Question

Алгебра 24.3
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Answer 1

$\displaystyle\bf\\1)\\\\Cos\frac{8\pi }{15} \cdot Cos\frac{\pi }{5}+Sin\frac{8\pi }{15} \cdot Sin\frac{\pi }{5} =Cos\Big(\frac{8\pi }{15} -\frac{\pi }{5} \Big)=Cos\frac{5\pi }{15} =\\\\\\=Cos\frac{\pi }{3} =\frac{1}{2} \\\\\\2)\\\\Cos\frac{1}{10}\pi \cdot Cos\frac{2 }{5}\pi -Sin\frac{1}{10} \pi \cdot Sin\frac{2}{5} \pi =Cos\Big(\frac{\pi }{10} +\frac{2\pi }{5} \Big)=Cos\frac{5\pi }{10} =\\\\\\=Cos\frac{\pi }{2} =0$

$\displaystyle\bf\\3)\\\\Sin\frac{\pi }{6} \cdot Cos\frac{\pi }{12}+Cos\frac{\pi }{6} \cdot Sin\frac{\pi }{12} =Sin\Big(\frac{\pi }{6} +\frac{\pi }{12} \Big)=Sin\frac{3\pi }{12} =\\\\\\=Sin\frac{\pi }{4} =\frac{\sqrt{2} }{2} \\\\\\4)\\\\Sin\frac{1 }{9} \pi \cdot Cos\frac{4}{9}\pi -Cos\frac{1 }{9} \pi \cdot Sin\frac{4 }{9}\pi =Sin\Big(\frac{\pi }{9} -\frac{4\pi }{9} \Big)=-Sin\frac{3\pi }{9} =\\\\\\=-Sin\frac{\pi }{3} =-\frac{\sqrt{3} }{2}$