от Добромир Глухаров » 23 Мар 2020, 15:39
$cos20^\circ+cos60^\circ=2cos\frac{60^\circ-20^\circ}{2}cos\frac{60^\circ+20^\circ}{2}=2cos20^\circ cos40^\circ$
$A=\frac{1}{2}sin10^\circ(cos20^\circ+cos60^\circ)=\frac{1}{2}sin10^\circ.2cos20^\circ.cos40^\circ=\frac{sin10^\circ cos10^\circ cos20^\circ cos40^\circ}{cos10^\circ}=\\=\frac{\frac{1}{2}sin20^\circ cos20^\circ cos40^\circ}{cos10^\circ}=\frac{\frac{1}{4}sin40^\circ cos40^\circ}{cos10^\circ}=\frac{\frac{1}{8}sin80^\circ}{cos10^\circ}=\frac{1}{8}\cdot\frac{cos(90^\circ-80^\circ)}{cos10^\circ}=\frac{1}{8}\cdot\frac{cos10^\circ}{cos10^\circ}=\frac{1}{8}$