0
$\begingroup$

While playing with hypergeometric functions, I numerically stumbled upon the identity: $$\mathrm{cos}\left(\dfrac{\pi}{6} - \dfrac{1}{6} \mathrm{arctan}\left( \dfrac{3\sqrt{15}}{11} \right) \right) = \dfrac{\sqrt{10} + 3\sqrt{2}}{8}$$ Is this identity true? Well-known?

$\endgroup$
1
  • $\begingroup$ well, sine and cosine of the arctan piece is straightforward, then some serious effort to include the $1/6.$ So, the trig functions are, well, cosmetic. Also, recall formula for $\cos (a-b)$ $\endgroup$
    – Will Jagy
    Mar 11, 2020 at 23:52

1 Answer 1

1
$\begingroup$

It's correct, $$ \cos \arctan\left( \frac{3\sqrt{15}}{11} \right) = \frac{11}{16} \; , \; $$ $$ \cos \frac{1}{6} \arctan\left( \frac{3\sqrt{15}}{11} \right) = \frac{\sqrt {18 + 6 \sqrt 5}}{4 \sqrt 2} \; , \; $$ $$ \sin \frac{1}{6} \arctan\left( \frac{3\sqrt{15}}{11} \right) = \frac{{3 - \sqrt 5}}{4 \sqrt 2} \; , \; $$ Takes a while. The cosine with the 1/6 is a root of $$ 512 c^6 - 768 c^4 + 288 c^2 - 27 = (8 c^2-3)(64 c^4 - 72 c^2 +9) $$

$\endgroup$
1
  • $\begingroup$ great! thank you! $\endgroup$
    – Libli
    Mar 12, 2020 at 8:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .