Here is a reference for the slope of a tangent with polar coordinates

The follow is the equation for #dy/dx# taken from the reference:

#dy/dx = (r'(theta)sin(theta) + r(theta)cos(theta))/(r'(theta)cos(theta) - r(theta)sin(theta)#

The slope, m, of the tangent line is the above evaluted at #theta = (-3pi)/8#

#m = (r'((-3pi)/8)sin((-3pi)/8) + r((-3pi)/8)cos((-3pi)/8))/(r'((-3pi)/8)cos((-3pi)/8) - r((-3pi)/8)sin((-3pi)/8)#

Compute #r'(theta)# and then evaluate it at #theta = (-3pi)/8#

#r'(\theta) = (d[cos(2\theta-\pi/4)/sin(\theta) - sin(\theta-pi/8)])/(d\theta)#

Using the quotient rule we can simplify this:

#r'(\theta)=(sin(\theta)d/(d\theta)[cos(2\theta-\pi/4)]- cos(2\theta-\pi/4)d/(d\theta)[sin(\theta)])/(sin(\theta)^2) -d/(d\theta)[sin(\theta-pi/8)]#

Using the derivatives of trig functions we can simplify this:

#r'(\theta)=(sin(\theta)(-sin(2\theta-\pi/4)\cdot2)- cos(2\theta-\pi/4)(cos(\theta)))/(sin(\theta)^2) -cos(\theta-pi/8)#

Now, evaluate it at #\theta=(-3\pi)/8#: (there is too much simplifying and the answer is irrational, so I am just going to give the decimal version) #r'((-3\pi)/8)~~0.448#

We need the value of #r((-3pi)/8)#:

#r((-3pi)/8) = cos(2((-3pi)/8)-\pi/4)/sin((-3pi)/8) - sin((-3pi)/8 -pi/8)#

#r((-3pi)/8) ~~ 2.082#

Thus far we have

#m = (0.448sin((-3pi)/8) + 2.082cos((-3pi)/8))/(0.448cos((-3pi)/8) - 2.082sin((-3pi)/8)#

Evaluating the sines and cosines:

#m ~~ 0.183#

Compute the #(x_1,y_1)# point:

#x_1 = rcos(theta) = 2.082cos((-3pi)/8) ~~ 0.797#

#y_1 = rsin(theta) = 2.082sin((-3pi)/8) ~~ -1.924#

Use the point-slope form of the equation of a line:

#y = m(x - x_1) + y_1#

The equation is:

#y = 0.183(x - 0.797) - 1.924#