Calculating nth Derivative of e^ax cos(bx+c)

[AAQIB IQBAL]

My question is about the nth derivative of e^ax cos(bx+c). Though i can calculate it easily but i am confused at one point.
When we calculate the first derivative we put a = r.cos(theta), b = r.sin(theta) (every thing is ok till here)
My confusion starts when we use (theta) = tan^-1(b/a) [tan inverse]
the reason for my confusion can be understood by:
suppose we have a = -1, b = 1
we put a = sqrt{2}*cos(3*pi/4)
b = sqrt{2}*sin(3*pi/4)
but the tan^-1(b/a) = tan^-1(-1) = -pi/4
but our theta is 3*pi/4
according to this theta our a will be 1 and b will be -1 which is different from our values of a and b

 

[mathman]

tan is periodic with period π, so arctan (-1) = -π/4 + kπ, for any integer k.