cos A + cos (240^circ + A) + cos (240^circ - | vedclass

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(B) Using the sum-to-product formula $\cos(X + Y) + \cos(X - Y) = 2 \cos X \cos Y$,we have:$\cos(240^\circ + A) + \cos(240^\circ - A) = 2 \cos 240^\ci

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(B) Using the sum-to-product formula $\cos(X + Y) + \cos(X - Y) = 2 \cos X \cos Y$,we have:$\cos(240^\circ + A) + \cos(240^\circ - A) = 2 \cos 240^\circ \cos A$Since $\cos 240^\circ = \cos(180^\circ + 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$,Substituting this value:$2 \cos 240^\circ \cos A = 2 \left( -\frac{1}{2} \right) \cos A = -\cos A$Now,substituting back into the original expression:$\cos A + (-\cos A) = 0$.

$\cos A + \cos (240^\circ + A) + \cos (240^\circ - A) = $

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