Question

помогите плиизз срочно 40 баллов​

Answer 1

Ответ:

a

$\sin( \frac{17\pi}{60} ) \cos( \frac{\pi}{20} ) + \cos( \frac{17\pi}{60} ) \sin( \frac{\pi}{20} ) = \\ = \sin( \frac{17\pi}{60} + \frac{\pi}{20} ) = \sin( \frac{17\pi + 3\pi}{60} ) = \sin( \frac{20\pi}{60} ) = \\ = \sin( \frac{\pi}{3} ) = \frac{ \sqrt{3} }{2}$

б

$\cos( \frac{\pi}{10} ) \cos( \frac{7\pi}{20} ) + \sin( \frac{\pi}{10} ) \sin( \frac{7\pi}{20} ) = \\ = \cos( \frac{\pi}{10} - \frac{7\pi}{20} ) = \cos( - \frac{5\pi}{20} ) = \cos( - \frac{\pi}{4} ) = \frac{ \sqrt{2} }{2}$

в

$\sin( \frac{139\pi}{90} ) \cos( \frac{17\pi}{45} ) - \cos( \frac{139\pi}{90} ) \sin( \frac{17\pi}{45} ) = \\ = \sin( \frac{139\pi}{90} - \frac{17\pi}{45} ) = \sin( \frac{139\pi - 34\pi}{90} ) = \\ = \sin( \frac{105\pi}{90} ) = \sin( \frac{7\pi}{6} ) = \sin(\pi + \frac{\pi}{6} ) = \\ = - \sin( \frac{\pi}{6} ) = - \frac{1}{2}$

г

$\cos( \frac{8\pi}{45} ) \cos( \frac{29\pi}{90} ) - \sin( \frac{8\pi}{45} ) \sin( \frac{29\pi}{90} ) = \\ = \cos( \frac{8\pi}{45} + \frac{29\pi}{90} ) = \cos( \frac{16\pi + 29\pi}{90} ) = \\ = \cos( \frac{45\pi}{90} ) = \cos( \frac{\pi}{2} ) = 0$