Question

Найти tg a, если sin a = -4√41/41 , [π; 3π/2]

Answer 1

$sin a = -frac{4sqrt{41}}{41}\cosa=бsqrt{1-sin^2a}=бsqrt{1- (-frac{4sqrt{41}}{41})^2}=бsqrt{1- frac{16}{41}}=бfrac{5}{sqrt{41}}=\=бfrac{5sqrt{41}}{41}$

т.к. a в 3 четв. и cos в 3 чет отриц то

$cosa=-frac{5sqrt{41}}{41}\tga=frac{ -frac{4sqrt{41}}{41}}{-frac{5sqrt{41}}{41}}=frac{4}{5}$