Дано: sinA= -21/29, 3n/2 < A < 2n Найти: CosA, tgA, ctgA!

Question

Дано: sinA= -21/29, 3n/2 < A < 2n Найти: CosA, tgA, ctgA!

Answer 1

cosL=√(1-〖sin〗^2 )=√(1-441/841)=√(400/841)=20/29 tgL=sinL/cosL=-21/29×29/20=-21/20 ctgL=-20/21