Không dùng máy, tính:
$\sin \dfrac{\pi}{16}.\sin \dfrac{3\pi}{16}.\sin \dfrac{5\pi}{16}.\sin \dfrac{7\pi}{16}$
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1 câu trả lời 2487
\(\begin{array}{l}
\sin \frac{\pi }{{16}}.\sin \frac{{3\pi }}{{16}}.\sin \frac{{5\pi }}{{16}}.\sin \frac{{7\pi }}{{16}}\\
= (\sin \frac{\pi }{{16}}.\sin \frac{{7\pi }}{{16}}).(\sin \frac{{3\pi }}{{16}}.\sin \frac{{5\pi }}{{16}})\\
= \sin \frac{\pi }{{16}}.\sin (\frac{\pi }{2} - \frac{\pi }{{16}}).\sin \frac{{3\pi }}{{16}}.\sin (\frac{\pi }{2} - \frac{{3\pi }}{{16}})\\
= \sin \frac{\pi }{{16}}.\cos \frac{\pi }{{16}}.\sin \frac{{3\pi }}{{16}}\cos \frac{{3\pi }}{{16}}\\
= \frac{1}{2}\sin (\frac{{2\pi }}{{16}}).\frac{1}{2}\sin (\frac{{2.3\pi }}{{16}})\\
= \frac{1}{4}\sin \frac{\pi }{8}.\sin \frac{{3\pi }}{8}\\
= \frac{1}{4}.(\frac{{ - 1}}{2}).[\cos (\frac{\pi }{8} + \frac{{3\pi }}{8}) - \cos (\frac{{3\pi }}{8} - \frac{\pi }{8})]\\
= - \frac{1}{8}.(\cos \frac{\pi }{2} - \cos \frac{\pi }{4})\\
= \frac{{ - 1}}{8}.(0 - \frac{{\sqrt 2 }}{2})\\
= \frac{{\sqrt 2 }}{{16}}
\end{array}\)
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