How do you graph $r = 8 sin θ$ $r=8sintheta$ ?

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Answer 1 · 2016-12-29T14:11:27

Graph is inserted. See explanation.

$r = 2 a cos (θ - α)$ $r=2a cos(theta-alpha)$

represents the circle through the pole r = 0, with radius a and center

at $(a, α)$ $(a, alpha)$ .

Here, a = 4 and $α = \frac{π}{2}$ $alpha = pi/2$ '

For graphing here, convert to cartesian form using

$r (cos θ, sin θ) = (x, y)$ $r(cos theta, sin theta ) = (x, y)$ that gives,

$sin θ = \frac{y}{r} and r = \sqrt{x^{2} + y^{2}} \geq 0 .$ $sin theta= y/r and r=sqrt(x^2+y^2)>=0.$

So, the cartesian form is

$\sqrt{x^{2} + y^{2}} = 8 \frac{y}{\sqrt{x^{2} + y^{2}}}$ $sqrt(x^2+y^2)=8y/sqrt(x^2+y^2)$ that gives the standard form

$x^{2} + {(y - 4)}^{2} = 4^{2}$ $x^2+(y-4)^2=4^2$

graph{x^2+y^2-6y=0 [-20, 20, -10, 10]}

How do you graph r=8sinthetar=8sinθr=8sintheta?