Found this equality
$$\tan 15^{\circ} = \tan 27^{\circ} \tan 33^{\circ} \tan 39^{\circ}$$
with a random search. It checks with WA, and these kind of equalities can be proved automatically.
I am looking for an elementary proof.
Some similarly looking equalities are using the identity;
$$\tan 3 x= \tan x\cdot \tan(\frac{\pi}{3}-x)\cdot \tan (\frac{\pi}{3}+x)$$
We notice that $27^{\circ} + 33^{\circ} = 60^{\circ}$, but does not seem enough.
Thank you for interest!
$\bf{Added:}$ Let me add the mechanical proof that works like here. One considers the basic angle $x = 3^{\circ}= \frac{2 \pi}{120}$, and we need to show that $\tan 5 x= \tan 9 x \tan 11 x \tan 13 x$, which is equivalent to: $z= e^{\frac{2 \pi i}{120}}$ is a root of the equation
$$\frac{z^{10}-1}{z^{10}+1} + \frac{z^{18} -1}{z^{18} + 1}\cdot \frac{z^{22} -1}{z^{22} + 1}\cdot \frac{z^{26} -1}{z^{26} + 1}=0$$
LHS factors and we see the cyclotomic polynomial $\Phi_{120}(z)$ in the numerator.
$\bf{Added:}$ Since in general we have to show some equality for functions of multiples of an angle $x$, where $x$ is a rational multiple of $2\pi$, we could prove an equivalent equality for the angle $m x$, where $m$ is relatively prime to the denominator of $\frac{x}{2 \pi}$, so we might reduce to other known identities. Perhaps this one works with say $m = 7$? Since $7\times 15^{\circ} = 90^{\circ} + 15^{\circ}$.
Best Answer
My original solution wasn't quite accurate. However, this one is, and the answer uses some nice elementary number theory.
Claim: $\cos(108^{\circ}) = \cos(48^{\circ})+\cos(168^{\circ})$.
In fact, for any $\theta$, we have $$ \cos(\theta)=\cos(\theta+60^{\circ})+\cos(\theta-60^{\circ})$$
Proof:
Clear denominators and multiply each side by $4$: $$4\sin(15{}^{\circ})\cos(27{}^{\circ})\cos(33{}^{\circ})\cos(39{}^{\circ})=4\sin(27{}^{\circ})\cos(15{}^{\circ})\sin(33{}^{\circ})\sin(39{}^{\circ})$$ Product-to-sum: $$(\sin(42{}^{\circ})+\sin(-12{}^{\circ}))(\cos(72{}^{\circ})+\cos(-6{}^{\circ}))=(\sin(42{}^{\circ})+\sin(12{}^{\circ}))(\cos(6{}^{\circ})-\cos(72{}^{\circ}))$$ Correct the signs using the parity of the functions and expand. After cancelling terms and rearranging, we are left with: $$2\sin(42^{\circ})\cos(72^{\circ})=2\sin(12^{\circ})\cos(6^{\circ})$$Turn sines into cosines, product-to-sum, and use complementary angles: $$2\cos(48^{\circ})\cos(72^{\circ})=2\cos(78^{\circ})\cos(6^{\circ})$$ $$\cos(24^{\circ})+\cos(120^{\circ})=\cos(72^{\circ})+\cos(84^{\circ})$$ $$\cos(24^{\circ})-\cos(72^{\circ})-\cos(84^{\circ})=\frac{1}{2}$$ $$\cos(24^{\circ})+\cos(96^{\circ})+\cos(108^{\circ})=\frac{1}{2}$$
By the claim, we can rewrite this as: $$\cos(24^{\circ})+ \cos(48^{\circ})+\cos(96^{\circ})+\cos(168^{\circ})=\frac{1}{2}$$ Rewrite the arguments in radians: $$\cos\left(2\pi \cdot \frac{1}{15}\right)+ \cos\left(2\pi \cdot \frac{2}{15}\right)+\cos\left(2\pi \cdot \frac{4}{15}\right)+\cos\left(2\pi \cdot \frac{7}{15}\right)=\frac{1}{2}$$ Here I paraphrase from the Wikipedia article on trig-identities:
If you don't want to rely on number theory, it is possible to keep using angle-addition to transform the LHS into $1/2$ using double-angle and a few other tricks; exercise.
I'm sure you could reverse-engineer additional identities by starting with similar evaluations of the Möbius function.
https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Identities_without_variables