# Revision history [back]

This plot :

var('r,l,m')
ex = l/sin(m/2/r)-r
plot(ex.subs(l=36/6, m=10), (r, -5, 5), ymin=-5, ymax=5, detect_poles=True, plot_points=1000)


suggests that ex has no (real) roots around 0. Proving it may require a non-trivial analytical exploration, possibly requiring the non-trivial use of non-trivial tools...

However, trying the suggested numerical solution :

print("Value of ex at suggested root r=%f is %f"%(3.68357, ex.subs([l==36/5,m==10]).subs(r==3.68357)))
Value of ex at suggested root r=3.683570 is 3.683570


suggests a typo in the question : at this value, ex approximately equals r. Which suggests thar this value is (ano approximation of) a root of ex-r, which can be found more elegantly by :

print("Root of ex-r : %f"%find_root(ex.subs(l=36/5, m=10)-r, 2, 5))
Root of ex-r : 3.683570


See plot and computations in Sagecell

This plot :

var('r,l,m')
ex = l/sin(m/2/r)-r
plot(ex.subs(l=36/6, m=10), (r, -5, 5), ymin=-5, ymax=5, detect_poles=True, plot_points=1000)


suggests that ex has no (real) roots around 0. Proving it may require a non-trivial analytical exploration, possibly requiring the non-trivial use of non-trivial tools...

However, trying the suggested numerical solution :

print("Value of ex at suggested root r=%f is %f"%(3.68357, ex.subs([l==36/5,m==10]).subs(r==3.68357)))
Value of ex at suggested root r=3.683570 is 3.683570


suggests a typo in the question : at this value, ex approximately equals r. Which suggests thar that this value is (ano approximation of) a root of ex-r, which can be found more elegantly by :

print("Root of ex-r : %f"%find_root(ex.subs(l=36/5, m=10)-r, 2, 5))
Root of ex-r : 3.683570


See plot and computations in Sagecell