If cos(x) = (1/4), where x lies in quadrant 4, how do you find sin(x - pi/6)?

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Answer 1 · 2016-06-17T16:41:05

#-(3sqrt(5)+1)/8#

in quadrant 4, sin(x)<0

so you can find
#sinx=-sqrt(1-cos^2x)#

#=-sqrt(1-1/16)#

#=-sqrt(15/16)#

#=-sqrt(15)/4#

Then
#sin(x-pi/6)=sinxcos(pi/6)-cosxsin(pi/6)#

#sinx*sqrt(3)/2-cosx*1/2#

and, by substituting sinx and cosx values, you have:

#-sqrt(15)/4*sqrt(3)/2-1/4*1/2#

#-sqrt(45)/8-1/8#

#-(3sqrt(5)+1)/8#