Question

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

Answer 1

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