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Let $\alpha $ be an acute angle such that $\sin \alpha = \dfrac{3 \sqrt 3}{14}$. Prove that $$\frac{2\cdot7^{n}}{\sqrt 3}\sin \left(n\alpha + \dfrac{\pi}{3} \right) \in \mathbb{Z} \qquad\forall n>0. $$

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What is $A_m$ ? – lab bhattacharjee Dec 4 '12 at 15:46
    
I suspect that it's $2\cdot7^{n-1}$ rather than $2.7^{n-1}$. – FrenzY DT. Dec 4 '12 at 15:49
    
What is to be proven? – lab bhattacharjee Dec 4 '12 at 15:57
up vote 6 down vote accepted

This is equivalent to saying that for all $n>0$ $$ \frac{2\cdot7^{n}}{\sqrt{3}}\sin\left(n\alpha+\frac\pi3\right)\in\mathbb{Z} $$ Note that $\cos(\alpha)=\frac{13}{14}$, so that $$ e^{i\alpha}=\frac{13}{14}+i\frac{3\sqrt3}{14} $$ therefore $$ \begin{align} e^{i(n\alpha+\pi/3)} &=\left(\frac{13}{14}+i\frac{3\sqrt3}{14}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right)\\[6pt] &=\frac1{7^n}\left(\frac{13}{2}+i\frac{3\sqrt3}{2}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right) \end{align} $$ Note that $$ a_n=\left(\frac{13}{2}+i\frac{3\sqrt3}{2}\right)^{\large\!n}\left(\frac12+i\frac{\sqrt{3}}{2}\right) $$ and $\frac2{\sqrt{3}}$ times its imaginary part are solutions to $$ a_n=13a_{n-1}-49a_{n-2} $$ $\dfrac2{\sqrt{3}}\mathrm{Im}(a_0)=1$ and $\dfrac2{\sqrt{3}}\mathrm{Im}(a_1)=8$. Thus, $\dfrac2{\sqrt{3}}\mathrm{Im}(a_n)\in\mathbb{Z}$. Substituting back, we get the desired result.

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