Prove: $\tan\frac{\pi}{24}=2\sqrt{2+\sqrt{3}}-\sqrt{3}-2$

Question

How to prove that $$\tan\frac{\pi}{24}=2\sqrt{2+\sqrt{3}}-\sqrt{3}-2$$

I get $$\tan\frac{\pi}{24}=\sqrt\frac{2\sqrt{2}-\sqrt{3}-1}{ 2\sqrt{2}+\sqrt{3}+1}$$ but i can't transform it.

Answer 1

$tan2\theta = 2tan\theta/(1-tan\theta^2)$

Let $tan\pi/12 = a, tan\pi/24 = x$, use formula, we get $(x = -1 - \sqrt{1+a^2})/a = (-1+ \sqrt{2-\sqrt{3}})/(2-\sqrt{3}) = -1 / (2-\sqrt{3}) + 2 / \sqrt{(2-\sqrt{3})} = -2 - \sqrt{3} + 2\sqrt{(2+\sqrt{3})}$

Answer 2

I will assume that you have proven up to the point that:

$\tan\frac{\pi}{24}=\sqrt\frac{2\sqrt{2}-\sqrt{3}-1}{ 2\sqrt{2}+\sqrt{3}+1}$

and that your working up to this point was correct.

Note that:

$\sqrt\frac{2\sqrt{2}-\sqrt{3}-1}{ 2\sqrt{2}+\sqrt{3}+1}$=$\sqrt\frac{(2\sqrt{2}-\sqrt{3}-1)^2}{ (2\sqrt{2}+\sqrt{3}+1)(2\sqrt{2}-\sqrt{3}-1)}=\sqrt{(3-2\sqrt{2})(5-2\sqrt{6})}$

And that last expression:

$\sqrt{(3-2\sqrt{2})(5-2\sqrt{6})}$=$\sqrt{15-6\sqrt{6}-10\sqrt{2}+8\sqrt{3}}$

I am not sure if that form is any more helpful. One more thing might be worthwhile to do. Numerically, these all check out. The tangent is equal to your QED expression which is equal to what we have right here.

So, why don't you try equating this form with your own and trying to reason to an identity. I.e.:

$\sqrt{15-6\sqrt{6}-10\sqrt{2}+8\sqrt{3}}$=$2\sqrt{2+\sqrt{3}}-\sqrt{3}-2$

$15-6\sqrt{6}-10\sqrt{2}+8\sqrt{3}$=$(2\sqrt{2+\sqrt{3}}-\sqrt{3}-2)^2$

...

$0=0$

You might be able to backwards-engineer a proof that way. (I.e. start from the identity and prove that these two expressions are equal, consequently showing that the tangent is equal to what you want it to be.)

EDIT Attempt at a proof 1:

$392-160\sqrt{6}=392-160\sqrt{6}$

$200+192-160\sqrt{6}=8(49-20\sqrt{6}$

$(10\sqrt{2}-8\sqrt{3})^2=4(2(49-20\sqrt{6}))$

Take the principal square root on both sides:

$10\sqrt{2}-8\sqrt{3}=2\sqrt{2(49-20\sqrt{6})}$

$-10\sqrt{2}+8\sqrt{3}=-2\sqrt{2(49-20\sqrt{6})}$

$15-6\sqrt{6}=15-6\sqrt{6}$

$15-6\sqrt{6}-10\sqrt{2}+8\sqrt{3}=15-6\sqrt{6}-2\sqrt{2(49-20\sqrt{6})}$

And of course, you majestically realize that the RHS factorizes into:

$15-6\sqrt{6}-10\sqrt{2}+8\sqrt{3}$=$(2\sqrt{2+\sqrt{3}}-\sqrt{3}-2)^2$

And taking the principal square root again, you deduce that:

$\sqrt{15-6\sqrt{6}-10\sqrt{2}+8\sqrt{3}}$=$2\sqrt{2+\sqrt{3}}-\sqrt{3}-2$

We have already shown that the LHS is equal to $\sqrt\frac{2\sqrt{2}-\sqrt{3}-1}{ 2\sqrt{2}+\sqrt{3}+1}$, which I presume you have proven to be $\tan\frac{π}{24}$