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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 $$\cos A+\cos B+\cos C$$

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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.

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  • $\begingroup$ Nice answer! (+1) $\endgroup$
    – mathproof
    Feb 21 at 7:21
  • $\begingroup$ (+1) .. a nice answer. $\endgroup$
    – Narasimham
    Feb 21 at 7:31
  • $\begingroup$ beautiflu derivation. thanks very much!! $\endgroup$ Feb 21 at 17:41

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