How do you graph $r = 5 - 2 sin θ$ $r=5-2sintheta$ ?

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Answer 1 · 2016-11-03T15:24:29

See explanation.

$r (θ) = 5 - 2 sin θ$ $r(theta)=5-2 sin theta$ is periodic in $θ$ $theta$ , with period $2 π$ $2pi$ .

This graph is oscillatory about the circle $r = 5$ $r = 5$ , with relative

amplitude 2 and period $2 π$ $2pi$ . A short Table for one period $[0, 2 π]$ $[0, 2pi]$

gives the whole graph that is simply repeat of this graph for one

period.

$(r, θ)$ $(r, theta)$ :

$(5, 0) (4, \frac{π}{6}) (5 - \sqrt{3}, \frac{π}{3}) (3, \frac{π}{2})$ $(5, 0) (4, pi/6) (5-sqrt 3, pi/3) (3, pi/2)$

In $[\frac{π}{2}, π]$ $[pi/2, pi]$ , use symmetry about $θ = \frac{π}{2}$ $theta=pi/2$ .

Then, in $[π . 2 π]$ $[pi. 2pi]$ , use $sin (π + θ) = - sin θ$ $sin (pi+theta)=-sin theta$ .

The whole curve is laid between the circles r = 3 and r = 7, touching

them at $θ = \frac{π}{2} and \frac{3}{2} π$ $theta=pi/2 and 3/2pi$ , respectively.

How do you graph r=5−2sinθr=5-2sintheta?