1)Как по формулам приведения найти sin 420градусовВычислите ([latex] \sqrt{200} - \sqrt{32} [/latex]/[latex] \sqrt{2} [/latex]

2) [latex]\frac{\sqrt{200}-\sqrt{32}}{\sqrt{2}}=\frac{\sqrt{2*100}-\sqrt{2*16}}{\sqrt{2}}=[/latex] [latex]=\frac{\sqrt{2*100}-\sqrt{2*16}}{\sqrt{2}}=\frac{\sqrt{2}*\sqrt{100}-\sqrt{2}\sqrt{16}}{\sqrt{2}}=[/latex] [latex]=\frac{\sqrt{2}*10-\sqrt{2}*4}{\sqrt{2}}=\frac{\sqrt{2}*(10-4)}{\sqrt{2}}=10-4=6[/latex] 1) [latex]\sin(420^\circ)=\sin(360^\circ+(420^\circ-360^\circ))=[/latex] По формуле приведения [latex]\sin(\alpha+360^\circ)=\sin\alpha[/latex], получаем [latex]=\sin(420^\circ-360^\circ)=\sin 60^\circ=\frac{\sqrt{3}}{2}[/latex]