Đáp án:
\[D = \frac{1}{2}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x = - \cos x\left( {\pi - x} \right)\\
\sin x = - \sin \left( { - x} \right)\\
\sin x = \sin \left( {\pi - x} \right)\\
\sin x.\cos y = \frac{1}{2}\left( {\sin \left( {x + y} \right) + \sin \left( {x - y} \right)} \right)\\
D = \cos \frac{\pi }{7} - \cos \frac{{2\pi }}{7} + \cos \frac{{3\pi }}{7}\\
\Leftrightarrow D = \cos \frac{\pi }{7} + \cos \left( {\pi - \frac{{2\pi }}{7}} \right) + \cos \frac{{3\pi }}{7}\\
\Leftrightarrow D = \cos \frac{\pi }{7} + \cos \frac{{3\pi }}{7} + \cos \frac{{5\pi }}{7}\\
\Leftrightarrow \sin \frac{\pi }{7}.D = \sin \frac{\pi }{7}.\cos \frac{\pi }{7} + \sin \frac{\pi }{7}.\cos \frac{{3\pi }}{7} + \sin \frac{\pi }{7}.\cos \frac{{5\pi }}{7}\\
\Leftrightarrow \sin \frac{\pi }{7}.D = \frac{1}{2}\sin \frac{{2\pi }}{7} + \frac{1}{2}.\left( {\sin \frac{{4\pi }}{7} + \sin \frac{{ - 2\pi }}{7}} \right) + \frac{1}{2}\left( {\sin \frac{{6\pi }}{7} + \sin \frac{{ - 4\pi }}{7}} \right)\\
\Leftrightarrow \sin \frac{\pi }{7}.D = \frac{1}{2}\left( {\sin \frac{{2\pi }}{7} + \sin \frac{{4\pi }}{7} - \sin \frac{{2\pi }}{7} + \sin \frac{{6\pi }}{7} - \sin \frac{{4\pi }}{7}} \right)\\
\Leftrightarrow \sin \frac{\pi }{7}.D = \frac{1}{2}\sin \frac{{6\pi }}{7}\\
\Leftrightarrow \sin \frac{\pi }{7}.D = \frac{1}{2}\sin \left( {\pi - \frac{{6\pi }}{7}} \right)\\
\Leftrightarrow D = \frac{1}{2}
\end{array}\)
Vậy \(D = \frac{1}{2}\)
