Question

можете решить хотя бы первое​

Answer 1

$\cos(60 + \alpha ) = \\ = \cos(60) \cos( \alpha ) - \sin(60) \sin( \alpha ) = \\ = \frac{1}{2} \cos( \alpha ) - \frac{ \sqrt{3} }{2} \sin( \alpha )$

$\sin( \alpha ) = \sqrt{1 - { \cos }^{2 } \alpha } \\ \sin( \alpha ) = \sqrt{1 - \frac{3}{4} } = \frac{1}{2} \\$

$\cos(60 \alpha ) = - \frac{1}{2} \times \frac{ \sqrt{3} }{2} - \frac{ \sqrt{3} }{2} \times \frac{1}{2} = - \frac{ \sqrt{3} }{2} \\$

2.

$\frac{ \sin( \frac{3\pi}{20} ) \times \cos(2\pi + \frac{\pi}{10} ) + \cos( \frac{3\pi}{20} ) \sin( \frac{\pi}{10} ) }{ \cos( \frac{7\pi}{24} ) \cos( \frac{\pi}{8} ) + \sin(\pi - \frac{\pi}{8} ) \sin( \frac{7\pi}{24} ) } = \\ = \frac{ \sin( \frac{3\pi}{20}) \cos(\frac{\pi}{10} ) + \cos( \frac{3\pi}{20} ) \sin( \frac{\pi}{10} ) }{ \cos( \frac{7\pi}{24} ) \cos( \frac{\pi}{8} ) + \sin( \frac{\pi}{8} ) \sin( \frac{7\pi}{24} ) } = \\ = \frac{ \sin( \frac{3\pi}{20} + \frac{\pi}{10} ) }{ \cos( \frac{7\pi}{24} - \frac{\pi}{8} ) } = \frac{ \sin( \frac{\pi}{4} ) }{ \cos( \frac{\pi}{6} ) } = \\ = \frac{ \sqrt{2} }{2} \times \frac{2}{ \sqrt{3} } = \frac{ \sqrt{2} }{ \sqrt{3} } = \frac{ \sqrt{6} }{3}$