Question

Найдите sin$alpha$, если $ctgalpha=frac{3}{4} ;180а<alpha<270а$

Answer 1

$1+ctg^2alpha=frac1{sin^2alpha}\ 1+left(frac34right)^2=frac1{sin^2alpha}\ 1+frac9{16}=frac1{sin^2alpha}\ frac{25}{16}=frac1{sin^2alpha}\ {sin^2alpha}=frac{16}{25}\ sinalpha=pmfrac45\ 180^o<alpha<270^oRightarrowsinalpha=-frac45$