1)Как по формулам приведения найти sin 420градусов
Вычислите ( $sqrt{200} - sqrt{32}$ / $sqrt{2}$

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
$sin(alpha+360^circ)=sinalpha$ , получаем

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