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

$$letQ=\frac{1+tan\left(\frac{3\pi }{8}\right).tan\left(\frac{\pi }{10}\right)}{1-tan\left(\frac{\pi }{8}\right).tan\left(\frac{\pi }{10}\right)}$$$$$$$$provethat$$$$$$$$\frac{Q-1}{Q+1}=\sqrt{7-3\sqrt{5}-\sqrt{85-38\sqrt{5}}}$$$$$$

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

$$\frac{\mathrm{Q}-1}{\mathrm{Q}+1}=\frac{\tan \left(\frac{\pi }{10}\right)(\tan \left(\frac{3\pi }{8}\right)+\tan \left(\frac{\pi }{8}\right))}{2+\tan \left(\frac{\pi }{10}\right)(\tan \left(\frac{3\pi }{8}\right)-\tan \left(\frac{\pi }{8}\right)}$$$$\text{Missing left or extra right}$$$$\cos \left(\mathrm{a}\right)\cos \left(\mathrm{b}\right)=\frac{1}{2}(\cos (\mathrm{a}-\mathrm{b})+\cos (\mathrm{a}+\mathrm{b}))$$$$\Rightarrow \mathrm{tg}\left(\mathrm{a}\right)+\mathrm{tg}\left(\mathrm{b}\right)=\frac{2\sin (\mathrm{a}+\mathrm{b})}{\cos (\mathrm{a}+\mathrm{b})+\cos (\mathrm{a}-\mathrm{b})}$$$$\Rightarrow \tan \left(\frac{3\pi }{8}\right)+\tan \left(\frac{\pi }{8}\right)=\frac{2\sin \left(\frac{\pi }{2}\right)}{\cos \left(\frac{\pi }{4}\right)}=2\sqrt{2}$$$$\tan \left(\frac{3\pi }{8}\right)-\tan \left(\frac{\pi }{8}\right)=\frac{2\sin \left(\frac{\pi }{4}\right)}{\cos \left(\frac{\pi }{4}\right)}=2$$$$\frac{\mathrm{Q}-1}{\mathrm{Q}+1}=\frac{2\sqrt{2}\tan \left(\frac{\pi }{10}\right)}{2+2\tan \left(\frac{\pi }{10}\right)}=\frac{\sqrt{2}\tan \left(\frac{\pi }{10}\right)}{1+\tan \left(\frac{\pi }{10}\right)}=\sqrt{\frac{{2\mathrm{tg}}^2\left(\frac{\pi }{10}\right)}{{(1+\mathrm{tg}\left(\frac{\pi }{10}\right))}^2}}$$$$\frac{{\mathrm{tg}}^2\left(\mathrm{x}\right)}{{(1+\mathrm{tg}\left(\mathrm{x}\right))}^2}=\mathrm{f}\left(\mathrm{x}\right)=\frac{{2\sin }^2\left(\mathrm{x}\right)}{(1+\sin \left(2\mathrm{x}\right))}=\frac{1+\cos \left(2\mathrm{x}\right)}{(1+\sin \left(2\mathrm{x}\right)}$$$$\mathrm{x}=\frac{\pi }{10}\Rightarrow \mathrm{f}\left(\frac{\pi }{10}\right)=\frac{1+\cos \left(\frac{\pi }{5}\right)}{(1+\sin \left(\frac{\pi }{5}\right))}$$$$\cos \left(\frac{2\pi }{5}\right)=\frac{-1+\sqrt{5}}{4}={2\cos }^2\left(\frac{\pi }{5}\right)-1=-{2\sin }^2\left(\frac{\pi }{5}\right)+1$$$$\mathrm{too}\mathrm{bee}\mathrm{continued}$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by aliesam last updated on 27/Oct/19

$$perfectsir$$

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