A Nice Trig Formula

Grégoire Nicollier
University of Applied Sciences of Western Switzerland
November 26, 2013

$\tan50^\circ+\tan60^\circ+\tan70^\circ=\tan50^\circ\cdot\tan60^\circ\cdot\tan70^\circ=\tan80^\circ$

The first equality is well-known: iterate the identity

(1) $\displaystyle\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

to get

(2) $\displaystyle\tan(\alpha+\beta+\gamma)=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\alpha\tan\beta-\tan\alpha\tan\gamma-\tan\beta\tan\gamma}.$

Set $\alpha=50^\circ$, $\beta=60^\circ$, and $\gamma=70^\circ$ in (2), noticing that these angles sum up to $180^\circ$.

The second equality is less known: using (1) for $60^\circ\pm\delta$, $\tan60^\circ=\sqrt3$, and (2) for $\alpha=\beta=\gamma=\delta$, one gets $\tan (60^\circ-\delta)\tan (60^\circ+\delta)\tan\delta=\tan3\delta.$ Take then $\delta=10^\circ$.

Remark

The identity (2) shows that the true reason why $\arctan 1 + \arctan 2 + \arctan 3 = 180^\circ$ (see https://www.cut-the-knot.org/pythagoras/PaperFolding/arctan123.shtml) is that $1 + 2 + 3 = 1 \times 2 \times 3.$

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