How do you graph $r = 4 sin 7 θ$ $r=4sin7theta$ ?

Question

2838 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-13T03:15:34

Graph is inserted. See explanation.

For using Socratic graphic utility, use the cartesian equivalent

graph{(x^2+y^2)^4=4(7x^6y-35x^4y^3+21x^2y^5-y^7) [-10, 10, -5, 5]}

Glory to De Moivre.

$sin 7 θ$ $sin 7theta$ is periodic, with period $\frac{2 π}{7}$ $(2pi)/7$ .

So, one rotation $θ \in [0, 2 π]$ $theta in [0, 2pi]$ creates 7 petals.

As $r = 4 sin 7 θ \geq 0, 7 θ \in Q_{1} or Q_{2}$ $r = 4 sin 7theta >= 0 , 7theta in Q_1 or Q_2$ , and this gives

$θ \in [0, \frac{π}{7}]$ $theta in [0, pi/7]$ .

r becomes negative, in the other half of the period, $[\frac{π}{7}, \frac{2}{7} π]$ $[pi/7, 2/7pi]$ .

This happens for every loop.

The graphing algorithm ignores $r < 0$ $r < 0$ .

I take this opportunity to let others see the grandeur in the graph

of the counterpart $r = sin (\frac{θ}{7})$ $r = sin (theta/7)$ :

graph{y-(x^2+y^2)(7+(x^2+y^2)(-56+(x^2+y^2)(112-64(x^2+y^2))))=0[-2 2 -1.2 1.2]}

How do you graph r=4sin7thetar=4sin7θr=4sin7theta?