# 5. Variant 1

It is known that $\cos \alpha+\cos \beta+\cos \gamma=\sqrt{\frac{1}{5}}, \sin \alpha+\sin \beta+\sin \gamma=\sqrt{\frac{4}{5}}$. Find $\cos (\alpha-\beta)+\cos (\beta-$ $\gamma)+\cos (\gamma-\alpha)$
Answer: -1.

Solution. Consider the expression $(\cos \alpha+\cos \beta+\cos \gamma)^{2}+(\sin \alpha+\sin \beta+\sin \gamma)^{2}$ and expand the brackets: $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma+2(\cos \alpha \cdot \cos \beta+\cos \alpha \cdot \cos \gamma+\cos \beta \cdot \cos \gamma)+\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma+2(\sin \alpha \cdot \sin \beta+\sin \alpha \cdot \sin \gamma+\sin \beta \cdot \sin \gamma)$. We transform, then apply the fundamental trigonometric identity and the cosine difference formula: $\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)+\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+\left(\cos ^{2} \gamma+\sin ^{2} \gamma\right)+2(\cos \alpha \cdot \cos \beta+\sin \alpha \cdot \sin \beta)+2(\cos \alpha \cdot \cos \gamma+\sin \alpha \cdot \sin \gamma)+2(\cos \beta \cdot \cos \gamma+\sin \beta \cdot \sin \gamma)=3+2(\cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\gamma-\alpha))$. Substituting the initial expression $\cos \alpha+\cos \beta+\cos \gamma=\sqrt{\frac{1}{5}}, \sin \alpha+\sin \beta+\sin \gamma=\sqrt{\frac{4}{5}}$ we get the equation: $\left(\sqrt{\frac{1}{5}}\right)^{2}+\left(\sqrt{\frac{4}{5}}\right)^{2}=3+2(\cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\gamma-\alpha))$, from which we find the answer.

## Variant 2

It is known that $\cos \alpha+\cos \beta+\cos \gamma=\sqrt{\frac{10}{7}}, \sin \alpha+\sin \beta+\sin \gamma=\sqrt{\frac{11}{7}}$. Find $\cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\gamma-\alpha)$

## Answer: 0.

## Variant 3

It is known that $\cos \alpha+\cos \beta+\cos \gamma=\sqrt{\frac{10}{3}}, \sin \alpha+\sin \beta+\sin \gamma=\sqrt{\frac{17}{3}}$. Find $\cos (\alpha-\beta)+\cos (\beta-\gamma)+\cos (\gamma-\alpha)$

Answer: 3.