Three sides of a right-angled triangle are $2\sqrt{3}$ (hypotenuse), $\sqrt{6}+1$, & $\sqrt{5-2\sqrt{6}}$. How do I transform the third side's length to $\sqrt{3}-\sqrt{2}$ for making my life easier in a trigonometric problem?
This is the problem if you're interested:
We have to prove left-hand side = right-hand side.
HINT
The argument of the square root is a perfect square:
\begin{align*} 5 - 2\sqrt{6} & = \sqrt{3^{2}} - 2\sqrt{3}\sqrt{2} + \sqrt{2^{2}}\\\\ & = (\sqrt{3} - \sqrt{2})^{2} \end{align*}
One: Is $(\sqrt 3 - \sqrt 2)^2 = 5 -2\sqrt 6$?
If so, then $\sqrt 3-2$ is one of the square roots of $5-2\sqrt 6$.
Two: Is $\sqrt 3-\sqrt 2 \ge 0$?
If so, then $\sqrt 3- \sqrt 2$ is the $\sqrt{5-2\sqrt 6}$.
==========
To show One.... just do it
$(\sqrt 3 - \sqrt 2)^2 = \sqrt 3^2 - 2\sqrt 3\sqrt 2 + \sqrt 2^2 =$
$3 - 2\sqrt 6 +2 = 5 -2\sqrt 6$.
To show Two.
If $0 < a <b$ then $\sqrt a < \sqrt b$. (If not then $0< \sqrt b \le\sqrt a$ and $b =\sqrt b\sqrt b \le \sqrt b\sqrt a \le \sqrt a\sqrt a = a$ which contradicts $a < b$)
So as $0 < 2 < 3$ we have $\sqrt 2 < \sqrt 3$ and $\sqrt 3 - \sqrt 2 > 0$.
That is all there is to it.
.......
But if you want to do it the hard way:
$\sqrt {5 - 2\sqrt 6} = \sqrt{3+2 -2\sqrt{3}\sqrt 2}=$
$\sqrt{\sqrt 3^2 - 2\sqrt{3}\sqrt 2 + \sqrt 2^2} =$
$\sqrt{(\sqrt 3 - \sqrt 2)^2} = $
$|\sqrt 3 - \sqrt 2| = \sqrt 3 - \sqrt 2$.
Let $\sqrt{5-2\sqrt{6}}=\sqrt{x}-\sqrt{y}$, where $x,y$ are rationals. Squaring we get $$5-2\sqrt{6}=x+y-2\sqrt{xy} \implies x+y=5, \sqrt{xy}=6 \implies x=3, y=2$.$$ Hence $$\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}$$
The thing about numbers that are sums and differences of integers and integer multiples of irrational numbers is that the irrational part persists in arithmetic operations like multiplication and exponentiation. So the square-root can be used to trace the number's "origin".
If we didn't sort of know the answer already, we could try $ \ (a + b\sqrt6)^2 \ = \ 5 - 2 \sqrt6 \ \ , $ which gives us the correspondences $ \ a^2 + 6b^2 \ = \ 5 \ \ $ and $ \ 2ab \sqrt6 \ = \ -2 \sqrt6 \ \ . $ The first equation does not have a real solution, so we would next try removing a factor of $ \ 2 \ $ or $ \ 3 \ $ from the argument of the radical. If we choose $ \ a + b \sqrt3 \ \ , $ the equations become $ \ a^2 + 3b^2 \ = \ 5 \ \ $ and $ \ 2ab \sqrt3 \ = \ -2 \sqrt6 \ = \ (-2 \sqrt2) · \sqrt3 \ \ . $ At this point, we could "eyeball" values for $ \ a \ $ and $ \ b \ \ , $ although they are no longer both integers.
If one doesn't trust "solution by inspection" (or has a more difficult problem), one could proceed by subtracting $ \ a^2 + (\sqrt3·b)^2 \ = \ 5 \ \ $ from $ \ (a + b\sqrt3)^2 \ = \ 5 - (2 · \sqrt2 · \sqrt3) \ \ , $ leaving $ \ 2a· ( \sqrt3·b) \ = \ -2 · \sqrt2 · \sqrt3 \ \Rightarrow \ ab \ = \ -\sqrt2 \ \ , $ Returning to $ \ a^2 + (\sqrt3·b)^2 \ = \ 5 \ \ $ indicates that we have $ \ a = \pm \sqrt2 \ $ and $ \ b = \mp 1 \ \ . $ Since we wish $ \ a + b \sqrt3 \ > \ 0 \ \ , $ since it represents one side of a right triangle, that side must be $ \ -\sqrt2 \ + \ 1·\sqrt3 \ \ . $
[I had worked out the relation to be proved in the main problem by using the cosine of a difference of two angles; it is possible to defer resolving this square-root until the penultimate step.]