Cho \(\cot \frac{\pi }{{14}} = a\). Khi đó giá trị biểu thức \(K = \sin }}{7} + \sin }}{7} + \sin }}{7}\) là:

* Hướng dẫn giải

\(\begin{array}{l} \tan \frac{\pi }{{14}} = \frac{1}{a} \Rightarrow \sin \frac{\pi }{7} = \frac{{\frac{2}{a}}}{{1 + \frac{1}{{{a^2}}}}} = \frac{{2a}}{{{a^2} + 1}} = \sin \frac{{6\pi }}{7};\cos \frac{\pi }{7} = \frac{{1 - \frac{1}{{{a^2}}}}}{{1 + \frac{1}{{{a^2}}}}} = \frac{{{a^2} - 1}}{{{a^2} + 1}}\\ \Rightarrow \sin \frac{{2\pi }}{7} = 2\sin \frac{\pi }{7}\cos \frac{\pi }{7} = \frac{{4a\left( {{a^2} - 1} \right)}}{{{{\left( {{a^2} + 1} \right)}^2}}}\\ \sin \frac{{4\pi }}{7} = \sin \frac{{3\pi }}{7} = 3\sin \frac{\pi }{7} - 4{\sin ^3}\frac{\pi }{7}\\ \Rightarrow \sin \frac{{4\pi }}{7} + \sin \frac{{6\pi }}{7} = 4\sin \frac{\pi }{7} - 4{\sin ^3}\frac{\pi }{7} = 4\sin \frac{\pi }{7}\left( {1 - {{\sin }^2}\frac{\pi }{7}} \right)\\ = 4\sin \frac{\pi }{7}.{\cos ^2}\frac{\pi }{7} = 2\sin \frac{{2\pi }}{7}\cos \frac{\pi }{7}\\ K = \sin \frac{{2\pi }}{7}\left( {2\cos \frac{\pi }{7} + 1} \right) = \frac{{4a\left( {{a^2} - 1} \right)}}{{{{\left( {{a^2} + 1} \right)}^2}}}.\frac{{2{a^2} - 2 + {a^2} + 1}}{{{a^2} + 1}} = \frac{{4a\left( {{a^2} - 1} \right)\left( {3{a^2} - 1} \right)}}{{{{\left( {{a^2} + 1} \right)}^3}}} \end{array}\)