Trigonometry problem

Solution 1:

HINT:

$$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2$$

$$=3+2\sum\cos(A-B)=0$$

Now use the fact that sum of squares of two real numbers is zero

Solution 2:

$\cos(a−b)+\cos(b−c)+\cos(c−a)=−3/2$ $\cos a \cos b+\sin a \sin b+\cos b \cos c+\sin b \sin c+\cos c \cos a+\sin c \sin a=-3/2 $

$\frac{1}{2}\cos a(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2a+\frac{1}{2}\sin a(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 a+ \frac{1}{2}\cos b(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2b+\frac{1}{2}\sin b(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 b+ \frac{1}{2}\cos c(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2c+\frac{1}{2}\sin c(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 c=-\frac{3}{2}$

$\cos a(\cos a + \cos b +\cos c)+\sin a(\sin a+\sin b+\sin c) - (sin^2 a+\cos^2a)+ \cos b(\cos a + \cos b +\cos c)+\sin b(\sin a+\sin b+\sin c) - (sin^2 b+\cos^2b)+ \cos c(\cos a + \cos b +\cos c)+\sin a(\sin a+\sin b+\sin c) - (sin^2 c+\cos^2c)=-3$

$(\cos a + \cos b +\cos c)^2+(\sin a+\sin b+\sin c)^2=0$

then $\cos a + \cos b +\cos c=0$ and $\sin a+\sin b+\sin c=0$