c) $$ \cos \frac{11 \pi}{12} $$ d) $$ \sin \frac{13 \pi}{12} $$

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$$
\textbf{c) Evaluating } \cos \frac{11\pi}{12}:
$$

1. Express the angle as
$$
\frac{11\pi}{12}=\pi - \frac{\pi}{12}.
$$
2. Use the identity
$$
\cos\left(\pi-x\right)=-\cos x.
$$
Thus,
$$
\cos \frac{11\pi}{12}=-\cos \frac{\pi}{12}.
$$
3. Recognize that $$\frac{\pi}{12}$$ corresponds to $$15^\circ$$. Express $$\cos 15^\circ$$ using the formula for the cosine of a difference:
$$
\cos 15^\circ=\cos(45^\circ-30^\circ)=\cos 45^\circ\cos 30^\circ+\sin 45^\circ\sin 30^\circ.
$$
4. Substitute the known values:
$$
\cos 45^\circ=\frac{\sqrt{2}}{2}, \quad \cos 30^\circ=\frac{\sqrt{3}}{2}, \quad \sin 45^\circ=\frac{\sqrt{2}}{2}, \quad \sin 30^\circ=\frac{1}{2}.
$$
Thus,
$$
\cos 15^\circ=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2} =\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}.
$$
5. Therefore,
$$
\cos \frac{11\pi}{12}=-\frac{\sqrt{6}+\sqrt{2}}{4}.
$$

$$
\textbf{d) Evaluating } \sin \frac{13\pi}{12}:
$$

1. Express the angle as
$$
\frac{13\pi}{12}=\pi+\frac{\pi}{12}.
$$
2. Use the identity
$$
\sin\left(\pi+x\right)=-\sin x.
$$
Thus,
$$
\sin \frac{13\pi}{12}=-\sin \frac{\pi}{12}.
$$
3. Recognize that $$\frac{\pi}{12}$$ corresponds to $$15^\circ$$. Express $$\sin 15^\circ$$ using the formula for the sine of a difference:
$$
\sin 15^\circ=\sin(45^\circ-30^\circ)=\sin 45^\circ\cos 30^\circ-\cos 45^\circ\sin 30^\circ.
$$
4. Substitute the known values:
$$
\sin 45^\circ=\frac{\sqrt{2}}{2}, \quad \cos 30^\circ=\frac{\sqrt{3}}{2}, \quad \cos 45^\circ=\frac{\sqrt{2}}{2}, \quad \sin 30^\circ=\frac{1}{2}.
$$
Thus,
$$
\sin 15^\circ=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}.
$$
5. Therefore,
$$
\sin \frac{13\pi}{12}=-\frac{\sqrt{6}-\sqrt{2}}{4}.
$$
$$ \textbf{c) } \cos \frac{11\pi}{12} = -\frac{\sqrt{6} + \sqrt{2}}{4} $$ $$ \textbf{d) } \sin \frac{13\pi}{12} = -\frac{\sqrt{6} - \sqrt{2}}{4} $$

c)
d)

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c) d)