How do you find the other trigonometric functions given #sin(pi/2-x)=sqrt2/2, sinx=-sqrt2/2#?
1 Answer
Aug 21, 2017
Explanation:
#"given "sin(pi/2-x)=sqrt2/2#
#rArrcosx=sqrt2/2" and "sinx=-sqrt2/2#
#rArrx" is in the fourth quadrant"#
#•color(white)(x)secx=1/cosx=1/(sqrt2/2)=2/sqrt2=sqrt2#
#•color(white)(x)cscx=1/sinx=1/(-sqrt2/2)=-2/sqrt2=-sqrt2#
#•color(white)(x)tanx=sinx/cosx=(-sqrt2/2)/(sqrt2/2)=-1#
#•color(white)(x)cotx=1/tanx=-1#
