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}}


\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).


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