Determining the Slope and Tangent Lines for a Polar Curve
Key Questions

By converting into parametric equations,
#{(x(theta)=r(theta)cos theta=cos2theta cos theta), (y(theta)=r(theta)sin theta=cos2theta sin theta):}# By Product Rule,
#x'(theta)=sin2theta cos thetacos2theta sin theta# #x'(pi/2)=sin(pi)cos(pi/2)cos(pi)sin(pi/2)=1# #y'(theta)=sin2thetasin theta+cos2theta cos theta# #y'(pi/2)=sin(pi)sin(pi/2)+cos(pi)cos(pi/2)=0# So, the slope
#m# of the curve can be found by#m={dy}/{dx}_{theta=pi/2}={y'(pi/2)}/{x'(pi/2)}=0/1=0# I hope that this was helpful.

Answer:
The equation of the tangent line is
#y7/2={11sqrt{3}}/5(x{7sqrt{3}}/2)# Explanation:
In order to find the equation of a line, we need two pieces of information:
#{(1. "Point: " (x_1,y_1)),(2. "Slope: " m):}# Let us find
#(x_1,y_1)# .Since
#{(x(theta)=rcos theta=(3+8sin theta)cos theta),(y(theta)=rsin theta=(3+8sin theta)sin theta):}# ,#x_1=x(pi/6)=[3+8sin(pi/6)]cos(pi/6)={7sqrt{3}}/2# #y_1=y(pi/6)=[3+8sin(pi/6)]sin(pi/6)=7/2# Now, let us find
#m# .By differentiating with respect to theta#,
#{dx}/{d theta}=8cos theta cdot cos theta+(3+8sin theta)cdot(sin theta)# #=8(cos^2thetasin^2theta)3sin theta# #=8cos(2theta)3sin theta# #{dy}/{d theta}=8cos theta cdot sin theta+(3 + 8sin theta)cdot cos theta# #=8(2sin theta cos theta)+3cos theta# #=8sin(2theta)+3cos theta# So,
#{dy}/{dx}={{dy}/{d theta}}/{{dx}/{d theta}}={8sin(2theta)+3cos theta}/{8cos(2theta)3sin theta}# Now, we can find
#m# .#m={dy}/{dx}_{theta=pi/6} ={8({sqrt{3}}/2)+3({sqrt{3}}/2)}/{8(1/2)3(1/2)}={11sqrt{3}}/5# By PointSlope Form:
#yy_1=m(xx_1)# ,#y7/2={11sqrt{3}}/5(x{7sqrt{3}}/2)# 
A polar equation of the form
#r=r(theta)# can be converted into a pair of parametric equations#{(x(theta)=r(theta)cos theta),(y(theta)=r(theta)sin theta):}# .The slope
#m# of the tangent line at#theta=theta_0# can be expressed as#m={dy}/{dx}_{theta=theta_0}={{dy}/{d theta}_{theta=theta_0}}/{{dx}/{d theta}_{theta=theta_0}}={y'(theta_0)}/{x'(theta_0)}# .I hope that this was helpful.