Given: #sec(theta)=3#
#cos(theta) = 1/sec(theta) = 1/3#
#sin(theta) = -sqrt(1-cos^2(theta)) larr# we use the negative because we are told that it is the 4th quadrant.
#sin(theta)=-sqrt(1-(1/3)^2)#
#sin(theta)=-sqrt(9/9-1/9)#
#sin(theta)=-sqrt(8/9)#
#sin(theta)=(-2sqrt(2))/3#
#sin(2theta) = 2sin(theta)cos(theta)#
#sin(2theta) = 2(-2sqrt(2))/3(1/3)#
#sin(2theta) = (-4sqrt(2))/9#
#cos(2theta) = 1-2cos^2(theta)#
#cos(2theta) = 1-2(1/3)^2#
#cos(2theta) = 7/9#
#sin(pi/6+theta) = sin(pi/6)cos(theta)+cos(pi/6)sin(theta)#
#sin(pi/6+theta) = (1/2)(1/3)+(sqrt3/2)((-2sqrt(2))/3)#
#sin(pi/6+theta) = (1-2sqrt(6))/6#
#tan(x/2) = sin(x)/(1+cos(x)#
#tan(x/2) = ((-2sqrt(2))/3)/(1+ 1/3)#
#tan(x/2) = -sqrt(2)/2#
