If sin (x)= 3/8, where x is in quadrant 2, how would you find cos (x)?
1 Answer
Aug 6, 2017
Explanation:
#•color(white)(x)sin^2x+cos^2x=1#
#rArrcosx=+-sqrt(1-sin^2x)#
#"x is in the second quadrant hence "cosx<0#
#cosx=-sqrt(1-(3/8)^2)#
#color(white)(cosx)=-sqrt(1-9/64)=-sqrt(55/64#
#color(white)(cosx)=-sqrt55/8#
