Question

Народ помогите решить номер 2 по алгебре, а за помощь дам 100 балов

Answer 1

Відповідь:

Пояснення:

Answer 2

$\displaystyle\bf\\Cos(-150^\circ)=Cos150^\circ=Cos(180^\circ-30^\circ)=-Cos30^\circ=-\frac{\sqrt{3} }{2} \\\\\\Cos210^\circ=Cos(180^\circ+30^\circ)=-Cos30^\circ=-\frac{\sqrt{3} }{2} \\\\\\Sin315^\circ=Sin(270^\circ+45^\circ)=-Cos45^\circ=-\frac{\sqrt{2} }{2} \\\\\\Sin\Big(-\frac{5\pi }{3} \Big)=-Sin\Big(2\pi -\frac{\pi }{3} \Big)=-Sin\Big(-\frac{\pi }{3} \Big)=Sin\frac{\pi }{3} =\frac{\sqrt{3} }{2}$

$\displaystyle\bf\\\frac{Cos64^\circ Cos4^\circ-Cos86^\circ Cos26^\circ}{Cos71^\circ Cos41^\circ-Cos49^\circ Cos19^\circ} =\\\\\\=\frac{Cos(90^\circ-26^\circ) \cdot Cos4^\circ-Cos(90^\circ-4^\circ)\cdot Cos26^\circ}{Cos(90^\circ-19^\circ)\cdot Cos41^\circ-Cos(90^\circ-41^\circ)\cdot Cos19^\circ} =\\\\\\=\frac{Sin26^\circ Cos4^\circ-Sin4^\circ Cos26^\circ}{Sin19^\circ Cos41^\circ-Sin41^\circ Cos19^\circ} =\frac{Sin(26^\circ-4^\circ)}{Sin(19^\circ-41^\circ)} =$

$\displaystyle\bf\\=\frac{Sin22^\circ}{-Sin22^\circ} =-1$