Correct Answer: Option D

Explanation

To find \(\cos 165^\circ\), we use the cosine of the difference formula:

\(\cos(a - b) = \cos a \cos b + \sin a \sin b\)

Expressing \(165^\circ\) as \(180^\circ - 15^\circ\):

\(\cos 165^\circ = \cos(180^\circ - 15^\circ) = -\cos 15^\circ\)

Now, finding \(\cos 15^\circ\) using \(45^\circ - 30^\circ\):

\(\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ\)

\(\cos 45^\circ = \frac{\sqrt{2}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\sin 45^\circ = \frac{\sqrt{2}}{2}\), \(\sin 30^\circ = \frac{1}{2}\):

\(\cos 15^\circ = \left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2} \cdot \frac{1}{2}\right)\)

\(= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

Thus, \(\cos 165^\circ = -\cos 15^\circ = -\frac{\sqrt{6} + \sqrt{2}}{4}\) = - \(\frac{1}{4}\)(\(\sqrt{6}\) + \(\sqrt{2}\))

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