If sin (x)= 3/8, where x is in quadrant 2, how would you find cos (x)?

1 Answer
Aug 6, 2017

cosx=-sqrt55/8

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