Question

Найдите cosx,если sinx=-0,8     -pi/2<x<0

Answer 1

$sin^2x+cos^2x=1 to cosx=pmsqrt{1-sin^2x}= pmsqrt{1-(-0.8)^2}= \ pmsqrt{1-0.64}= pmsqrt{0.36}=pm0.6; x in [-pi/2;0] to cosx=0.6$