If ` A B C` having vertices `A(acostheta_1,asintheta_1),B(acostheta_2asintheta_2),a n dC(acostheta_3,asintheta_3)` is equilateral, then prove that `costheta_1+costheta_2+costheta_3=sintheta_1+sintheta_2+sintheta_3=0.`

Explanation

The distance of given vertices` A(acostheta_1,a sintheta_1), B(acostheta_2,a sintheta_2)`, and `C(acostheta_3,a sintheta_3)` from the origin (0,0) is a .
Hence, the circumcenter of the triangle is (1,0). Also,in an equilateral triangle, the controid coincides with the circumcenter. We have `(acostheta-1+acostheta_2+acostheta_3)/(3)=0`
`(asintheta-1+asintheta_2+asintheta_3)/(3)=0`
or `costheta_1+costheta_2+costheta_3=sintheta_1+sintheta_2+sintheta=0`