Question

Тригонометрия, помогите

Answer 1

Ответ:

$\cos( \beta ) = \frac{ \sqrt{6} }{3}$

$\cos( \frac{\pi}{4} + \beta ) = \cos( \frac{\pi}{4} ) \cos( \beta ) - \sin( \frac{\pi}{4} ) \sin( \beta ) = \frac{ \sqrt{2} }{2} \cos( \beta ) - \frac{ \sqrt{2} }{2} \sin( \beta ) = \frac{ \sqrt{2} }{2} ( \cos( \beta ) - \sin( \beta ))$

$\cos( \frac{\pi}{3} - \beta ) = \cos( \frac{\pi}{3} ) \cos( \beta ) + \sin( \frac{\pi}{3} ) \sin( \beta ) = \frac{1}{2} \cos( \beta ) + \frac{ \sqrt{3} }{2} \sin( \beta )$

$\sin( \beta ) = \sqrt{1 - { \cos( \beta ) }^{2} } = \sqrt{1 - \frac{6}{9} } = \sqrt{ \frac{3}{9} } = \frac{ \sqrt{3} }{3}$

$\cos( \frac{\pi}{4} + \beta ) = \frac{ \sqrt{2} }{2} ( \frac{ \sqrt{6} }{3} - \frac{ \sqrt{3} }{3} ) = \frac{ \sqrt{2}( \sqrt{6} - \sqrt{3}) }{6}$

$\cos( \frac{\pi}{3} - \beta ) = \frac{1}{2} \times \frac{ \sqrt{6} }{3} + \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{3} = \frac{ \sqrt{6} + 3}{6}$