Giá trị đúng của biểu thức M=cos pi 15cos 2pi 15cos 3pi 15cos 4pi 15co
Câu hỏi
Nhận biếtGiá trị đúng của biểu thức \(M= \cos \frac{ \pi }{15} \cos \frac{2 \pi }{15} \cos \frac{3 \pi }{15} \cos \frac{4 \pi }{15} \cos \frac{5 \pi }{15} \cos \frac{6 \pi }{15} \cos \frac{7 \pi }{15} \) bằng:
Đáp án đúng: D
Lời giải của Tự Học 365
Giải chi tiết:
\(\begin{array}{l}\,\,\,\,\,\,M = \cos \dfrac{\pi }{{15}}\cos \dfrac{{2\pi }}{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{4\pi }}{{15}}\cos \dfrac{{5\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\cos \dfrac{{7\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{\pi }{{15}}M = 2\sin \dfrac{\pi }{{15}}\cos \dfrac{\pi }{{15}}\cos \dfrac{{2\pi }}{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{4\pi }}{{15}}\cos \dfrac{{5\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\cos \dfrac{{7\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{\pi }{{15}}M = \sin \dfrac{{2\pi }}{{15}}\cos \dfrac{{2\pi }}{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{4\pi }}{{15}}\cos \dfrac{{5\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\cos \dfrac{{7\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{\pi }{{15}}M = \dfrac{1}{2}\sin \dfrac{{4\pi }}{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{4\pi }}{{15}}\cos \dfrac{{5\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\cos \dfrac{{7\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{\pi }{{15}}M = \dfrac{1}{4}\sin \dfrac{{8\pi }}{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{5\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\cos \left( {\pi - \dfrac{{7\pi }}{{15}}} \right)\\ \Leftrightarrow 2\sin \dfrac{\pi }{{15}}M = - \dfrac{1}{4}\sin \dfrac{{8\pi }}{{15}}cos\dfrac{{8\pi }}{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{\pi }{3}\cos \dfrac{{6\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{\pi }{{15}}M = - \dfrac{1}{{16}}\sin \dfrac{{16\pi }}{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{\pi }{{15}}M = - \dfrac{1}{{16}}\sin \left( {\pi + \dfrac{\pi }{{15}}} \right)\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{\pi }{{15}}M = \dfrac{1}{{16}}\sin \dfrac{\pi }{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\\ \Leftrightarrow M = \dfrac{1}{{32}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{{3\pi }}{{15}}M = \dfrac{1}{{32}}2\sin \dfrac{{3\pi }}{{15}}\cos \dfrac{{3\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{{3\pi }}{{15}}M = \dfrac{1}{{32}}\sin \dfrac{{6\pi }}{{15}}\cos \dfrac{{6\pi }}{{15}}\\ \Leftrightarrow 2\sin \dfrac{{3\pi }}{{15}}M = \dfrac{1}{{64}}\sin \dfrac{{12\pi }}{{15}} \Leftrightarrow 2\sin \dfrac{{3\pi }}{{15}}M = \dfrac{1}{{64}}\sin \left( {\pi - \dfrac{{3\pi }}{{15}}} \right)\\ \Leftrightarrow 2\sin \dfrac{{3\pi }}{{15}}M = \dfrac{1}{{64}}\sin \dfrac{{3\pi }}{{15}} \Leftrightarrow M = \dfrac{1}{{128}}\end{array}\)
Chọn D.
