Prove that the value $\frac{1}{2}\sqrt{2 - \sqrt{3}} = \left(\sqrt{6} - \sqrt{2}\right)/4$

Question

Using half-angle formula, the simplest intuitive ‘exact’ answer to $\sin(15^{\circ}) = \frac{1}{2}\sqrt{2 - \sqrt{3}} $. However, using instead angle sum-addition properties the most available reduced exact form is $\frac{\sqrt6-\sqrt2}{4}$. Using a calculator to check confirms they are the same value. Why is this so?, And which form is more pure or ideal, the one without a radical within a radical? How can you convert from the other form to that one, or vice-versa?

Answer 1

There's a mistake in your question. It should be $\dfrac{\sqrt{2-\sqrt3}}{2}$.

$\dfrac{\sqrt{2-\sqrt3}}{2} =\dfrac{\sqrt{(2-\sqrt3)\cdot\color{red}{4}}}{2\cdot\color{red}{2}} = \dfrac{\sqrt{8-4\sqrt3}}{4} = \dfrac{\sqrt{6+2-2\cdot\sqrt2\sqrt6}}{4}=\dfrac{\sqrt{(\sqrt6-\sqrt2)^2}}{4}=\dfrac{\sqrt6-\sqrt2}{4}$

---

Which form is more pure or ideal? It depends on where you end with the above value. For example, on solving a particular question you have $x= \dfrac{\sqrt{2-\sqrt3}}{2}$ you may reduce it to $\dfrac{\sqrt6-\sqrt2}{4}$.
Let's consider a case where you end up with $\sqrt x = \dfrac{\sqrt{2-\sqrt3}}{2}$. You can just square both sides to get $x= \dfrac{2-\sqrt3}{4}$. Here, squaring $\dfrac{\sqrt6-\sqrt2}{4}$ will also yield the same result but requires evaluating $(\sqrt6-\sqrt2)^2 = 6+2-2\sqrt2\sqrt6 = 8-4\sqrt3$, then simplifying the fraction.

Answer 2

Alternate approach

In your query's title, both the LHS and the RHS are positive.

If $0 < A,B$ and $A^2 = B^2$, then $A$ must $= B$.

LHS squared is

$$\frac{1}{4} \left(2 - \sqrt{3}\right)$$

RHS squared is

$$\frac{1}{16} \left(6 - 2\sqrt{12} + 2\right)$$

$$ = \frac{1}{4} \left(\frac{8}{4} - \frac{2\sqrt{12}}{4}\right)$$

Since $\sqrt{12} = 2\sqrt{3}$, the LHS squared = RHS squared.

Answer 3

You mean:

$$\sqrt{2-\sqrt3}=\frac12(\sqrt6-\sqrt2)$$

For these cases use identity:

$$\sqrt{A ±\sqrt B}=\sqrt{\frac{A+\sqrt{A^2-B}}{2}} ±\sqrt{\frac{A-\sqrt{A^2-B}}{2}}$$

Puting A=2 and B=3 you get:

$$\sqrt{2-\sqrt3}=\sqrt{\frac{3}2}-\sqrt{\frac 12}=\frac12(\sqrt6-\sqrt2)$$

Answer 4

There is no hard and fast rule, but it is generally preferred to simplify (where possible) so that you have simple radicals and not nested ones. So the right hand side looks better to me.

To show they are equal, start with the right hand side:

$$\frac{\sqrt 6 - \sqrt 2} {4} = \sqrt{\frac{(\sqrt 6 - \sqrt 2)^2} {16}}= \sqrt{\frac{(6 -2\sqrt{12} + 2} {16}} = \sqrt{\frac{8 - 4\sqrt 3} {16}} = \frac 12{\sqrt{2 - \sqrt 3}}$$