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Finite trigonometric product

By Tolaso in Uncategorized on June 10, 2017.

Evaluate the product

$\Pi = \left( \tan 61^\circ - \sqrt{3} \right) \left( \tan 62^\circ - \sqrt{3} \right) \cdots \left( \tan 89^\circ - \sqrt{3} \right)$

Solution

One good way to begin is by invoking the identity

$\tan \alpha + \tan \beta = \frac{\sin \left ( \alpha + \beta \right )}{\cos \alpha \cos \beta}$

Then noting that $\sqrt{3}=\tan 60^\circ$ we have that

$\begin{align*} \Pi &= \prod_{n=61}^{89} \left ( \tan n^\circ - \sqrt{3} \right ) \\ &= \prod_{n=61}^{89} \left ( \tan n^\circ - \tan 60^\circ \right )\\ &= \prod_{n=1}^{29} \frac{\sin n^\circ}{ \cos 60^\circ \cos \left ( 90^\circ-n^\circ \right )} \\ &=\frac{1}{\left ( \cos 60^\circ \right )^{29}}\prod_{n=1}^{29} \frac{\cancel{\sin n^\circ}}{ \cancel{\sin n^\circ}} \\ &=\frac{1}{ \left( \cos 60^\circ \right)^{29} } \prod_{n=1}^{29} 1 \\ &= 2^{29} \end{align*}$