polar form of decagon by using r = cos( 1+sqrt(theta))

[apple2357]

In this article ( http://www.foster77.co.uk/ATM-MT224-10-11.pdf)

It states you can generate a decagon by using r = cos( 1+sqrt(theta)) as below.
When i try this equation in graph plotter i dont get it. What am i doing wrong?

 

[Dr.Peterson]

In this article ( http://www.foster77.co.uk/ATM-MT224-10-11.pdf)

It states you can generate a decagon by using r = cos( 1+sqrt(theta)) as below.
When i try this equation in graph plotter i dont get it. What am i doing wrong?

View attachment 10660

It doesn't make much sense to me. The page says at one point,

Is there anything about the curve that you think
is not true to the mathematical form of the
equation but is an artefact of the way the
software works? Why? Would the equation(s)
display the same on a graphical calculator or on
different graph-drawing software?
​

I suspect that may be the issue here; they are using some particular software that does something odd; or else they are deliberately doing something more than what the equation shows, and not telling us. (Some of the features of your own graph are probably overflow artifacts.)

 

[apple2357]

Your suspicion is right

He has made them look straight by fiddling with the step length when plotting.
This can be done on Autograph too.