We want to figure out two things: the slope of the curve and the position of the point. Therefore, we can find a line that crosses perpendicular to that point.

The point is easy. Plugging in #theta = pi/2#, we get #f(pi/2) = cos(0) + pi/2 sin(2pi) = 1#, i.e. a point at a radius of 1 and an angle of #pi/2#, which is the point (0, 1). Therefore, this is our y-intercept.

Next, we find the slope. We can do this by calculating #(df)/(d theta)# and using it in the following:

#y = rsintheta -> dy = sintheta dr + rcostheta d theta = y/r dr + xd theta#

#x = rcostheta -> dx = costheta dr - rsintheta d theta = x/r dr - yd theta #

In other words,

#(dy)/(dx) = (dy)/(d theta) / (dx)/(d theta) = (y/r cdot (dr)/(d theta) - x)/(x/r cdot (dr)/(d theta) - y)#

Since our point is at (0, 1), this simplifies significantly to

#(dy)/(dx) =((dr)/(d theta))/(- 1) = -(dr)/(d theta)#

Now what is #(dr)/(d theta)#? We calculate using the formula given:

#f(theta) = cos(pi - 2 theta) + theta sin(5theta - pi/2)#

#f'(theta) = 2 sin(pi - 2 theta) + sin(5theta - pi/2) + 5thetacos(5theta - pi/2) #

#f'(pi/2) = 2 sin(0) + sin(2pi) + 5 cdot pi/2 cdot cos(2pi) = (5pi)/2 #

Therefore, the slope of the curve (in Cartesian coordinates) is #-(5pi)/2#.

Orthogonal slopes are determined using the negative reciprocal, i.e. the slope for the line is #2/(5pi)#. From the slope and the intercept, we get the equation

#y = 2/(5pi)x + 1#