Question

Найдите синус косинус тангенс котангенс 22³30

Answer 1

$22^030'=22,5^0=frac{45^0}{2};\ a=22,5^0;\ sin22^030', cos22^030', tg22^030'-?\ left { {{sin2a=2sin acos a} atop {cos2a=cos^2a-sin^2a=2cos^2a-1=1-2sin^2a}} right.\ sin^2a =frac{1-cos2a}{2};\ cos^2a=frac{1+cos2a}{2};\ sin a=pmsqrt{frac{1-cos2a}{2}};\ cos a=pmsqrt{frac{1+cos2a}{2}};\ a=22,5^0==>sin a, cos a, tga>0;\ sin a=sqrt{frac{1-cos2a}{2}};\ cos a=sqrt{frac{1+cos2a}{2}};\ tg a=frac{sin a}{cos a}=sqrt{frac{1-cos2a}{1+cos2a}};$
$cos45^0=frac{1}{sqrt2}=frac{sqrt{2}}{2}\ sin22^030'=sqrt{frac{1-cos45^0}{2}}=sqrt{frac{1-frac{sqrt2}{2}}{2}}=sqrt{frac{2-sqrt2}{4}}=frac{sqrt{2-sqrt{2}}}{2};\ cos22^030'=sqrt{frac{1+cos45^0}{2}}=sqrt{frac{1+frac{sqrt2}{2}}{2}}=sqrt{frac{2+sqrt2}{4}}=frac{sqrt{2+sqrt{2}}}{2};\ tg22^030'=frac{sin22^030'}{cos22^030'}=sqrt{frac{1-cos45^0}{1+cos45^0}}=sqrt{frac{1-frac{sqrt2}{2}}{1+frac{sqrt2}{2}}}=$
$=\ =sqrt{frac{1-frac{sqrt{2}}{2}}{1+frac{sqrt{2}}{2}}}=sqrt{frac{frac{2-sqrt{2}}{2}}{frac{2+sqrt{2}}{2}}}=sqrt{frac{2-sqrt2}{2+sqrt2}}$


Ответ:
$sin22^030'=frac{sqrt{2-sqrt{2}}}{2};\ cos22^030'=frac{sqrt{2+sqrt{2}}}{2};\ tg22^030'=frac{sqrt{2-sqrt{2}}}{sqrt{2+sqrt{2}}};$