Let A and B denote the statements:A: cos alph | vedclass

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(B) Given $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$.Multiplying by $2$,we get $2[\cos (\alpha - \beta) +

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both are true

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(B) Given $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$.Multiplying by $2$,we get $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] = -3$.Adding $3$ to both sides: $2[\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha)] + 3 = 0$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we can write $3 = (\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + (\sin^2 \gamma + \cos^2 \gamma)$.Substituting this into the equation:$(\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma + 2\sin \alpha \sin \beta + 2\sin \beta \sin \gamma + 2\sin \gamma \sin \alpha) + (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2\cos \alpha \cos \beta + 2\cos \beta \cos \gamma + 2\cos \gamma \cos \alpha) = 0$.This simplifies to $(\sin \alpha + \sin \beta + \sin \gamma)^2 + (\cos \alpha + \cos \beta + \cos \gamma)^2 = 0$.Since the sum of squares is zero,each term must be zero:$\sin \alpha + \sin \beta + \sin \gamma = 0$ and $\cos \alpha + \cos \beta + \cos \gamma = 0$.Therefore,both statements $A$ and $B$ are true.

Let $A$ and $B$ denote the statements:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then:

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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 $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then: