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This question is from SL Loney.

If $\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)+1=0$,
then show that $\alpha-\beta$ or $\beta-\gamma$ or $\gamma-\alpha$ is a multiple of $\pi$.

My try: Let $\alpha-\beta=A$, $\beta-\gamma=B$, $\gamma-\alpha=C$ so that $A+B+C=0$. So we have to prove that:

If $\cos A +\cos B+\cos C+1=0$, then show that $A,B$ or $C$ is a multiple of $\pi$.

$$2 \cos\frac{A+B}{2}\cos\frac{A-B}{2}+2\cos^2\frac{C}{2}=0\\ 2 \cos\frac{C}{2}\cos\frac{A-B}{2}+2\cos^2\frac{C}{2}=0\\ 4 \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}=0.$$ Either $\cos\frac{A}{2}=0$ or $\cos\frac{B}{2}=0$ or $\cos\frac{C}{2}$.

Either $\frac{A}{2}$=odd multiple of $\frac{\pi}{2}$ or $\frac{B}{2}$=odd multiple of $\frac{\pi}{2}$ or $\frac{C}{2}$=odd multiple of $\frac{\pi}{2}$.

Either $A$=odd multiple of $\pi$ or $B$=odd multiple of $\pi$ or $C$=odd multiple of $\pi$.

But the answer is:

either $A$=multiple of $\pi$ or $B$=multiple of $\pi$ or $C$=multiple of $\pi$

Is my approach correct? Or is there some other method to prove it.

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    $\begingroup$ In your last step it should be $2 \cos \frac{C}{2}$ instead of $2 \sin \frac{C}{2}$ because $A+B=-C$. Once that is there factor it and simplify further. $\endgroup$ – Anurag A Aug 5 '15 at 4:19
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using your abbreviations, since $$ \cos A + \cos B + \cos (A+B) + 1 = 0 $$ we have $$ (1+\cos A)(1+\cos B) = \sin A \sin B $$ this leads directly to $$ \cos \frac{A}2 \cos \frac{B}2 (\cos \frac{A}2 \cos \frac{B}2 -\sin \frac{A}2 \sin \frac{B}2)=0 $$ hence either $A=n\pi$ or $B=n\pi$ or $$ \cos \frac{A}2 \cos \frac{B}2 -\sin \frac{A}2 \sin \frac{B}2=0 $$ which gives $$ \tan \frac{A}2 = \cot \frac{B}2 $$ or $$ A+B = (2n+1)\pi $$

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