# Real numbers satisfing a trigonometric property

One of my students sent me the problem below and I am looking for solution. I tried simply to expand using complex numbers ($$Re(\sum e^{i(A-B)}$$) but to no avail. Does any body have idea how to proceeed?

If $$\cos (A-B)+\cos (B-C)+\cos (C-A)=-\frac32$$ find the value of $$\cos A+\cos B+\cos C$$

Consider the expression $$(\cos A + \cos B + \cos C)^2+(\sin A + \sin B + \sin C)^2\ .$$ Using the relations $$\sin^2 x + \cos^2 y=1$$ and $$\cos(x-y) = \cos x \cos y + \sin x \sin y$$, this expression simplifies to $$3 + 2 (\cos(A-B) + \cos(B-C)+\cos(C-A))\ .$$ Plugging in the provided value $$-\frac{3}{2}$$ for the part in parentheses, we find that the initial sum of squares is equal to zero. This means that $$\cos A + \cos B + \cos C = \sin A + \sin B + \sin C = 0$$.
Alternatively, using the language of complex numbers, we can consider $$e^{i A} + e^{i B} + e^{i C}$$. The norm is \begin{align} & \quad (e^{i A} + e^{i B} + e^{i C})(e^{-i A} + e^{-i B} + e^{-i C})\\ & = 3 + e^{i(A-B)} + e^{i(B-A)} + e^{i(B-C)} + e^{i(C-B)} + e^{i(C-A)} + e^{i(A-C)} \\ & = 3 + 2 (\cos(A-B) + \cos(B-C)+\cos(C-A))\ . \end{align} As before this is zero, so $$e^{i A} + e^{i B} + e^{i C}$$ must be zero. This means that its real part is also zero.