let t=reference angle of theta=asin(3/5) in Q1
sin(theta)=3/5 in Q2 =>
cos(theta)=cos(pi-t)=-cos(t)=-4/5
(a) sin(pi/6)=sin(30)=1/2, cos(pi/6)=cos(30)=sqrt(3)/2
sin(theta+pi/6)
=sin(theta)cos(pi/6)+cos(theta)sin(pi/6)
=(3/5)(sqrt(3)/2)+(-4/5)(1/2)
=(3sqrt(3)-4)/10 [ after a few simplifications]
(b)
cos(5pi/3)=cos(pi+2pi/3)=cos(pi)cos(2pi/3)-sin(pi)sin(2pi/3)=-1(-1/2)-0=1/2
sin(5pi/3)=sin(pi+2pi/3)=sin(pi)cos(2pi/3)+cos(pi)sin(2pi/3)=0+(-1)(sqrt(3)/2)=-sqrt(3)/2
cos(5pi/3-theta)
=cos(5pi/3)cos(theta)+sin(5pi/3)sin(theta)
=(1/2)(-4/5)+(-sqrt(3)/2)(3/5)
=-(3sqrt(3)+4)/10
(c)
cos(2theta)
=cos^2(theta)-sin^2(theta)
=(-4/5)^2-(3/5)^2
=7/25
(d)
csc(pi/2-theta)
=1/sin(pi/2-theta)
=1/(sin(pi/2)cos(theta)-cos(pi/2)sin(theta))
=1/[(1)(-4/5)-0(3/5)]
=1/[-4/5]
=-5/4