Tính A = cos 2pi/9.cos 4pi/9.cos 8pi/9. 1/2

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} A\sin \frac{2\pi }{9}=\sin \frac{2\pi }{9}cos\frac{2\pi }{9}cos\frac{4\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{2}.2\sin \frac{2\pi }{9}cos\frac{2\pi }{9}cos\frac{4\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{2}\sin \frac{4\pi }{9}cos\frac{4\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{2}.\frac{1}{2}.2\sin \frac{4\pi }{9}cos\frac{4\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{4}\sin \frac{8\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{4}.\frac{1}{2}.2\sin \frac{8\pi }{9}cos\frac{8\pi }{9} \\ =\frac{1}{8}\sin \frac{16\pi }{9}=\frac{1}{8}\sin \left(2\pi -\frac{2\pi }{9}\right) \\ =-\frac{1}{8}\sin \frac{2\pi }{9} \\ \Rightarrow A=-\frac{1}{8} \end{gathered}$$