Question

Помогите со вторым заданием

Answer 1

$sinalpha=frac{24}{25},;0 textless alpha textless fracpi2\sin^2alpha+cos^2alpha=1Rightarrowcosalpha=sqrt{1-sin^2alpha}\cos^2alpha=sqrt{1-frac{576}{625}}=sqrt{frac{49}{625}}=pmfrac7{25}\0 textless alpha textless fracpi2Rightarrowcosalpha textgreater 0\cosalpha=frac7{25}\sin2alpha=2sinalphacosalpha=2cdotfrac{24}{25}cdotfrac7{25}=frac{336}{625}\ctg2alpha=frac{cos2alpha}{sin2alpha}\cos2alpha=cos^2alpha-sin^2alpha=frac{49}{625}-frac{576}{625}=-frac{527}{625}$
$ctg2alpha=-frac{527}{625}:frac{24}{25}=-frac{527}{625}cdotfrac{25}{24}=-frac{527}{25}frac1{24}=-frac{527}{600}$