The value of the series cos 12^{circ} + cos 8 | vedclass

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(D) We use the formula $\cos C + \cos D = 2 \cos \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right)$.Given expression: $S = \cos 132^{\cir

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$-\frac{1}{2}$

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(D) We use the formula $\cos C + \cos D = 2 \cos \left( \frac{C+D}{2} \right) \cos \left( \frac{C-D}{2} \right)$.Given expression: $S = \cos 132^{\circ} + \cos 12^{\circ} + \cos 156^{\circ} + \cos 84^{\circ}$.Applying the formula to the pairs:$S = 2 \cos \left( \frac{132^{\circ}+12^{\circ}}{2} \right) \cos \left( \frac{132^{\circ}-12^{\circ}}{2} \right) + 2 \cos \left( \frac{156^{\circ}+84^{\circ}}{2} \right) \cos \left( \frac{156^{\circ}-84^{\circ}}{2} \right)$$S = 2 \cos 72^{\circ} \cos 60^{\circ} + 2 \cos 120^{\circ} \cos 36^{\circ}$.Substituting the known values $\cos 60^{\circ} = \frac{1}{2}$,$\cos 120^{\circ} = -\frac{1}{2}$,$\cos 72^{\circ} = \frac{\sqrt{5}-1}{4}$,and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$:$S = 2 \left( \frac{\sqrt{5}-1}{4} \right) \left( \frac{1}{2} \right) + 2 \left( -\frac{1}{2} \right) \left( \frac{\sqrt{5}+1}{4} \right)$$S = \frac{\sqrt{5}-1}{4} - \frac{\sqrt{5}+1}{4}$$S = \frac{\sqrt{5}-1-\sqrt{5}-1}{4} = \frac{-2}{4} = -\frac{1}{2}$.

The value of the series $\cos 12^{\circ} + \cos 84^{\circ} + \cos 132^{\circ} + \cos 156^{\circ}$ is

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