let-Q-1-tan-3pi-8-tan-pi-10-1-tan-pi-8-tan-pi-10-prove-that-Q-1-Q-1-7-3-5-85-38-5-

Question Number 72332 by aliesam last updated on 27/Oct/19

l e t Q = \frac{1 + t a n (\frac{3 π}{8}) . t a n (\frac{π}{10})}{1 - t a n (\frac{π}{8}) . t a n (\frac{π}{10})}

p r o v e t h a t

\frac{Q - 1}{Q + 1} = \sqrt{7 - 3 \sqrt{5} - \sqrt{85 - 38 \sqrt{5}}}

Answered by mind is power last updated on 27/Oct/19

\frac{Q - 1}{Q + 1} = \frac{\tan (\frac{π}{10}) (\tan (\frac{3 π}{8}) + \tan (\frac{π}{8}))}{2 + \tan (\frac{π}{10}) (\tan (\frac{3 π}{8}) - \tan (\frac{π}{8})}

Missing \left or extra \right

\cos (a) \cos (b) = \frac{1}{2} (\cos (a - b) + \cos (a + b))

\Rightarrow tg (a) + tg (b) = \frac{2 \sin (a + b)}{\cos (a + b) + \cos (a - b)}

\Rightarrow \tan (\frac{3 π}{8}) + \tan (\frac{π}{8}) = \frac{2 \sin (\frac{π}{2})}{\cos (\frac{π}{4})} = 2 \sqrt{2}

\tan (\frac{3 π}{8}) - \tan (\frac{π}{8}) = \frac{2 \sin (\frac{π}{4})}{\cos (\frac{π}{4})} = 2

\frac{Q - 1}{Q + 1} = \frac{2 \sqrt{2} \tan (\frac{π}{10})}{2 + 2 \tan (\frac{π}{10})} = \frac{\sqrt{2} \tan (\frac{π}{10})}{1 + \tan (\frac{π}{10})} = \sqrt{\frac{{2 tg}^{2} (\frac{π}{10})}{{(1 + tg (\frac{π}{10}))}^{2}}}

\frac{{tg}^{2} (x)}{{(1 + tg (x))}^{2}} = f (x) = \frac{{2 \sin}^{2} (x)}{(1 + \sin (2 x))} = \frac{1 + \cos (2 x)}{(1 + \sin (2 x)}

x = \frac{π}{10} \Rightarrow f (\frac{π}{10}) = \frac{1 + \cos (\frac{π}{5})}{(1 + \sin (\frac{π}{5}))}

\cos (\frac{2 π}{5}) = \frac{- 1 + \sqrt{5}}{4} = {2 \cos}^{2} (\frac{π}{5}) - 1 = - {2 \sin}^{2} (\frac{π}{5}) + 1

too bee continued

Commented by aliesam last updated on 27/Oct/19

p e r f e c t s i r

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