How do you convert r=5.47193-.51793sin(1.5154x+90)r=5.47193.51793sin(1.5154x+90) into cartesian form?

1 Answer
Jun 9, 2016

y=(b Cos(c (k pi + arcTan(y/x))+a)sin(k pi + arcTan(y/x)) y=(bcos(c(kπ+arctan(yx))+a)sin(kπ+arctan(yx))

with

a =5.47193 a=5.47193,
b = -0.51793b=0.51793
c = 1.5154c=1.5154

for {k = 0, pm 1, pm 2,...}

Explanation:

Given r = a + b xx sin(c theta) and using the pass equations

{ (x = r cos(theta)), (y = r sin(theta)) :}

From those equations we obtain

theta = arctan(y/x)+k pi, with {k = 0, pm 1, pm 2,...}
r = y/sin(theta)

and substituting we obtain

y /sin(k pi + arcTan(y/x)) - b Cos(c (k pi + arcTan(y/x))) =a

so

y=(b Cos(c (k pi + arcTan(y/x))+a)sin(k pi + arcTan(y/x))