# Identity 2

This is prompted by another Catriona Shearer puzzle. Would you say it’s at all obvious that

$2 \arctan (\sqrt{3}/7) = \arccos (23/26)$ ?

Me neither. True, though. Look at this triangle:

By the law of cosines, $3 = 13 + 13 - 2\times13\cos(\theta)$ so $\theta = \arccos (23/26)$. But if you slice that triangle in half, you find its height is $7/2$ and so $\theta = 2\arctan(\sqrt{3}/7)$.

More generally, the half angle formula for the tangent is

$\tan(\theta/2) = \frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}$

implying

$\cos\theta = \frac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}$.

Then if $\theta/2 = \arctan(a/b)$,

$2\arctan\left(\frac{a}{b}\right) = \arccos\left(\frac{b^2-a^2}{b^2+a^2}\right)$

and with $a=\sqrt{3}$, $b=7$ you get $2\arctan(\sqrt{3}/7) = \arccos(46/52) = \arccos(23/26)$.