Question

Вычислить : $cos22^{0} 30; tg22^{0} 30$

Answer 1

$1)Cos22^{0}30^{'}=\sqrt{\frac{1+Cos45^{0}}{2}}=\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}=\sqrt{\frac{2+\sqrt{2}}{4}}=\frac{\sqrt{2+\sqrt{2} }}{2}\\\\2)tg22^{0}30^{'}=\frac{Sin45^{0}}{1+Cos45^{0}}=\frac{\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}=\frac{\sqrt{2}*2}{2*(2+\sqrt{2})}=\frac{\sqrt{2}}{\sqrt{2}+1)}=\frac{\sqrt{2}*(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=2-\sqrt{2}$