The exact value of cos frac{2pi}{28} csc frac | vedclass

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(D) Let $x = \frac{\pi}{28}$. The expression is $S = \cos(2x) \csc(3x) + \cos(6x) \csc(9x) + \cos(18x) \csc(27x)$.Consider the general term $T_k = \co

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(D) Let $x = \frac{\pi}{28}$. The expression is $S = \cos(2x) \csc(3x) + \cos(6x) \csc(9x) + \cos(18x) \csc(27x)$.Consider the general term $T_k = \cos(2 \cdot 3^{k-1} x) \csc(3^k x) = \frac{\cos(2 \cdot 3^{k-1} x)}{\sin(3^k x)}$.Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have $T_k = \frac{2 \sin(3^{k-1} x) \cos(2 \cdot 3^{k-1} x)}{2 \sin(3^{k-1} x) \sin(3^k x)} = \frac{\sin(3^k x) - \sin(3^{k-1} x)}{2 \sin(3^{k-1} x) \sin(3^k x)} = \frac{1}{2} (\csc(3^{k-1} x) - \csc(3^k x))$.Summing for $k=1, 2, 3$:$S = \frac{1}{2} [(\csc x - \csc 3x) + (\csc 3x - \csc 9x) + (\csc 9x - \csc 27x)]$.This is a telescoping sum: $S = \frac{1}{2} (\csc x - \csc 27x)$.Since $x = \frac{\pi}{28}$,$27x = \frac{27\pi}{28} = \pi - \frac{\pi}{28} = \pi - x$.Thus,$\csc 27x = \csc(\pi - x) = \csc x$.Therefore,$S = \frac{1}{2} (\csc x - \csc x) = 0$.

The exact value of $\cos \frac{2\pi}{28} \csc \frac{3\pi}{28} + \cos \frac{6\pi}{28} \csc \frac{9\pi}{28} + \cos \frac{18\pi}{28} \csc \frac{27\pi}{28}$ is equal to

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