From the reference Tangents with Polar Coordinates , we obtain the equation for #dy/dx#:

#dy/dx = ((dr(theta))/(d theta)sin(theta)+r(theta)cos(theta))/((dr(theta))/(d theta)cos(theta)-r(theta)sin(theta))" [1]"#

The slope, m, of the normal the polar curve will be the negative of the reciprocal of equation [1] evaluated at #theta=pi/3#:

#m = (r(pi/3)sin(pi/3)-(dr(pi/3))/(d theta)cos(pi/3))/((dr(pi/3))/(d theta)sin(pi/3)+r(pi/3)cos(pi/3))" [2]"#

I need to backtrack a bit and explain that:

#r(theta) = f(theta) = sin(theta)-theta/2#

We must compute #(dr(theta))/(d theta)# and evaluate at #theta = pi/3#

#(dr(theta))/(d theta) = cos(theta) - 1/2#

#(dr(pi/3))/(d theta) = cos(pi/3)-1/2 = 0#

This greatly simplifies equation [2]:

#m = (r(pi/3)sin(pi/3)-(0)cos(pi/3))/((0)sin(pi/3)+r(pi/3)cos(pi/3))" [2.1]"#

#m = (r(pi/3)sin(pi/3))/(r(pi/3)cos(pi/3))" [2.2]"#

#m = (sin(pi/3))/(cos(pi/3))" [2.3]"#

#m = tan(pi/3)" [2.4]"#

#m = sqrt3" [2.5]"#

Compute the Cartesian point for #(r(pi/3),pi/3)#:

#x = r(theta)cos(theta)#

#x = (sin(pi/3)-pi/6)cos(pi/3)#

#x = (3sqrt3-pi)/12#

#x = r(theta)cos(theta)#

#y = (sin(pi/3)-pi/6)sin(pi/3)#

#y = (9-sqrt3pi)/12#

Using the point slope form of the equation of a line:

#y = sqrt3(x - (3sqrt3-pi)/12)+(9-sqrt3pi)/12#

This simplifies to:

#y = sqrt3x#