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

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Answer 1 · 2016-06-09T23:01:17

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

with

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

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

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)) #