<s>[INST] How to prove $2\sqrt{2+\sqrt{3}}=\sqrt{2}+\sqrt{6}$? My calculator and I were arguing one day about the cosine of some number.
The calculator said "$\cos(\frac x2)=\sqrt{2}+\sqrt{6}$".
I said "That's absurd because $\cos(\frac x2)=\sqrt{\frac{1+\cos(x)}2}$, which evaluates to $2\sqrt{2+\sqrt{3}}$ for this particular $x$!"
So I finally decided to do this the non-radical way and found the decimal approximations to be equivalent.
Which is weird and probably not a coincidence.  So why is this, and does this have other values that work out like this?
 [/INST] $$\color{gray}{2+\sqrt 3=\frac 1 2+\sqrt 3+\frac 3 2=\frac{1+2\sqrt3 +3}{2}=\frac{1+2\sqrt 3+(\sqrt 3)^2}{2}=\frac{(1+\sqrt3)^2}{2}}\\2\sqrt{\frac{(1+\sqrt 3)^2}{2}}\\ \color{gray}{\sqrt{\frac 1 2(1+\sqrt 3)^2}=\frac{\sqrt{(1+\sqrt 3)}}{\sqrt 2}}\\ 2\frac{\sqrt{(1+\sqrt 3)^2}}{\sqrt 2}\\ \color{gray}{\sqrt{(1+\sqrt 3)^2}=1+\sqrt 3:}\\ \frac{2}{\sqrt 2}1+\sqrt 3\\ \color{gray}{\frac{2(1+\sqrt 3)}{\sqrt 2}=\frac{2(1+\sqrt 3)}{\sqrt 2}\times \frac{\sqrt 2}{\sqrt 2}=\frac{1(1+\sqrt 3)\sqrt 2}{2}:}\\ \frac{2(1+\sqrt 3)\sqrt 2}{2}\\ \color{gray}{\frac{2(1+\sqrt 3)\sqrt 2}{2}=\frac 2 2\times(1+\sqrt 3)\sqrt 2=(1+\sqrt3)\sqrt 2}\\ (1+\sqrt 3)\sqrt 2\\ \color{gray}{\sqrt 2(1+\sqrt 3)}=\boxed{\color{blue}{\sqrt 2+\sqrt 6}}$$
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