How to prove the following trignometric identity?
$$ \cot7\frac12 ^\circ = \sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$$
Using half angle formulas, I am getting a number for $\cot7\frac12 ^\circ $, but I don't know how to show it to equal the number $\sqrt2 + \sqrt3 + \sqrt4 + \sqrt6$.
I would however like to learn the technique of dealing with surds such as these, especially in trignometric problems as I have a lot of similar problems and I don't have a clue as to how to deal with those.
Hints please!
EDIT:
What I have done using half angles is this: (and please note, for convenience, I am dropping the degree symbols. The angles here are in degrees however).
I know that
$$ \cos 15 = \dfrac{\sqrt3+1}{2\sqrt2}$$
So,
$$\sin7.5 = \sqrt{\dfrac{1-\cos 15} {2}}$$
$$\cos7.5 = \sqrt{\dfrac{1+\cos 15} {2}} $$
$$\implies \cot 7.5 = \sqrt{\dfrac{2\sqrt2 + \sqrt3 + 1} {2\sqrt2 – \sqrt3 + 1}} $$
Best Answer
Here is an elementary (almost without words) proof that does not rely explicitly on half-angle or double-angle formulas. What is used is the fact that $\cot(30^\circ) = \sqrt{3}$, the exterior angle theorem, isosceles triangle theorems and Pythagorean theorem of Euclidean geometry, and the fact that $8 + 4\sqrt{3} = \left(\sqrt{2}+\sqrt{6}\right)^2$ (cf. Robert Israel's answer). The crude, not-to-scale, diagram below is, I hope, self-explanatory especially if you start from the right side and work your way to the left.
The length of the base of the triangle is $$\cot(7.5^\circ) = \sqrt{8+4\sqrt{3}} +2+\sqrt{3} = \sqrt{2}+\sqrt{6} + \sqrt{4}+\sqrt{3}$$