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If DeltaABC having vertices A(acostheta1...

If `DeltaABC` having vertices `A(acostheta_1, asintheta_1), B(acostheta_2, asintheta_2), and C(acostheta_3, asintheta_3)` are equilateral triangle, then prove that `cos theta_1 + costheta_2 + cos theta_3 =0 and sintheta_1 + sintheta_2 + sintheta_3 =0`

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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`
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