Let ${A_{1,}}{A_2},{A_3},{A_4},{A_5},{A_6},{A_1}$ be regular hexagon .Write the x- component of the vectors represented by the taken six sides taken in order . Use the fact that the resultant of these six vector is zero , to prove that
$\cos 0 + \cos \dfrac{\pi }{3} + \cos \dfrac{{2\pi }}{3} + \cos \dfrac{{3\pi }}{3} + \cos \dfrac{{4\pi }}{3} + \cos \dfrac{{5\pi }}{3} = 0$
Use know cosine value to verify the result
               

Hint: In this The resultant of these six vector is equal to zero hence the vectors hence the horizontal component also become equal to zero mean that ${A_1}{A_2}\cos {0^\circ } + {A_2}{A_3}\cos {60^\circ } + {A_3}{A_4}\cos {120^\circ } + {A_4}{A_5}\cos {180^\circ } + {A_5}{A_6}\cos {240^\circ } + {A_6}{A_1}\cos {300^\circ } = 0$ and it is regular hexagon then $\left| {{A_1}{A_2}} \right| = \left| {{A_2}{A_3}} \right| = \left| {{A_3}{A_4}} \right| = \left| {{A_4}{A_5}} \right| = \left| {{A_5}{A_6}} \right| = \left| {{A_6}{A_1}} \right| = a$ from these we will prove the given condition

Complete step-by-step answer:
Let us take the magnitude of vector ${A_1}{A_2}$ or $\left| {{A_1}{A_2}} \right| = a$
As in the question it is given that the Hexagon is regular hence the magnitude of
$\left| {{A_1}{A_2}} \right| = \left| {{A_2}{A_3}} \right| = \left| {{A_3}{A_4}} \right| = \left| {{A_4}{A_5}} \right| = \left| {{A_5}{A_6}} \right| = \left| {{A_6}{A_1}} \right| = a$
Hence
The resultant of these six vector is equal to zero hence the vectors
\[{A_1}{A_2} + {A_2}{A_3} + {A_3}{A_4} + {A_4}{A_5} + {A_5}{A_6} + {A_6}{A_1} = 0\]
So its horizontal component is also equal to zero and vertical component is also equal to zero
Therefore,
               
Hence for the horizontal component
${A_1}{A_2}\cos {0^\circ } + {A_2}{A_3}\cos {60^\circ } + {A_3}{A_4}\cos {120^\circ } + {A_4}{A_5}\cos {180^\circ } + {A_5}{A_6}\cos {240^\circ } + {A_6}{A_1}\cos {300^\circ } = 0$
we know that $\left| {{A_1}{A_2}} \right| = \left| {{A_2}{A_3}} \right| = \left| {{A_3}{A_4}} \right| = \left| {{A_4}{A_5}} \right| = \left| {{A_5}{A_6}} \right| = \left| {{A_6}{A_1}} \right| = a$
$a\cos {0^\circ } + a\cos {60^\circ } + a\cos {120^\circ } + a\cos {180^\circ } + a\cos {240^\circ } + a\cos {300^\circ } = 0$
Hence a is common in the equation it is cancel out remaining equation become ,
$\cos {0^\circ } + \cos {60^\circ } + \cos {120^\circ } + \cos {180^\circ } + \cos {240^\circ } + \cos {300^\circ } = 0$
Now in radian form we can write it as ,
$\cos 0 + \cos \dfrac{\pi }{3} + \cos \dfrac{{2\pi }}{3} + \cos \dfrac{{3\pi }}{3} + \cos \dfrac{{4\pi }}{3} + \cos \dfrac{{5\pi }}{3} = 0$
hence proved .
For cross checking
Now by putting the value of $\cos 0 = 1,\cos \pi = - 1$
$1 + \cos \dfrac{\pi }{3} + \cos \dfrac{{2\pi }}{3} - 1 + \cos \dfrac{{4\pi }}{3} + \cos \dfrac{{5\pi }}{3}$
$\cos \dfrac{\pi }{3} + \cos \dfrac{{2\pi }}{3} + \cos \dfrac{{4\pi }}{3} + \cos \dfrac{{5\pi }}{3}$
Now we know that $\cos \dfrac{{2\pi }}{3}$ can we written as ,
$\cos \left( {\pi - \dfrac{\pi }{3}} \right) = - \cos \dfrac{\pi }{3}$
Or
$\cos \dfrac{{4\pi }}{3} = \cos \left( {\pi + \dfrac{\pi }{3}} \right) = - \cos \dfrac{\pi }{3}$
Or
$\cos \dfrac{{5\pi }}{3} = \cos \left( {2\pi - \dfrac{\pi }{3}} \right) = \cos \dfrac{\pi }{3}$ As cos is positive in the fourth quadrant hence it is in positive ,
Now by putting the value of it
$\cos \dfrac{\pi }{3} - \cos \dfrac{\pi }{3} - \cos \dfrac{\pi }{3} + \cos \dfrac{\pi }{3}$
= $0$
Hence it is correct

Note: If in the question it is given that to prove
$\sin 0 + \sin \dfrac{\pi }{3} + \sin \dfrac{{2\pi }}{3} + \sin \dfrac{{3\pi }}{3} + \sin \dfrac{{4\pi }}{3} + \sin \dfrac{{5\pi }}{3} = 0$ then we only do that the vertical component of the vectors is zero . Hence from this we will prove that this equation and we will also cross check it by putting the values. Always take the only from the positive x-axis otherwise the question becomes complicated.