n}{\frac{1}{2} \left(e^{\frac{i\pi}{3n}}+e^{\frac{-i\pi}{3n}}\right)^n}$$ $$S=\sqrt3$$ So the end result is; $$T+2\sqrt 3 S = \sqrt 3+6=\frac{6\sqrt3+3}{\sqrt3}$$