I have this (not necessarily optimum) method for finding the value of sin $15^{\circ}$.
(Meant to be an application of Angle Bisector Theorem).
In a equilateral triangle with side length 1, draw the altitude to get two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles
Then, in one of the $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, draw the angle bisector again to get a $15^{\circ}-90^{\circ}-75^{\circ}$ triangle.
Apply the Angle bisector theorem to find that the side opposite $15$ degrees is $\dfrac{(2\sqrt{3}-3)}{2}$.
Combine with height $\dfrac {\sqrt{3}}{2}$ to get the hypotenuse squared as $6 \sqrt{3}$, which then has as its square root $\dfrac {(3-\sqrt{3})}{\sqrt{2}}$.
Apply sin = opp/hyp to get the usual formula for $\sin 15^{\circ}$.
I got $\sqrt{6-3\sqrt {3}}$ as $\dfrac{(3-\sqrt{3})}{\sqrt{2}}$ through trial and error.
Is there a systematic way of finding it?
Best Answer
Presumably you mean
$\sqrt{6-\color{blue}{3}\sqrt3}=(3-\sqrt3)/\sqrt2.$
Suppose we are given two positive rational numbers $a,b$ and we seek $x,y$ such that
$\sqrt{a-\sqrt{b}}=\sqrt{x}-\sqrt{y}$
Then we will have the conjugate relation
$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}$
And upon multiplication
$\color{blue}{\sqrt{a^2-b}=x-y}$
We can also square each of the conjugate equations and add those squared results. The radicals cancel and we end with
$\color{blue}{a=x+y}$
Now, if $a^2-b$ is a perfect square, then the left sides of both blue equations are rational numbers and we can simply solve these equations as a linear system for $x$ and $y$. For the problem at hand we have $a=6,b=27,\therefore a^2-b=9=3^2$ and this process will work. We get $x=9/2,y=3/2$ which matches the claim.