To find the equation of the line, we need a #color(red)(poi nt)#, and we need a #color(blue)(slope)#.
First, let's find the #color(red)(poi nt)#.
Plug in #pi/4# into #f(theta)#.
#f(pi/4)=2sin(3(pi/4)+pi/3)-2(pi/4)(cos(pi/4))#
#f(pi/4)=(sqrt2(2-2sqrt3-pi))/4#
We have the polar coordinate #(pi/4, (sqrt2(2-2sqrt3-pi))/4)#.
We have to convert this to Cartesian form. Using the formulas:
#x = rcostheta#
#y = rcostheta#
We find that the Cartesian coordinate is #color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4)#.
Good! We have found the #color(red)(poi nt)#.
Now, we need to find the #color(blue)(slope)#.
The formula for the tangent line of a polar function is:
#dy/dx = ((dr)/(d theta)sintheta+rcostheta)/((dr)/(d theta)costheta-rsintheta)#
If you notice, #sintheta# and #costheta# will be the same in this case, because #sin(pi/4)# and #cos(pi/4)# are the same, so for this case, our #dy/dx# is simply:
#dy/dx = ((dr)/(d theta)+r)/((dr)/(d theta)-r)#
We solved #r# already previously, which was #(sqrt2(2-2sqrt3-pi))/4#.
Now, we need to find #(dr)/(d theta)#.
#f'(theta)=(dr)/(d theta)=6cos(3theta+pi/3)+2thetasintheta-2costheta#
Now, we plug in #pi/4# and get:
#f'(pi/4)=(dr)/(d theta)=(sqrt2(-10-6sqrt3+pi))/4#
Going back to the tangent line formula:
#dy/dx = ((sqrt2(-10-6sqrt3+pi))/4+(sqrt2(2-2sqrt3-pi))/4)/((sqrt2(-10-6sqrt3+pi))/4-(sqrt2(2-2sqrt3-pi))/4)#
After some tedious simplification, we find this reduces to #color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)#
We have our #color(red)(poi nt)#:
#color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4)#
We have our #color(blue)(slope)#:
#color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)#
All that is left is to plug in these values into the point slope formula:
#(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#
We get:
#y - ((2-2sqrt3-pi)/4) = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4))#
So the equation is:
#y = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4)) +((2-2sqrt3-pi)/4)#
