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?

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    $\begingroup$ square each and see if the squares are equal $\endgroup$ – Mirko Dec 23 '15 at 15:14
  • $\begingroup$ @Mirko Oh. :/ That's weird. Hm... $\endgroup$ – Simply Beautiful Art Dec 23 '15 at 15:16
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    $\begingroup$ As answers show, it's relatively straightforward to verify this particular equality. In general, however, the process of "de-nesting" radicals can be something of an art ... though not a simple one. ;) The answers and comments on this question have some interesting links. $\endgroup$ – Blue Dec 23 '15 at 15:18
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    $\begingroup$ @Blue Thank you for the links. Very helpful. $\endgroup$ – Simply Beautiful Art Dec 23 '15 at 15:28
  • $\begingroup$ The angle $\frac{x}{2}$ seems to be an imaginary angle. $\endgroup$ – André Nicolas Dec 23 '15 at 15:43

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

$$\sqrt{ a + \sqrt{b}}$$.

One tries to write it in the form $$\sqrt{ a + \sqrt{b}} = \sqrt{x} + \sqrt{y}$$

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 $$\sqrt{ a + \sqrt{b}} = \sqrt{ \frac{a + \sqrt{a^2 - b} }{2}} + \sqrt{ \frac{a - \sqrt{a^2 - b} }{2}}$$

Hardly anything simpler... except when $a^2 - b$ is a square, and we get the nested radical as a sum 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.

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...

  • $\begingroup$ Oh, great. This explains the solution to the general equation. Thank you. It even hints at the solution to the third degree equation, which now that you've mentioned it, the solution to the third degree does look like a nested radical. Thank you! $\endgroup$ – Simply Beautiful Art Dec 23 '15 at 16:28
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    $\begingroup$ You probably know this, but for others who might not I'll mention that this method is a standard topic in 1800s algebra textbooks that cover radical manipulations, and there are probably over 100 different such books (from the 1800s) digitized by google that include this topic. However, the topic was slowly written out of such books so that by the early 1930s very few college algebra textbooks included it anymore. $\endgroup$ – Dave L. Renfro Dec 23 '15 at 16:38
  • $\begingroup$ @Simple Art: You are very welcome! It turned out to be an inspiring question for me. $\endgroup$ – Orest Bucicovschi Dec 23 '15 at 16:38
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    $\begingroup$ @DaveL.Renfro Wow. That's kind of disappointing, math being taken out of books. $\endgroup$ – Simply Beautiful Art Dec 23 '15 at 16:40
  • $\begingroup$ @DaveL.Renfro: Yes, it used to be in older books. Still find it very interesting. I remember there was once a nested radical $\sqrt[3]{a + c\sqrt{d}}$ that in fact was $p + q \sqrt{d}$ in a non-obvious way...All my Galois theory did not help at that time. $\endgroup$ – Orest Bucicovschi Dec 23 '15 at 16:45

Just note that $$2\sqrt{2+\sqrt{3}}=\sqrt{8+4\sqrt{3}}=\sqrt{(\sqrt{6})^2+(\sqrt{2})^2+2(\sqrt2)(\sqrt6)}=\sqrt{(\sqrt{2}+\sqrt{6})^2}=\sqrt{2}+\sqrt{6}.$$

  • $\begingroup$ This method seems to rely on pure chance that you can get a perfect square in the radical. $\endgroup$ – Simply Beautiful Art Dec 23 '15 at 15:28
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    $\begingroup$ Yes, but it does the trick just fine. $\endgroup$ – Jack Frost Dec 23 '15 at 15:28

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 $\;2\sqrt{2 \pm \sqrt{3}}.$

  • $\begingroup$ Very nice! Never seen this before... $\endgroup$ – Orest Bucicovschi Dec 23 '15 at 16:41




$$\color{gray}{2+\sqrt 3=\frac 1 2+\sqrt 3+\frac 3 2=\frac{1+2\sqrt3 +3}{2}=\frac{1+2\sqrt 3+(\sqrt 3)^2}{2}=\frac{(1+\sqrt3)^2}{2}}\\2\sqrt{\frac{(1+\sqrt 3)^2}{2}}\\ \color{gray}{\sqrt{\frac 1 2(1+\sqrt 3)^2}=\frac{\sqrt{(1+\sqrt 3)}}{\sqrt 2}}\\ 2\frac{\sqrt{(1+\sqrt 3)^2}}{\sqrt 2}\\ \color{gray}{\sqrt{(1+\sqrt 3)^2}=1+\sqrt 3:}\\ \frac{2}{\sqrt 2}1+\sqrt 3\\ \color{gray}{\frac{2(1+\sqrt 3)}{\sqrt 2}=\frac{2(1+\sqrt 3)}{\sqrt 2}\times \frac{\sqrt 2}{\sqrt 2}=\frac{1(1+\sqrt 3)\sqrt 2}{2}:}\\ \frac{2(1+\sqrt 3)\sqrt 2}{2}\\ \color{gray}{\frac{2(1+\sqrt 3)\sqrt 2}{2}=\frac 2 2\times(1+\sqrt 3)\sqrt 2=(1+\sqrt3)\sqrt 2}\\ (1+\sqrt 3)\sqrt 2\\ \color{gray}{\sqrt 2(1+\sqrt 3)}=\boxed{\color{blue}{\sqrt 2+\sqrt 6}}$$


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