Phương pháp giải:

+) Nhân cả 2 vế với \(2\sin \dfrac{\pi }{{15}}\) và áp dụng công thức \(\sin 2a = 2\sin a\cos a\).

+) Sử dụng tính chất của các góc bù nhau và hơn kém nhau \(\pi \).

Lời 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.