<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] A way of showing this that has not yet been given by anyone is to obtain a "nice equation" (i.e. with integer coefficients) having $\;\sqrt{2}+\sqrt{6}\;$ as a solution and then solving this equation by standard methods.
Start by setting $\;x = \sqrt{2}+\sqrt{6}.\;$ Subtracting $\sqrt 2$ from both sides and then squaring both sides gives $\;x^2 - 2\sqrt{2}x + 2 = 6.\;$ Isolate the radical term and square again, to get $\;x^2 - 4 = 2\sqrt{2}x,\;$ followed by $\;x^4 - 8x^2 + 16 = 8x^2,\;$ or $\;x^4 - 16x^2 + 16 = 0.$
This equation is quadratic in $x^2,$ so using the quadratic formula gives
$$x^2 \; =\; \frac{16 \pm \sqrt{256 - 64}}{2} \;=\; \frac{16 \pm 8\sqrt{4 - 1}}{2} \;=\; 4(2 \pm \sqrt{3})$$
Therefore, we get
$$x \; = \; \pm2\sqrt{2 \pm \sqrt{3}}$$
Since $\;\sqrt{2} + \sqrt{6}\;$ is clearly greater than $3$ (note that $\sqrt{2} > 1$ and $\sqrt{6} > 2$), it follows that the value of $x$ that corresponds to $\;\sqrt{2} + \sqrt{6}\;$ is the expression $\;2\sqrt{2 \pm \sqrt{3}}.$
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