sin-pi-7-cos-pi-14-tan-3pi-14-2cos-pi-7-1-

Question Number 135604 by liberty last updated on 14/Mar/21

\frac{\sin (π / 7) . \cos (π / 14)}{\tan (3 π / 14) (2 \cos (π / 7) - 1)} = ?

Answered by EDWIN88 last updated on 14/Mar/21

Remark {\begin{cases} \sin 3 x = 3 \sin x - 4 \sin^{3} x \\ \cos 3 x = 4 \cos^{3} x - 3 \cos x \end{cases}

let x = \frac{π}{14} \Rightarrow \frac{\sin (π / 7) \cos (π / 14)}{\tan (3 π / 14) (2 \cos (π / 7) - 1)}

= \frac{\sin 2 x \cos x}{\tan 3 x (2 \cos 2 x - 1)} = \frac{2 \sin x \cos^{2} x \cos 3 x}{\sin 3 x ({4 \cos}^{2} x - 3)}

= \frac{2 \sin x \cos^{2} x (4 \cos^{3} x - 3 \cos x)}{(3 \sin x - 4 \sin^{3} x) (4 \cos^{2} x - 3)}

= \frac{2 \cos^{3} x (4 \cos^{2} x - 3)}{(3 - 4 \sin^{2} x) (4 \cos^{2} x - 3)}

= \frac{2 \cos^{3} x}{4 \cos^{2} x - 1}

according \cos 2 x = \cos (π / 7) = Q is one

of the roots of {8 Q}^{3} - {4 Q}^{2} - 4 Q + 1 = 0

therefore 8 {(1 + Q)}^{3} = 7 {(1 + 2 Q)}^{2}

\frac{7}{8} = \frac{{(1 + \cos 2 x)}^{3}}{{(1 + 2 \cos 2 x)}^{2}} = \frac{{(2 \cos^{2} x)}^{3}}{{(4 \cos^{2} x - 1)}^{2}}

\frac{7}{8} = \frac{8 \cos^{6} x}{{(4 \cos^{2} x - 1)}^{2}}; \frac{7}{16} = \frac{{(2 \cos^{3} x)}^{2}}{{(4 \cos^{2} x - 1)}^{2}}

∴ \frac{2 \cos^{3} x}{4 \cos^{2} x - 1} = \frac{\sqrt{7}}{4}

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