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Let $A$ and $B$ denote the statements
A: $\cos \alpha+\cos \beta+\cos \gamma=0$
B: $\sin \alpha+\sin \beta+\sin \gamma=0$
If $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$, then
Options:
A: $\cos \alpha+\cos \beta+\cos \gamma=0$
B: $\sin \alpha+\sin \beta+\sin \gamma=0$
If $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$, then
Solution:
2958 Upvotes
Verified Answer
The correct answer is:
both $A$ and $B$ are true
both $A$ and $B$ are true
$$
\begin{aligned}
& \cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2} \\
& \Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+3=0 \\
& \Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+\sin ^2 \alpha+\cos ^2 \alpha+\sin ^2 \beta+\cos ^2 \beta+\sin ^2 \gamma+\cos ^2 \gamma=0 \\
& \Rightarrow(\sin \alpha+\sin \beta+\sin \gamma)^2+(\cos \alpha+\cos \beta+\cos \gamma)^2=0
\end{aligned}
$$
\begin{aligned}
& \cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2} \\
& \Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+3=0 \\
& \Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+\sin ^2 \alpha+\cos ^2 \alpha+\sin ^2 \beta+\cos ^2 \beta+\sin ^2 \gamma+\cos ^2 \gamma=0 \\
& \Rightarrow(\sin \alpha+\sin \beta+\sin \gamma)^2+(\cos \alpha+\cos \beta+\cos \gamma)^2=0
\end{aligned}
$$
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