First of all this simplifies to
#r = 2 sin theta +sin 2theta#
the tangent vector is the velocity vector so we differentiate wrt time as a parameter
#vec r = (2 sin theta +sin 2theta) hat r#
#vec r' = vec v #
by product rule
# = d/dt (2 sin theta +sin 2theta) hat r + (2 sin theta +sin2 theta) d/dt hat r#
where #d/dt hat r = d/dt ((cos theta),(sin theta)) = ((-sin theta),(cos theta)) dot theta = dot theta hat theta#
#implies (2 cos theta + 2 cos 2theta) \ dot theta \ hat r + (2 sin theta +sin 2theta) \ dot theta \ hat theta#
dropping the #dot theta# scalar...
#= (2 cos theta + 2 cos 2theta) ((cos theta),(sin theta))+ (2 sin theta +sin 2theta) ((-sin theta),(cos theta)) #
#= ((2 cos^2 theta + 2 cos 2 theta cos theta - 2 sin^2 theta - sin theta sin 2 theta),(2 cos theta sin theta + 2 cos 2 theta sin theta + 2 sin theta cos theta + sin 2 theta cos theta))#
#= ((1 + 0 - 1 - 1/sqrt 2),(1 + 0 + 1+ 1/sqrt 2))#
#= (( - 1),(2sqrt 2+ 1))#
the slope is therefore
#m = -(1+ 2sqrt 2)#
#r (pi/4) = 1 + sqrt 2#
#x = r cos theta = 1/sqrt 2+ 1 #
#y = r sin theta = 1/sqrt 2+ 1 #
So we have
#(y - 1/sqrt 2 - 1)/(x - 1/sqrt 2 - 1) = -(1+ 2sqrt 2)#
Or
#y = -(1+ 2sqrt 2)(x - 1/sqrt 2 - 1) + 1 + 1/sqrt 2#
that could be further simplified of course
