1)Как по формулам приведения найти sin 420градусовВычислите (/

2)
$\frac{\sqrt{200}-\sqrt{32}}{\sqrt{2}}=\frac{\sqrt{2*100}-\sqrt{2*16}}{\sqrt{2}}=$

$=\frac{\sqrt{2*100}-\sqrt{2*16}}{\sqrt{2}}=\frac{\sqrt{2}*\sqrt{100}-\sqrt{2}\sqrt{16}}{\sqrt{2}}=$

$=\frac{\sqrt{2}*10-\sqrt{2}*4}{\sqrt{2}}=\frac{\sqrt{2}*(10-4)}{\sqrt{2}}=10-4=6$

1)
$\sin(420^\circ)=\sin(360^\circ+(420^\circ-360^\circ))=$
По формуле приведения
$\sin(\alpha+360^\circ)=\sin\alpha$ , получаем

$=\sin(420^\circ-360^\circ)=\sin 60^\circ=\frac{\sqrt{3}}{2}$