Question

723. Используя формулы полушария, вычислите: 1) $\tan( \frac{\pi}{8} )$
нужно выполнить​

Answer 1

$\tan(\frac{\pi}{8}) = \frac{\sin(\frac{\pi}{8})}{\cos(\frac{\pi}{8})}$

$\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - \sin^2(x) - \sin^2(x) = 1 - 2\sin^2(x)$

$\sin^2(x) = \frac{1 - \cos(2x)}{2}$

$\sin^2(\frac{\pi}{8}) = \frac{1 - \cos(\frac{\pi}{4})}{2}$

Из геометрии известно, что

$\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$

$\sin^2(\frac{\pi}{8}) = \frac{1 - \frac{\sqrt{2}}{2}}{2} = \frac{2 - \sqrt{2}}{4}$

$0 < \frac{\pi}{8} < \frac{\pi}{2}$

$\sin(\frac{\pi}{8}) 0$

$\sin(\frac{\pi}{8}) = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{ \sqrt{2 - \sqrt{2}} }{2}$

$\cos(2x) = \cos^2(x) - \sin^2(x) = \cos^2(x) - (1 - \cos^2(x)) = 2\cos^2(x) - 1$

$\cos^2(x) = \frac{1 + \cos(2x)}{2}$

$\cos^2(\frac{\pi}{8}) = \frac{1 + \cos(\frac{\pi}{4})}{2} =$

$= \frac{1 + \frac{\sqrt{2}}{2}}{2} = \frac{2 + \sqrt{2}}{4}$

$0 < \frac{\pi}{8} < \frac{\pi}{2}$

$\cos(\frac{\pi}{8}) 0$

$\cos(\frac{\pi}{8}) = \sqrt{\frac{2+\sqrt{2}}{4}} = \frac{\sqrt{2+\sqrt{2}}}{2}$

$\tan(\frac{\pi}{8}) = \frac{\frac{\sqrt{2 - \sqrt{2}}}{2}}{\frac{\sqrt{2+\sqrt{2}}}{2}} =$

$= \sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}} = \sqrt{\frac{(2 - \sqrt{2})^2}{(2 + \sqrt{2})\cdot (2 - \sqrt{2})}} =$

$= \sqrt{\frac{(2 - \sqrt{2})^2}{4 - 2}} = \frac{|2 - \sqrt{2}|}{\sqrt{2}} =$

$= \frac{2 - \sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1$