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?
Just note that $$2\sqrt{2+\sqrt{3}}=\sqrt{8+4\sqrt{3}}=\sqrt{(\sqrt{6})^2+(\sqrt{2})^2+2(\sqrt2)(\sqrt6)}=\sqrt{(\sqrt{2}+\sqrt{6})^2}=\sqrt{2}+\sqrt{6}.$$
I must confess I've seen a while ago the method that takes care of these problems but failed to appreciated its usefulness, and choose altogether to ignore the result and the method. Till I was put in front of a concrete example that gave me some trouble..
It's the method of "simplifying" nested radicals of the form
$$\sqrt{ a \pm \sqrt{b}}$$.
One tries to write it in the form $$\sqrt{ a \pm \sqrt{b}} = \sqrt{x} \pm \sqrt{y}$$
From experience : in many cases it is much simpler to solve the system that gets $a$ and $b$ then to guess them.
For, by squaring the above we get $$a + \sqrt{b} = x + y + 2 \sqrt{x y}$$ and so we want $$ x+y= a \\ 4 x y = b$$ from where we get $(x-y)^2 = a^2 - b$ and therefore $$x = \frac{a + \sqrt{a^2 - b} }{2} \\ y = \frac{a - \sqrt{a^2 - b} }{2} $$ and so $$\sqrt{ a \pm \sqrt{b}} = \sqrt{ \frac{a + \sqrt{a^2 - b} }{2}} \pm \sqrt{ \frac{a - \sqrt{a^2 - b} }{2}}$$
Hardly anything simpler... except when $a^2 - b$ is a square, and we get the nested radical as a sum/difference of two radicals!
Side note: but what if not? Notice that in general the newly obtained radicals are again nested. What if we apply to each this transformation? Well, one should check that we simply get back the original expression...
Anyhow, whenever one is faced with this kind of nested radicals, there is a method, or a formula, that will take care of it.
Note: This formula also shows how to take square roots of complex numbers.
Obs: What about radicals of the form $\sqrt[3]{a + \sqrt{b}}$? Yes, there is a method even here... It deals with solution of equations of third degree...
Square both expressions.
$$(2\sqrt{2+\sqrt3})^2=4(2+\sqrt3)=8+4\sqrt3$$ $$(\sqrt2+\sqrt6)^2=2+2\sqrt{2\cdot 6}+6=8+4\sqrt3.$$
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 $\;2\sqrt{2 \pm \sqrt{3}}.$
$$\color{gray}{2+\sqrt 3=\frac 1 2+\sqrt 3+\frac 3 2=\frac{1+2\sqrt3 +3}{2}=\frac{1+2\sqrt 3+(\sqrt 3)^2}{2}=\frac{(1+\sqrt3)^2}{2}}\\2\sqrt{\frac{(1+\sqrt 3)^2}{2}}\\ \color{gray}{\sqrt{\frac 1 2(1+\sqrt 3)^2}=\frac{\sqrt{(1+\sqrt 3)}}{\sqrt 2}}\\ 2\frac{\sqrt{(1+\sqrt 3)^2}}{\sqrt 2}\\ \color{gray}{\sqrt{(1+\sqrt 3)^2}=1+\sqrt 3:}\\ \frac{2}{\sqrt 2}1+\sqrt 3\\ \color{gray}{\frac{2(1+\sqrt 3)}{\sqrt 2}=\frac{2(1+\sqrt 3)}{\sqrt 2}\times \frac{\sqrt 2}{\sqrt 2}=\frac{1(1+\sqrt 3)\sqrt 2}{2}:}\\ \frac{2(1+\sqrt 3)\sqrt 2}{2}\\ \color{gray}{\frac{2(1+\sqrt 3)\sqrt 2}{2}=\frac 2 2\times(1+\sqrt 3)\sqrt 2=(1+\sqrt3)\sqrt 2}\\ (1+\sqrt 3)\sqrt 2\\ \color{gray}{\sqrt 2(1+\sqrt 3)}=\boxed{\color{blue}{\sqrt 2+\sqrt 6}}$$
