Dear Uncle Colin,

How can I prove that $\frac{\sqrt{6}-\sqrt{2}}{4} = \frac{\sqrt{2 - \sqrt{3}}}{2}$?

- Help! I’d Probably Put A $\sin$ Under Something

Hi, HIPPAsUS, and thanks for your message! I can see several ways to approach this.

### Very directly

The left hand side is clearly positive; if you square it, you get $\frac{6 - 2\sqrt{12} + 2}{16}$, which simplifies:

• $\dots = \frac{8 - 4\sqrt{3}}{16}$
• $\dots = \frac{2-\sqrt{3}}{4}$

The right hand side is also positive, and its square is $\frac{2- \sqrt{3}}{4}$, the same as the left. The two numbers have the same sign and the same square, so they’re equal.

### A Pythagorean Approach

If you’ve followed the blog assiduously for many years and have an almost photographic memory, or otherwise, you might know that the left hand side is $\sin(15^o)$.

Let’s have a maths disco and throw some shapes. Specifically, triangles.

Let’s pick a triangle with an angle ($\theta$) that the left hand side is the sine of – and I’m going to pick one with a hypotenuse of 2 and an opposite side of $x = \frac{\sqrt{6} - \sqrt{2}}{2}$. Why not a hypotenuse of 4 and a less cumbersome opposite side? Because I did that first and it came out a little bit messier.

What’s the adjacent side? Well, it’s the square root of $2^2 - x^2$. It turns out that $x^2 = 2 + \sqrt{3}$, so $\cos(\theta) = \frac{\sqrt{2+\sqrt{3}}}{2}$.

Now, a (supposedly) different triangle with an angle $\phi$ that the right hand side is the sine of: again, a hypotenuse of 2 and an opposite side of $y = \sqrt{2 - \sqrt{3}}$. The adjacent side in this triangle is $4 - (2-\sqrt{3})$, or $2+\sqrt{3}$ – and $\cos(\phi) = \frac{\sqrt{2+\sqrt{3}}}{2}$.

That’s the same as $\cos(\theta)$, and they’re both in the same quadrant, so $\theta = \phi$. The two hypotenuses are the same, so the side lengths are the same – and in particular, $x=y$, as required.

Dear readers, do you have any other ways to tackle it? I’d love to hear about them.

And as for you, HIPPAsUS, I hope that helps!

- Uncle Colin