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

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

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