sin⁡(arccos⁡(13))=23 proof this equation

Let ${\displaystyle \theta =\arccos \left(\frac{1}{\sqrt{3}}\right)}$, so ${\displaystyle \cos \theta =\frac{1}{\sqrt{3}}}$
You want to find ${\displaystyle \sin \theta }$
Use ${\displaystyle {\sin }^2\theta +{\cos }^2\theta =1}$

${\displaystyle {\sin }^2\left(\arccos \left(\frac{1}{\sqrt{3}}\right)\right)+{\cos }^2\left(\arccos \left(\frac{1}{\sqrt{3}}\right)\right)={\sin }^2\left(\arccos \left(\frac{1}{\sqrt{3}}\right)\right)+\frac{13}{=}1}$
or
Solution sketch: Draw a right triangle, with 1 as the hypotenuse and ${\displaystyle \frac{1}{\sqrt{3}}}$ as one of the legs (or ${\displaystyle \sqrt{3}}$ as the hypotenuse and 1 as one of the legs, same thing). Then see what ${\displaystyle \arccos \left(\frac{1}{\sqrt{3}}\right)}$ corresponds to in that triangle, then see what sin of that represents. Once you know what you are actually after, you do the calculations.