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Question Number 186741 by cortano12 last updated on 09/Feb/23

$$\cos \left(\frac{\pi }{18}\right).\cos \left(\frac{3\pi }{18}\right).\cos \left(\frac{5\pi }{18}\right).\cos \left(\frac{7\pi }{18}\right)=?$$

Answered by pablo1234523 last updated on 09/Feb/23

$$\frac{1}{2}[\cos \frac{8\pi }{18}+\cos \frac{6\pi }{18}]\cdot \frac{1}{2}[\cos \frac{8\pi }{18}+\cos \frac{2\pi }{18}]$$$$=\frac{1}{4}({\cos }^2\frac{4\pi }{9}+\cos \frac{4\pi }{9}\cos \frac{\pi }{9}+\cos \frac{3\pi }{9}\cos \frac{4\pi }{9}+\cos \frac{3\pi }{9}\cos \frac{\pi }{9})$$$$=\frac{1}{8}(1+\cos \frac{8\pi }{9}+\cos \frac{5\pi }{9}+\cos \frac{3\pi }{9}+\cos \frac{7\pi }{9}+\cos \frac{\pi }{9}+\cos \frac{4\pi }{9}+\cos \frac{2\pi }{9})$$$$=\frac{1}{8}(1-\cos \frac{\pi }{9}-\cos \frac{4\pi }{9}+\cos \frac{3\pi }{9}-\cos \frac{2\pi }{9}+\cos \frac{\pi }{9}+\cos \frac{4\pi }{9}+\cos \frac{2\pi }{9})$$$$=\frac{1}{8}(1+\cos \frac{3\pi }{9})$$$$=\frac{1}{8}(1+\cos \frac{\pi }{3})$$$$=\frac{1}{8}(1+\frac{1}{2})=\frac{1}{8}\left(\frac{3}{2}\right)$$$$=\frac{3}{16}$$

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