Evaluate the determinant: left|begin{array}{c | vedclass

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(C) The given determinant is $D = \sin \frac{11 \pi}{36} \cos \frac{2 \pi}{9} - \cos \frac{11 \pi}{36} \sin \frac{2 \pi}{9}$.Using the trigonometric i

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$\cos \frac{5 \pi}{12}$

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(C) The given determinant is $D = \sin \frac{11 \pi}{36} \cos \frac{2 \pi}{9} - \cos \frac{11 \pi}{36} \sin \frac{2 \pi}{9}$.Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we set $A = \frac{11 \pi}{36}$ and $B = \frac{2 \pi}{9}$.First,convert $B$ to a common denominator: $B = \frac{2 \pi}{9} = \frac{8 \pi}{36}$.Now,$D = \sin \left( \frac{11 \pi}{36} - \frac{8 \pi}{36} ight) = \sin \left( \frac{3 \pi}{36} ight) = \sin \left( \frac{\pi}{12} ight)$.Since $\sin 	heta = \cos \left( \frac{\pi}{2} - 	heta ight)$,we have $\sin \left( \frac{\pi}{12} ight) = \cos \left( \frac{\pi}{2} - \frac{\pi}{12} ight) = \cos \left( \frac{6 \pi - \pi}{12} ight) = \cos \frac{5 \pi}{12}$.Thus,the correct option is $C$.

Evaluate the determinant: $\left|\begin{array}{cc}\sin \frac{11 \pi}{36} & \cos \frac{11 \pi}{36} \\\sin \frac{2 \pi}{9} & \cos \frac{2 \pi}{9}\end{array}\right|$.

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